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Solução de parte do livro Linear Algebra, do Hoffman e Kunze

Tipologia: Exercícios

2020

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Baixe Solutions of Linear Algebra - Hoffman and Kunze e outras Exercícios em PDF para Matemática, somente na Docsity! Linear Algebra Hoffman & Kunze 2nd edition Answers and Solutions to Problems and Exercises Typos, comments and etc... Gregory R. Grant University of Pennsylvania email: [email protected] Julyl 2017 2 Note This is one of the ultimate classic textbooks in mathematics. It will probably be read for generations to come. Yet I cannot find a comprehensive set of solutions online. Nor can I find lists of typos. Since I’m going through this book in some detail I figured I’d commit some of my thoughts and solutions to the public domain so others may have an easier time than I have finding supporting matericals for this book. Any book this classic should have such supporting materials readily avaiable. If you find any mistakes in these notes, please do let me know at one of these email addresses: [email protected] or [email protected] or [email protected] The latex source file for this document is available at http://greg.grant.org/hoffman and kunze.tex. Use it wisely. And if you cannot manage to download the link because you included the period at the end of the sentence, then maybe you are not smart enough to be reading this book in the first place. Section 1.2: Systems of Linear Equations 3 Solution: We must check the nine conditions on pages 1-2: 1. An operation is commutative if the table is symmetric across the diagonal that goes from the top left to the bottom right. This is true for the addition table so addition is commutative. 2. There are eight cases. But if x = y = z = 0 or x = y = z = 1 then it is obvious. So there are six non-trivial cases. If there’s exactly one 1 and two 0’s then both sides equal 1. If there are exactly two 1’s and one 0 then both sides equal 0. So addition is associative. 3. By inspection of the addition table, the element called 0 indeed acts like a zero, it has no effect when added to another element. 4. 1 + 1 = 0 so the additive inverse of 1 is 1. And 0 + 0 = 0 so the additive inverse of 0 is 0. In other words −1 = 1 and −0 = 0. So every element has an additive inverse. 5. As stated in 1, an operation is commutative if the table is symmetric across the diagonal that goes from the top left to the bottom right. This is true for the multiplication table so multiplication is commutative. 6. As with addition, there are eight cases. If x = y = z = 1 then it is obvious. Otherwise at least one of x, y or z must equal 0. In this case both x(yz) and (xy)z equal zero. Thus multiplication is associative. 7. By inspection of the multiplication table, the element called 1 indeed acts like a one, it has no effect when multiplied to another element. 8. There is only one non-zero element, 1. And 1 · 1 = 1. So 1 has a multiplicative inverse. In other words 1−1 = 1. 9. There are eight cases. If x = 0 then clearly both sides equal zero. That takes care of four cases. If all three x = y = z = 1 then it is obvious. So we are down to three cases. If x = 1 and y = z = 0 then both sides are zero. So we’re down to the two cases where x = 1 and one of y or z equals 1 and the other equals 0. In this case both sides equal 1. So x(y + z) = (x + y)z in all eight cases. Exercise 6: Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent. Solution: Write the two systems as follows: a11x + a12y = 0 a21x + a22y = 0 ... am1x + am2y = 0 b11x + b12y = 0 b21x + b22y = 0 ... bm1x + bm2y = 0 Each system consists of a set of lines through the origin (0, 0) in the x-y plane. Thus the two systems have the same solutions if and only if they either both have (0, 0) as their only solution or if both have a single line ux + vy − 0 as their common solution. In the latter case all equations are simply multiples of the same line, so clearly the two systems are equivalent. So assume that both systems have (0, 0) as their only solution. Assume without loss of generality that the first two equations in the first system give different lines. Then a11 a12 , a21 a22 (1) We need to show that there’s a (u, v) which solves the following system: a11u + a12v = bi1 a21u + a22v = bi2 Solving for u and v we get u = a22bi1 − a12bi2 a11a22 − a12a21 v = a11bi2 − a21bi1 a11a22 − a12a12 By (1) a11a22 − a12a21 , 0. Thus both u and v are well defined. So we can write any equation in the second system as a combination of equations in the first. Analogously we can write any equation in the first system in terms of the second. 4 Chapter 1: Linear Equations Exercise 7: Prove that each subfield of the field of complex numbers contains every rational number. Solution: Every subfield of C has characterisitc zero since if F is a subfield then 1 ∈ F and n · 1 = 0 in F implies n · 1 = 0 in C. But we know n · 1 = 0 in C implies n = 0. So 1, 2, 3, . . . are all distinct elements of F. And since F has additive inverses −1,−2,−3, . . . are also in F. And since F is a field also 0 ∈ F. Thus Z ⊆ F. Now F has multiplicative inverses so ± 1n ∈ F for all natural numbers n. Now let mn be any element of Q. Then we have shown that m and 1 n are in F. Thus their product m · 1 n is in F. Thus mn ∈ F. Thus we have shown all elements of Q are in F. Exercise 8: Prove that each field of characteristic zero contains a copy of the rational number field. Solution: Call the additive and multiplicative identities of F 0F and 1F respectively. Define nF to be the sum of n 1F’s. So nF = 1F + 1F + · · · + 1F (n copies of 1F). Define −nF to be the additive inverse of nF . Since F has characteristic zero, if n , m then nF , mF . For m, n ∈ Z, n , 0, let mn F = mF · n −1 F . Since F has characteristic zero, if m n , m′ n′ then m n F , m′ n′ F . Therefore the map mn 7→ m n F gives a one-to-one map from Q to F. Call this map h. Then h(0) = 0F , h(1) = 1F and in general h(x + y) = h(x) + h(y) and h(xy) = h(x)h(y). Thus we have found a subset of F that is in one-to-one correspondence to Q and which has the same field structure as Q. Section 1.3: Matrices and Elementary Row Operations Page 10, typo in proof of Theorem 4. Pargraph 2, line 6, it says kr , k but on the next line they call it k′ instead of kr. I think it’s best to use k′, because kr is a more confusing notation. Exercise 1: Find all solutions to the systems of equations (1 − i)x1 − ix2 = 0 2x1 + (1 − i)x2 = 0. Solution: The matrix of coefficients is [ 1 − i −i 2 1 − i ] . Row reducing → [ 2 1 − i 1 − i −i ] → [ 2 1 − i 0 0 ] Thus 2x1 + (1 − i)x2 = 0. Thus for any x2 ∈ C, ( 12 (i − 1)x2, x2) is a solution and these are all solutions. Exercise 2: If A = 3 −1 22 1 11 −3 0 find all solutions of AX = 0 by row-reducing A. Solution: → 1 −3 02 1 13 −1 2 → 1 −3 00 7 10 8 2 → 1 −3 00 1 1/70 8 2 → 1 0 3/70 1 1/70 0 6/7 → 1 0 3/70 1 1/70 0 1 → 1 0 00 1 100 0 1 . Thus A is row-equivalent to the identity matrix. It follows that the only solution to the system is (0, 0, 0). Section 1.3: Matrices and Elementary Row Operations 5 Exercise 3: If A = 6 −4 04 −2 0 −1 0 3 find all solutions of AX = 2X and all solutions of AX = 3X. (The symbol cX denotes the matrix each entry of which is c times the corresponding entry of X.) Solution: The system AX = 2X is 6 −4 04 −2 0 −1 0 3 xyz = 2 xyz which is the same as 6x − 4y = 2x 4x − 2y = 2y −x + 3z = 2z which is equivalent to 4x − 4y = 0 4x − 4y = 0 −x + z = 0 The matrix of coefficients is 4 −4 04 −4 0 −1 0 1 which row-reduces to 1 0 −10 1 −10 0 0 Thus the solutions are all elements of F3 of the form (x, x, x) where x ∈ F. The system AX = 3X is 6 −4 04 −2 0 −1 0 3 xyz = 3 xyz which is the same as 6x − 4y = 3x 4x − 2y = 3y −x + 3z = 3z which is equivalent to 3x − 4y = 0 x − 2y = 0 −x = 0 8 Chapter 1: Linear Equations Exercise 8: Consider the system of equations AX = 0 where A = [ a b c d ] is a 2 × 2 matrix over the field F. Prove the following: (a) If every entry of A is 0, then every pair (x1, x2) is a solution of AX = 0. (b) If ad − bc , 0, the system AX = 0 has only the trivial solution x1 = x2 = 0. (c) If ad − bc = 0 and some entry of A is different from 0, then there is a solution (x01, x 0 2) such that (x1, x2) is a solution if and only if there is a scalar y such that x1 = yx01, x2 = yx 0 2. Solution: (a) In this case the system of equations is 0 · x1 + 0 · x2 = 0 0 · x1 + 0 · x2 = 0 Clearly any (x1, x2) satisfies this system since 0 · x = 0 ∀ x ∈ F. (b) Let (u, v) ∈ F2. Consider the system: a · x1 + b · x2 = u c · x1 + d · x2 = v If ad − bc , 0 then we can solve for x1 and x2 explicitly as x1 = du − bv ad − bc x2 = av − cu ad − bc . Thus there’s a unique solution for all (u, v) and in partucular when (u, v) = (0, 0). (c) Assume WOLOG that a , 0. Then ad − bc = 0⇒ d = cba . Thus if we multiply the first equation by c a we get the second equation. Thus the two equations are redundant and we can just consider the first one ax1 + bx2 = 0. Then any solution is of the form (− ba y, y) for arbitrary y ∈ F. Thus letting y = 1 we get the solution (−b/a, 1) and the arbitrary solution is of the form y(−b/a, 1) as desired. Section 1.4: Row-Reduced Echelon Matrices Exercise 1: Find all solutions to the following system of equations by row-reducing the coefficient matrix: 1 3 x1 + 2x2 − 6x3 = 0 −4x1 + 5x3 = 0 −3x1 + 6x2 − 13x3 = 0 − 7 3 x1 + 2x2 − 8 3 x3 = 0 Solution: The coefficient matrix is 1 3 2 −6 −4 0 5 −3 6 −13 − 73 2 − 8 3 Section 1.4: Row-Reduced Echelon Matrices 9 This reduces as follows: → 1 6 −18 −4 0 5 −3 6 −13 −7 6 −8 → 1 6 −18 0 24 −67 0 24 −67 0 48 −134 → 1 6 −18 0 24 −67 0 0 0 0 0 0 → 1 6 −18 0 1 −67/24 0 0 0 0 0 0 → 1 0 −5/4 0 1 −67/24 0 0 0 0 0 0 Thus x − 5 4 z = 0 y − 67 24 z = 0 Thus the general solution is ( 54 z, 67 24 z, z) for arbitrary z ∈ F. Exercise 2: Find a row-reduced echelon matrix which is row-equivalent to A = 1 −i2 2u 1 + i . What are the solutions of AX = 0? Solution: A row-reduces as follows: → 1 −i1 1i 1 + i → 1 −i0 1 + i0 i → 1 −i0 10 i → 1 00 10 0 Thus the only solution to AX = 0 is (0, 0). Exercise 3: Describe explicitly all 2 × 2 row-reduced echelon matrices. Solution: [ 1 0 0 1 ] , [ 1 x 0 0 ] , [ 0 1 0 0 ] , [ 0 0 0 0 ] Exercise 4: Consider the system of equations x1 − x2 + 2x3 = 1 2x1 + 2x3 = 1 x1 − 3x2 + 4x3 = 2 Does this system have a solution? If so, describe explicitly all solutions. Solution: The augmented coefficient matrix is 1 −1 2 12 0 2 11 −3 4 2 We row reduce it as follows: → 1 −1 2 10 2 −2 −10 −2 2 1 → 1 −1 2 10 1 −1 −1/20 0 0 0 → 1 0 1 1/20 1 −1 −1/20 0 0 0 10 Chapter 1: Linear Equations Thus the system is equivalent to x1 + x3 = 1/2 x2 − x3 = −1/2 Thus the solutions are parameterized by x3. Setting x3 = c gives x1 = 1/2 − c, x2 = c − 1/2. Thus the general solution is( 1 2 − c, c − 1 2 , c ) for c ∈ R. Exercise 5: Give an example of a system of two linear equations in two unkowns which has no solutions. Solution: x + y = 0 x + y = 1 Exercise 6: Show that the system x1 − 2x2 + x3 + 2x4 = 1 x1 + x2 − x3 + x4 = 2 x1 + 7x2 − 5x3 − x4 = 3 has no solution. Solution: The augmented coefficient matrix is as follows 1 −2 1 2 11 1 −1 1 21 7 −5 −1 3 This row reduces as follows: → 1 −2 1 2 10 3 −2 −1 10 9 −6 −3 2 → 1 −2 1 2 10 3 −2 −1 10 0 0 0 −1 At this point there’s no need to continue because the last row says 0x1 + 0x2 + 0x3 + 0x4 = −1. But the left hand side of this equation is zero so this is impossible. Exercise 7: Find all solutions of 2x1 − 3x2 − 7x3 + 5x4 + 2x5 = −2 x1 − 2x2 − 4x3 + 3x4 + x5 = −2 2x1 − 4x3 + 2x4 + x5 = 3 x1 − 5x2 − 7x3 + 6x4 + 2x5 = −7 Solution: The augmented coefficient matrix is 2 −3 −7 5 2 −2 1 −2 −4 3 1 −2 2 0 −4 2 1 3 1 −5 −7 6 2 −7 Section 1.5: Matrix Multiplication 13 must also be zero. Thus we have shown that no two Ri and R j have the same solutions unless i = j. NOTE: This fact is actually true in general not just for 2 × 3 (search for “1832109” on math.stackexchange). Section 1.5: Matrix Multiplication Page 18: Typo in the last paragraph where it says “the columns of B are the 1 × n matrices . . . ” it should be n × 1 not 1 × n. Page 20: Typo in the Definition, it should say “An m × m matrix is said to be an elementary matrix”... Otherwise it doesn’t make sense that you can obtain an m × n matrix from an m × m matrix by an elementary row operation unless m = n. Exercise 1: Let A = [ 2 −1 1 1 2 1 ] , B = 31 −1 , C = [1 − 1]. Compute ABC and CAB. Solution: AB = [ 4 4 ] , so ABC = [ 4 4 ] · [1 − 1] = [ 4 −4 4 −4 ] . and CBA = [1 − 1] · [ 4 4 ] = [0]. Exercise 2: Let A = 1 −1 12 0 13 0 1 , B = 2 −21 34 4 . Verify directly that A(AB) = A2B. Solution: A2 = 1 −1 12 0 13 0 1 · 1 −1 12 0 13 0 1 = 2 −1 15 −2 36 −3 4 . And AB = 1 −1 12 0 13 0 1 · 2 −21 34 4 = 5 −18 010 −2 . 14 Chapter 1: Linear Equations Thus A2B = 2 −1 15 −2 36 −3 4 · 2 −21 34 4 = 7 −320 −425 −5 . (9) And A(AB) = 1 −1 12 0 13 0 1 · 5 −18 010 −2 7 −320 −425 −5 . (10) Comparing (9) and (10) we see both calculations result in the same matrix. Exercise 3: Find two different 2 × 2 matrices A such that A2 = 0 but A , 0. Solution: [ 0 1 0 0 ] , [ 0 0 1 0 ] are two such matrices. Exercise 4: For the matrix A of Exercise 2, find elementary matrices E1, E2, . . . , Ek such that Ek · · · E2E1A = I. Solution: A = 1 −1 12 0 13 0 1 E1A = 1 0 0−2 1 00 0 1 1 −1 12 0 13 0 1 = 1 −1 10 2 −13 0 1 E2(E1A) = 1 0 00 1 0 −3 0 1 1 −1 10 2 −13 0 1 = 1 −1 10 2 −10 3 0 E3(E2E1A) = 1 0 00 1/2 00 0 1 1 −1 10 2 −10 3 0 = 1 −1 10 1 −1/20 3 0 E4(E3E2E1A) = 1 1 00 1 00 0 1 1 −1 10 1 −1/20 3 0 = 1 0 1/20 1 −1/20 3 0 E5(E4E3E2E1A) = 1 0 00 1 00 −3 1 1 0 1/20 1 −1/20 3 0 = 1 0 1/20 1 −1/20 0 3/2 Section 1.5: Matrix Multiplication 15 E6(E5E4E3E2E1A) = 1 0 00 1 00 0 2/3 1 0 1/20 1 −1/20 0 3/2 = 1 0 1/20 1 −1/20 0 1 E7(E6E5E4E3E2E1A) = 1 0 −1/20 1 00 0 1 1 0 1/20 1 −1/20 0 1 = 1 0 00 1 −1/20 0 1 E8(E7E6E5E4E3E2E1A) = 1 0 00 1 1/20 0 1 1 0 00 1 −1/20 0 1 = 1 0 00 1 00 0 1 Exercise 5: Let A = 1 −12 21 0 , B = [ 3 1 −4 4 ] . Is there a matrix C such that CA = B? Solution: To find such a C = [ a b c d e f ] we must solve the equation [ a b c d e f ] 1 −12 21 0 = [ 3 1 −4 4 ] . This gives a system of equations a + 2b + c = 3 −a + 2b = 1 d + 2e + f = −4 −d + 2e = 4 We row-reduce the augmented coefficient matrix 1 2 1 0 0 0 3 −1 2 0 0 0 0 1 0 0 0 1 2 1 −4 0 0 0 −1 2 0 4 → 1 0 1/2 0 0 0 1 0 1 1/4 0 0 0 1 0 0 0 1 0 1/2 −4 0 0 0 0 1 1/4 0 . Setting c = f = 4 gives the solution C = [ −1 0 4 −6 −1 4 ] . Checking: [ −1 0 4 −6 −1 4 ] 1 −12 21 0 = [ 3 1 −4 4 ] . Exercise 6: Let A be an m× n matrix and B an n× k matrix. Show that the columns of C = AB are linear combinations of the columns of A. If α1, . . . , αn are the columns of A and γ1, . . . , γk are the columns of C then γ j = n∑ r=1 Br jαr. 18 Chapter 1: Linear Equations Then AB − BA = [ 0 c12 −c21 0 ] [ −1 0 0 0 ] − [ −1 0 0 0 ] [ 0 c12 −c21 0 ] = [ 0 0 c21 0 ] − [ 0 −c12 0 0 ] = [ 0 c12 c21 0 ] = C. So we can assume going forward that c11 , 0. We want to show the system (12) can be solved. In other words we have to find a, b, c, d such that the system has a solution in x, y, z,w. If we assume b , 0 and c , 0 then this matrix row-reduces to the following row-reduced echelon form 1 0 (d − a)/c −1 d−abc c11 − c12 b 0 1 −d/c 0 −c11/c 0 0 0 0 − c11b (d − a) + c21 + c12c b 0 0 0 0 c11 + c22 We see that necessarily − c11 b (d − a) + c21 + c12c b = 0. Since c11 , 0, we can set a = 0, b = c = 1 and d = c12+c21c11 . Then the system can be solved and we get a solution for any choice of z and w. Setting z = w = 0 we get x = c21 and y = c11. Summarizing, if c11 , 0 then: a = 0 b = 1 c = 1 d = (c12 + c21)/c11 x = c21 y = c11 z = 0 w = 0 For example if C = [ 2 1 3 −2 ] then A = [ 0 1 1 2 ] and B = [ 3 −2 0 0 ] . Checking: [ 0 1 1 2 ] [ 3 −2 0 0 ] − [ 3 −2 0 0 ] [ 0 1 1 2 ] = [ 2 1 3 −2 ] . Section 1.6: Invertible Matrices 19 Section 1.6: Invertible Matrices Exercise 1: Let A = 1 2 1 0−1 0 3 51 −2 1 1 . Find a row-reduced echelon matrix R which is row-equivalent to A and an invertible 3 × 3 matrix P such that R = PA. Solution: As in Exercise 4, Section 1.5, we row reduce and keep track of the elementary matrices involved. It takes nine steps to put A in row-reduced form resulting in the matrix P = 3/8 −1/4 3/81/4 0 −1/41/8 1/4 1/8 . Exercise 2: Do Exercise 1, but with A = 2 0 i1 −3 −ii 1 1 . Solution: Same story as Exercise 1, we get to the identity matrix in nine elementary steps. Multiplying those nine elementary matrices together gives P = 1/3 − 29+3i30 1−3i 10 0 − 3+i10 1−3i 10 −i/3 3+i15 3+i 5 . Exercise 3: For each of the two matrices 2 5 −14 −1 26 4 1 , 1 −1 23 2 40 1 −2 use elementary row operations to discover whether it is invertible, and to find the inverse in case it is. Solution: For the first matrix we row-reduce the augmented matrix as follows: 2 5 −1 1 0 04 −1 2 0 1 06 4 1 0 0 1 → 2 5 −1 1 0 00 −11 4 −2 1 00 −11 4 −3 0 1 → 2 5 −1 1 0 00 −11 4 −2 1 00 0 0 −1 −1 1 At this point we see that the matrix is not invertible since we have obtained an entire row of zeros. For the second matrix we row-reduce the augmented matrix as follows: 1 −1 2 1 0 03 2 4 0 1 00 1 −2 0 0 1 20 Chapter 1: Linear Equations → 1 −1 2 1 0 00 5 −2 −3 1 00 1 −2 0 0 1 → 1 −1 2 1 0 00 1 −2 0 0 10 5 −2 −3 1 0 → 1 0 0 1 0 10 1 −2 0 0 10 0 8 −3 1 −5 → 1 0 0 1 0 10 1 −2 0 0 10 0 1 −3/8 1/8 −5/8 → 1 0 0 1 0 10 1 0 −3/4 1/4 −1/40 0 1 −3/8 1/8 −5/8 Thus the inverse matrix is 1 0 1−3/4 1/4 −1/43/8 1/8 −5/8 Exercise 4: Let A = 5 0 01 5 00 1 5 . For which X does there exist a scalar c such that AX = cX? Solution: 5 0 01 5 00 1 5 xyz = c xyz implies 5x = cx (13) x + 5y = cy (14) y + 5z = cz (15) Now if c , 5 then (13) implies x = 0, and then (14) implies y = 0, and then (15) implies z = 0. So it is true for 000 with c = 0. If c = 5 then (14) implies x = 0 and (15) implies y = 0. So if c = 5 any such vector must be of the form 00z and indeed any such vector works with c = 5. So the final answer is any vector of the form 00z . Section 1.6: Invertible Matrices 23 Now suppose that ad − bc = 0. We will show there is no solution. If a = b = c = d = 0 then obviously A has no inverse. So suppose WOLOG that a , 0 (because by elementary row and column operations we can move any of the four elements to be the top left entry, and elementary row and column operations do not change a matrix’s status as being invertible or not). Subtracting ca times the 3rd row from the 4th row of (16) gives a 0 b 0 1 c 0 d 0 0 0 a 0 b 0 0 c − ca a 0 d − c a b 1 . Now c − ca a = 0 and since ad − bc = 0 also d − c a b = 0. Thus we get a 0 b 0 1 c 0 d 0 0 0 a 0 b 0 0 0 0 0 1 . and it follows that A is not invertible. Exercise 9: An n×n matrix A is called upper-triangular if ai j = 0 for i > j, that is, if every entry below the main diagonal is 0. Prove that an upper-triangular (square) matrix is invertible if and only if every entry on its main diagonal is different from zero. Solution: Suppose that aii , 0 for all i. Then we can divide row i by aii to give a row-equivalent matrix which has all ones on the diagonal. Then by a sequence of elementary row operations we can turn all off diagonal elements into zeros. We can therefore row-reduce the matrix to be equivalent to the identity matrix. By Theorem 12 page 23, A is invertible. Now suppose that some aii = 0. If all aii’s are zero then the last row of the matrix is all zeros. A matrix with a row of zeros cannot be row-equivalent to the identity so cannot be invertible. Thus we can assume there’s at least one i such that aii , 0. Let i′ be the largest such index, so that ai′i′ = 0 and aii , 0 for all i > i′. We can divide all rows with i > i′ by aii to give ones on the diagonal for those rows. We can then add multiples of those rows to row i′ to turn row i′ into an entire row of zeros. Since again A is row-equivalent to a matrix with an entire row of zeros, it cannot be invertible. Exercise 10: Prove the following generalization of Exercise 6. If A is an m × n matrix and B is an n × m matrix and n < m, then AB is not invertible. Solution: There are n colunms in A so the vector space generated by those columns has dimension no greater than n. All columns of AB are linear combinations of the columns of A. Thus the vector space generated by the columns of AB is con- tained in the vector space generated by the columns of A. Thus the column space of AB has dimension no greater than n. Thus the column space of the m×m matrix AB has dimension less or equal to n and n < m. Thus the columns of AB generate a space of dimension strictly less than m. Thus AB is not invertible. Exercise 11: Let A be an n × m matrix. Show that by means of a finite number of elementary row and/or column operations one can pass from A to a matrix R which is both ‘row-reduced echelon’ and ‘column-reduced echelon,’ i.e., Ri j = 0 if i , j, Rii = 1, 1 ≤ i ≤ r, Rii = 0 if i > r. Show that R = PAQ, where P is an invertible m×m matrix and Q is an invertible n×n matrix. Solution: First put A in row-reduced echelon form, R′. So ∃ an invertible m ×m matrix P such that R′ = PA. Each row of R′ is either all zeros or starts (on the left) with zeros, then has a one, then may have non-zero entries after the one. Suppose row i has a leading one in the j-th column. The j-th column has zeros in all other places except the i-th, so if we add a multiple of this column to another column then it only affects entries in the i-th row. Therefore a sequence of such operations can turn this row into a row of all zeros and a single one. 24 Chapter 1: Linear Equations Let B be the n × n matrix such that Brr = 1 and Brs = 0 ∀ r , s except B jk , 0. Then AB equals A with B jk times column j added to column k. B is invertible since any such operation can be undone by another such operation. By a sequence of such operations we can turn all values after the leading one into zeros. Let Q be a product of all of the elementary matrices B involved in this transformation. Then PAQ is in row-reduced and column-reduced form. Exercise 12: The result of Example 16 suggests that perhaps the matrix 1 12 · · · 1 n 1 2 1 3 · · · 1 n+1 ... ... ... 1 n 1 n+1 · · · 1 2n−1 is invertible and A−1 has integer entries. Can you prove that? Solution: This problem seems a bit hard for this book. There are a class of theorems like this, in particular these are called Hilbert Matrices and a proof is given in this article on arxiv by Christian Berg called Fibonacci numbers and orthogonal polynomials (http://arxiv.org/pdf/math/0609283v2.pdf). See Theorem 4.1. Also there might be a more elementary proof in this discussion on mathoverflow.net where two proofs are given: http://mathoverflow.net/questions/47561/deriving-inverse-of-hilbert-matrix. Also see http://vigo.ime.unicamp.br/HilbertMatrix.pdf where a general formula for the i, j entry of the inverse is given explicitly as (−1)i+ j(i + j − 1) ( n + i − 1 n − j )( n + j − 1 n − i )( i + j − 1 i − 1 )2 Chapter 2: Vector Spaces Section 2.1: Vector Spaces Page 29, three lines into the first paragraph of text they refer to “two operations”. This could be confusing since they just said a vector space is a composite object consisting of a field and a set with a rule. There are two sets, the field F and the set of vectors V . Really there are four operations going around. There is addition and multiplication in the field F, and there is addition also in V , that’s three. But there’s also multiplication of an element of F and an element of V . That’s why there are two distributive rules. But keep in mind we do not multiply elements of V together - there’s no multiplication within V , only within F and between F and V . So anyway, why did they say “two” operations? They’re clearly ignoring the two operations in the field and just talking about operations involving vectors. You get the point. Exercise 1: If F is a field, verify that Fn (as defined in Example 1) is a vector space over the field F. Solution: Example 1 starts with any field and defines the objects, the addition rule and the scalar multiplication rule. We must show the set of n-tuples satisfies the eight properties required in the definition. 1) Addition is commutative. Let α = (x1, . . . , xn) and β = (y1, . . . , yn) be two n-tuples. Then α + β = (x1 + y1, . . . , xn + yn). And since F is commutative this equals (y1 + x1, . . . , yn + xn), which equals β + α. Thus α + β = β + α. 2) Addition is associative. Let α = (x1, . . . , xn), β = (y1, . . . , yn) and γ = (z1, . . . , zn) be three n-tuples. Then (α + β) + γ = ((x1 +y1)+z1, . . . , (xn +yn)+zn). And since F is associative this equals (x1 +(y1 +z1), . . . , xn +(yn +zn)), which equals α+(β+γ). 3) We must show there is a unique vector 0 in V such that α + 0 = α ∀ α ∈ V . Consider (0F , . . . , 0F) the vector of all 0’s of length n, where 0F is the zero element of F. Then this vector satisfies the property that (0F , . . . , 0F) + (x1, . . . , xn) = (0F + x1, . . . , 0F + xn) = (x1, . . . , xn) since 0F + x = x ∀ x ∈ F. Thus (0F , . . . , 0F) + α = α ∀α ∈ V . We must just show this vector is unique with respect to this property. Suppose β = (x1, . . . , xn) also satisfies the property that β+ α = α for all α ∈ V . Let α = (0F , . . . , 0F). Then (x1, . . . , xn) = (x1 + 0F , . . . , xn + 0F) = (x1, . . . , xn) + (0F , . . . , 0F) and by definition of β this equals (0F , . . . , 0F). Thus (x1, . . . , xn) = (0F , . . . , 0F). Thus β = α and the zero element is unique. 4) We must show for each vector α there is a unique vector β such that α + β = 0. Suppose α = (x1, . . . , xn). Let β = (−x1, . . . ,−xn). Then β has the required property α + β = 0. We must show β is unique with respect to this property. Suppose also β′ = (x′1, . . . , x ′ n) also has this property. Then α+β = 0 and α+β ′ = 0. So β = β+0 = β+(α+β′) = (β+α)+β′ = 0+β′ = β′. 5) Let 1F be the multiplicative identity in F. Then 1F · (x1, . . . , xn) = (1 · x1, . . . , 1 · xn) = (x1, . . . , xn) since 1F · x = x ∀ x ∈ F. Thus 1Fα = α ∀ α ∈ V . 6) Let α = (x1, . . . , xn). Then (c1c2)α = ((c1c2)x1, . . . , (c1c2)xn) and since multiplication in F is associative this equals (c1(c2x1), . . . , c1(c2xn)) = c1(c2x1, . . . c2xn) = c1 · (c2α). 7) Let α = (x1, . . . , xn) and β = (y1, . . . , yn). Then c(α+β) = c(x1 +y1, . . . , xn +yn) = (c(x1 +y1), . . . , c(xn +yn)) and since multi- plication is distributive over addition in F this equals (cx1+cy1, . . . , cxn+xyn). This then equals (cx1, . . . , cxn)+(cy1, . . . , cyn) = 25 28 Chapter 2: Vector Spaces This is true because ( f + g)(−t) = f (−t) + g(−t) = f (t) + g(t) = ( f (t) + g(t) = ( f + g)(t). Similary, if c ∈ R, (c f )(−t) = c f (−t) = c f (t) = c f (t) since c = c when c ∈ R. Thus the operations are well defined. We now show the eight properties hold: 1) Addition on functions in V is defined by adding in C to the values of the functions in C. Thus since C is commutative, addition in V inherits this commutativity. 2) Similar to 1, since C is associative, addition in V inherits this associativity. 3) The zero function g(t) = 0 is in V since −0 = 0. And g satisfies f +g = f for all f ∈ V . Thus V has a right additive identity. 4) Let g be the function g(t) = − f (t). Then g(−t) = − f (−t) = − f (t) = − f (t) = g(t). Thus g ∈ V . And ( f + g)(t) = f (t) + g(t) = f (t) − f (t) = 0. Thus g is a right additive inverse for f . 5) Clearly 1 · f = f since 1 is the multiplicative identity in R. 6) As before, associativity in C implies (c1c2) f = c1(c2 f ). 7) Similarly, the distributive property in C implies c( f + g) = c f + cg. 8) Similarly, the distributive property in C implies (c1 + c2) f = c1 f + c2 f . An example of a function in V which is not real valued is f (x) = ix. Since f (1) = i f is not real-valued. And f (−x) = −ix = ix since x ∈ R, so f ∈ V . Exercise 7: Let V be the set of pairs (x, y) of real numbers and let F be the field of real numbers. Define (x, y) + (x1, y1) = (x + x1, 0) c(x, y) = (cx, 0). Is V , with these operations, a vector space? Solution: This is not a vector space because there would have to be an additive identity element (a, b) which has the property that (a, b) + (x, y) = (x, y) for all (x, y) ∈ V . But this is impossible, because (a, b) + (0, 1) = (a, 0) , (0, 1) no matter what (a, b) is. Thus V does not satisfy the third requirement of having an additive identity element. Section 2.2: Subspaces Page 39, typo in Exercise 3. It says R5, should be R4. Exercise 1: Which of the following sets of vectors α = (a1, . . . , an) in Rn are subspaces of Rn (n ≥ 3)? (a) all α such that a1 ≥ 0; (b) all α such that a1 + 3a2 = a3; (c) all α such that a2 = a21; (d) all α such that a1a2 = 0; Section 2.2: Subspaces 29 (e) all α such that a2 is rational. Solution: (a) This is not a subspace because for (1, . . . , 1) the additive inverse is (−1, . . . ,−1) which does not satisfy the condition. (b) Suppose (a1, a2, a3, . . . , an) and (b1, b2, b3, . . . , bn) satisfy the condition and let c ∈ R. By Theorem 1 (page 35) we must show that c(a1, a2, a3, . . . , an)+(b1, b2, b3, . . . , bn) = (ca1+b1, . . . , can+bn) satisfies the condition. Now (ca1+b1)+3(ca2+b2) = c(a1 + 3a2) + (b1 + 3b2) = c(a3) + (b3) = ca3 + b3. Thus it does satisfy the condition so V is a vector space. (c) This is not a vector space because (1, 1) satisfies the condition since 12 = 1, but (1, 1, . . . ) + (1, 1, . . . ) = (2, 2, . . . ) and (2, 2, . . . ) does not satisfy the condition because 22 , 2. (d) This is not a subspace. (1, 0, . . . ) and (0, 1, . . . ) both satisfy the condition, but their sum is (1, 1, . . . ) which does not satisfy the condition. (e) This is not a subspace. (1, 1, . . . , 1) satisfies the condition, but π(1, 1, . . . , 1) = (π, π, . . . , π) does not satisfy the condition. Exercise 2: Let V be the (real) vector space of all functions f from R into R. Which of the following sets of functions are subspaces of V? (a) all f such that f (x2) = f (x)2; (b) all f such that f (0) = f (1); (c) all f such that f (3) = 1 + f (−5); (d) all f such that f (−1) = 0; (e) all f which are continuous. Solution: (a) Not a subspace. Let f (x) = x and g(x) = x2. Then both satisfy the condition: f (x2) = x2 = ( f (x))2 and g(x2) = (x2)2 = (g(x))2. But ( f + g)(x) = x + x2 and ( f + g)(x2) = x2 + x4 while [( f + g)(x)]2 = (x + x2)2 = x4 + 2x3 + x2. These are not equal polynomials so the condition does not hold for f + g. (b) Yes a subspace. Suppose f and g satisfy the property. Let c ∈ R. Then (c f +g)(0) = c f (0)+g(0) = c f (1)+g(1) = (c f +g)(1). Thus (c f + g)(0) = (c f + g)(1). By Theorem 1 (page 35) the set of all such functions constitute a subspace. (c) Not a subspae. Let f (x) be the function defined by f (3) = 1 and f (x) = 0 for all x , 3. Let g(x) be the function defined by g(−5) = 0 and g(x) = 1 for all x , −5. Then both f and g satisfy the condition. But ( f + g)(3) = f (3) + g(3) = 1 + 1 = 2, while 1 + ( f + g)(−5) = 1 + f (−5) + g(−5) = 1 + 0 + 0 = 1. Since 1 , 2, f + g does not satisfy the condition. (d) Yes a subspace. Suppose f and g satisfy the property. Let c ∈ R. Then (c f + g)(−1) = c f (−1) + g(−1) = c · 0 + 0 = 0. Thus (c f + g)(−1) = 0. By Theorem 1 (page 35) the set of all such functions constitute a subspace. (e) Yes a subspace. Let f and g be continuous functions from R to R and let c ∈ R. Then we know from basic results of real analysis that the sum and product of continuous functions are continuous. Since the function c 7→ c is continuous as well as f and g, it follows that c f +g is continuous. By Theorem 1 (page 35) the set of all cotinuous functions constitute a subspace. Exercise 3: Is the vector (3,−1, 0,−1) in the subspace of R5 (sic) spanned by the vectors (2,−1, 3, 2), (−1, 1, 1,−3), and (1, 1, 9,−5)? 30 Chapter 2: Vector Spaces Solution: I assume they meant R4. No, (3,−1, 0,−1) is not in the subspace. If we row reduce the augmented matrix 2 −1 1 3 −1 1 1 −1 3 1 9 0 2 −3 −5 −1 we obtain 1 0 2 2 0 1 3 1 0 0 0 −7 0 0 0 −2 . The two bottom rows are zero rows to the left of the divider, but the values to the right of the divider in those two rows are non-zero. Thus the system does not have a solution (see comments bottom of page 24 and top of page 25). Exercise 4: Let W be the set of all (x1, x2, x3, x4, x5) in R5 which satisfy 2x1 − x2 + 4 3 x3 − x4 = 0 x1 + 2 3 x3 − x5 = 0 9x1 − 3x2 + 6x3 − 3x4 − 3x5 = 0. Find a finite set of vectors which spans W. Solution: The matrix of the system is 2 −1 4/3 −1 01 0 2/3 0 −19 −3 6 −3 −3 . Row reducing to reduced echelon form gives 1 0 2/3 0 −10 1 0 1 −20 0 0 0 0 . Thus the system is equivalent to x1 + 2/3x3 − x5 = 0 x2 + x4 − 2x5 = 0. Thus the system is parametrized by (x3, x4, x5). Setting each equal to one and the other two to zero (as in Example 15, page 42), in turn, gives the three vectors (−2/3, 0, 1, 0, 0), (0,−1, 0, 1, 0) and (1, 2, 0, 0, 1). These three vectors therefore span W. Exercise 5: Let F be a field and let n be a positive integer (n ≥ 2). Let V be the vector space of all n × n matrices over F. Which of the following sets of matrices A in V are subspaces of V? (a) all invertible A; (b) all non-invertible A; (c) all A such that AB = BA, where B is some fixed matrix in V; (d) all A such that A2 = A. Section 2.3: Bases and Dimension 33 Solution: Suppose v1 and v2 are linearly dependent. If one of them, say v1, is the zero vector then it is a scalar multiple of the other one v1 = 0 · v2. So we can assume both v1 and v2 are non-zero. Then if ∃ c1, c2 such that c1v1 + c2v2 = 0, both c1 and c2 must be non-zero. Therefore we can write v1 = − c2c1 v2. Exercise 2: Are the vectors α1 = (1, 1, 2, 4), α2 = (2,−1,−5, 2) α3 = (1,−1,−4, 0), α4 = (2, 1, 1, 6) linearly independent in R4? Solution: By Corollary 3, page 46, it suffices to determine if the matrix whose rows are the αi’s is invertible. By Theorem 12 (ii) we can do this by row reducing the matrix 1 1 2 4 2 −1 −5 2 1 −1 −4 0 2 1 1 6 . 1 1 2 4 2 −1 −5 2 1 −1 −4 0 2 1 1 6 → 1 1 2 4 0 −3 −9 −6 0 −2 −6 −4 0 −1 −3 −2 swap→rows 1 1 2 4 0 −1 −3 −2 0 −3 −9 −6 0 −2 −6 −4 → 1 1 2 4 0 1 3 2 0 −3 −9 −6 0 −2 −6 −4 → 1 1 2 4 0 1 3 2 0 0 0 0 0 0 0 0 Thus the four vectors are not linearly independent. Exercise 3: Find a basis for the subspace of R4 spanned by the four vectors of Exercise 2. Solution: In Section 2.5, Theorem 9, page 56, it will be proven that row equivalent matrices have the same row space. The proof of this is almost immediate so there seems no easier way to prove it than to use that fact. If you multiply a matrix A on the left by another matrix P, the rows of the new matrix PA are linear combinations of the rows of the original matrix. Thus the rows of PA generate a subspace of the space generated by the rows of A. If P is invertible, then the two spaces must be contained in each other since we can go backwards with P−1. Thus the rows of row-equivalent matrices generate the same space. Thus using the row reduced form of the matrix in Exercise 2, it must be that the space is two dimensoinal and generated by (1, 1, 2, 4) and (0, 1, 3, 2). Exercise 4: Show that the vectors α1 = (1, 0,−1), α2 = (1, 2, 1), α3 = (0,−3, 2) form a basis for R3. Express each of the standard basis vectors as linear combinations of α1, α2, and α3. Solution: By Corollary 3, page 46, to show the vectors are linearly independent it suffices to show the matrix whose rows are the αi’s is invertible. By Theorem 12 (ii) we can do this by row reducing the matrix A = 1 0 −11 2 10 −3 2 . 1 0 −11 2 10 −3 2 → 1 0 −10 2 20 −3 2 → 1 0 −10 1 10 −3 2 → 1 0 −10 1 10 0 5 → 1 0 −10 1 10 0 1 → 1 0 00 1 00 0 1 . Now to write the standard basis vectors in terms of these vectors, by the discussion at the bottom of page 25 through page 26, we can row-reduce the augmented matrix 1 0 −1 1 0 01 2 1 0 1 00 −3 2 0 0 1 . 34 Chapter 2: Vector Spaces This gives 1 0 −1 1 0 01 2 1 0 1 00 −3 2 0 0 1 → 1 0 −1 1 0 00 2 2 −1 1 00 −3 2 0 0 1 → 1 0 −1 1 0 00 1 1 −1/2 1/2 00 −3 2 0 0 1 → 1 0 −1 1 0 00 1 1 −1/2 1/2 00 0 5 −3/2 3/2 1 → 1 0 −1 1 0 00 1 1 −1/2 1/2 00 0 1 −3/10 3/10 1/5 → 1 0 0 7/10 3/10 1/50 1 0 −1/5 1/5 −1/50 0 1 −3/10 3/10 1/5 . Thus if P = 7/10 3/10 1/5−1/5 1/5 −1/5 −3/10 3/10 1/5 then PA = I, so we have 7 10 α1 + 3 10 α2 + 1 5 α3 = (1, 0, 0) − 1 5 α1 + 1 5 α2 − 1 5 α3 = (0, 1, 0) − 3 10 α1 + 3 10 α2 + 1 5 α3 = (0, 0, 1). Exercise 5: Find three vectors in R3 which are linearly dependent, and are such that any two of them are linearly independent. Solution: Let v1 = (1, 0, 0), v2 = (0, 1, 0) and v3 = (1, 1, 0). Then v1 + v2 − v3 = (0, 0, 0) so they are linearly dependent. We know v1 and v2 are linearly independent as they are two of the standard basis vectors (see Example 13, page 41). Suppose av1 + bv3 = 0. Then (a + b, b, 0) = (0, 0, 0). The second coordinate implies b = 0 and then the first coordinate in turn implies a = 0. Thus v1 and v3 are linearly independent. Analogously v2 and v3 are linearly independent. Exercise 6: Let V be the vector space of all 2 × 2 matrices over the field F. Prove that V has dimension 4 by exhibiting a basis for V which has four elements. Solution: Let v11 = [ 1 0 0 0 ] , v12 = [ 0 1 0 0 ] v21 = [ 0 0 1 0 ] , v22 = [ 0 0 0 1 ] Section 2.3: Bases and Dimension 35 Suppose av11 + bv12 + cv21 + dv22 = [ 0 0 0 0 ] . Then [ a b c d ] = [ 0 0 0 0 ] , from which it follows immediately that a = b = c = d = 0. Thus v11, v12, v21, v22 are linearly independent. Now let [ a b c d ] be any 2 × 2 matrix. Then [ a b c d ] = av11 + bv12 + cv21 + dv22. Thus v11, v12, v21, v22 span the space of 2 × 2 matrices. Thus v11, v12, v21, v22 are both linearly independent and they span the space of all 2 × 2 matrices. Thus v11, v12, v21, v22 constitue a basis for the space of all 2 × 2 matrices. Exercise 7: Let V be the vector space of Exercise 6. Let W1 be the set of matrices of the form[ x −x y z ] and let W2 be the set of matrices of the form [ a b −a c ] . (a) Prove that W1 and W2 are subspaces of V . (b) Find the dimension of W1, W2, W1 + W2, and W1 ∩W2. Solution: (a) Let A = [ x −x y z ] and B = [ x′ −x′ y′ z′ ] be two elements of W1 and let c ∈ F. Then cA + B = [ cx + x′ −cx − x′ cy + y′ cz + z′ ] = [ a −a u v ] where a = cx + x′, u = cy + y′ and v = cz + z′. Thus cA + B is in the form of an element of W1. Thus cA + B ∈ W1. By Theorem 1 (page 35) W1 is a subspace. Now let A = [ a b −a d ] and B = [ a′ b′ −a′ d′ ] be two elements of W1 and let c ∈ F.Then cA + B = [ ca + a′ cb + b′ −ca − a′ cd + d′ ] = [ x y −x z ] where x = ca + a′, y = cb + b′ and z = cd + d′. Thus cA + B is in the form of an element of W2. Thus cA + B ∈ W2. By Theorem 1 (page 35) W2 is a subspace. (b) Let A1 = [ 1 −1 0 0 ] , A2 = [ 0 0 1 0 ] , A2 = [ 0 0 0 1 ] . Then A1, A2, A3 ∈ W1 and c1A1 + c2A2 + c3A3 = [ c1 −c1 c2 c3 ] = [ 0 0 0 0 ] 38 Chapter 2: Vector Spaces (b) We can write the general element of V as A = [ a + bi e + f i g + hi −a − bi ] . Let v1 = [ 1 0 0 −1 ] , v2 = [ i 0 0 −i ] , v3 = [ 0 1 0 0 ] , v4 = [ 0 i 0 0 ] , v5 = [ 0 0 1 0 ] , v6 = [ 0 0 i 0 ] . Then A = av1 + bv2 + ev3 + f v4 + gv5 + hv6 so v1, v2, v3, v4, v5, v6 span V . Suppose av1 + bv2 + ev3 + f v4 + gv5 + hv6 = 0. Then av1 + bv2 + ev3 + f v4 + gv5 + hv6 = [ a + bi e + f i g + hi −a − bi ] = [ 0 0 0 0 ] implies a = b = c = d = e = f = g = h = 0 because a complex number u + vi = 0⇔ u = v = 0. Thus v1, v2, v3, v4, v5, v6 are linearly independent. Thus {v1, . . . , v6} is a basis for V as a vector space over R, and dim(V) = 6. (c) Let A, B ∈ W and c ∈ R. By Theorem 1 (page 35) we must show cA + B ∈ W. Write A = [ x y −ȳ −x ] and B =[ x′ y′ −ȳ′ −x′ ] , where x, y, x′, y′ ∈ C. Then cA + B = [ cx + x′ cy + y′ −cȳ − ȳ′ −cx − x′ ] . Since −cȳ − ȳ′ = −(cy + y′), it follows that cA + B ∈ W. Note that we definitely need c ∈ R for this to be true. It remains to find a basis for W. We can write the general element of W as A = [ a + bi e + f i −e + f i −a − bi ] . Let v1 = [ 1 0 0 −1 ] , v2 = [ i 0 0 −i ] , v3 = [ 0 1 −1 0 ] , v4 = [ 0 i i 0 ] . Then A = av1 + bv2 + ev3 + f v4 so v1, v2, v3, v4 span V . Suppose av1 + bv2 + ev3 + f v4 = 0. Then av1 + bv2 + ev3 + f v4 = [ a + bi e + f i −e + f i −a − bi ] = [ 0 0 0 0 ] implies a = b = e = f = 0 because a complex number u + vi = 0⇔ u = v = 0. Thus v1, v2, v3, v4 are linearly independent. Thus {v1, . . . , v4} is a basis for V as a vector space over R, and dim(V) = 4. Exercise 12: Prove that the space of m × n matrices over the field F has dimension mn, by exhibiting a basis for this space. Solution: Let M be the space of all m × n matrices. Let Mi j be the matrix of all zeros except for the i, j-th place which is a one. We claim {Mi j | 1 ≤ i ≤ m, 1 ≤ j ≤ n} constitute a basis for M. Let A = (ai j) be an arbitrary marrix in M. Then Section 2.4: Coordinates 39 A = ∑ i j ai jMi j. Thus {Mi j} span M. Suppose ∑ i j ai jMi j = 0. The left hand side equals the matrix (ai j) and this equals the zero matrix if and only if every ai j = 0. Thus {Mi j} are linearly independent as well. Thus the nm matrices constitute a basis and M has dimension mn. Exercise 13: Discuss Exercise 9, when V is a vector space over the field with two elements described in Exercise 5, Section 1.1. Solution: If F has characteristic two then (α + β) + (β + γ) + (γ + α) = 2α + 2β + 2γ = 0 + 0 + 0 = 0 since in a field of characteristic two, 2 = 0. Thus in this case (α + β), (β + γ) and (γ + α) are linearly dependent. However any two of them are linearly independent. For example suppose a1(α + β) + a2(β + γ) = 0. The LHS equals a1α + a2γ + (a1 + a2)β. Since α, β, γ are linearly independent, this is zero only if a1 = 0, a2 = 0 and a1 + a2 = 0. In particular a1 = a2 = 0, so α + β and β + γ are linearly independent. Exercise 14: Let V be the set of real numbers. Regard V as a vector space over the field of rational numbers, with the usual operations. Prove that this vector space is not finite-dimensional. Solution: We know that Q is countable and R is uncountable. Since the set of n-tuples of things from a countable set is countable, Qn is countable for all n. Now, suppose {r1, . . . , rn} is a basis for R over Q. Then every element of R can be written as a1r1 + · · · + anrn. Thus we can map n-tuples of rational numbers onto R by (a1, . . . , an) 7→ a1r1 + · · · + anrn. Thus the cardinality of R must be less or equal to Qn. But the former is uncountable and the latter is countable, a contradiction. Thus there can be no such finite basis. Section 2.4: Coordinates Exercise 1: Show that the vectors α1 = (1, 1, 0, 0), α2 = (0, 0, 1, 1) α3 = (1, 0, 0, 4), α4 = (0, 0, 0, 2) form a basis for R4. Find the coordinates of each of the standard basis vectors in the ordered basis {α1, α2, α3, α4}. Solution: Using Theorem 7, page 52, if we calculate the inverse of P = 1 0 1 0 1 0 0 0 0 1 0 0 0 1 4 2 . then the columns of P−1 will give the coefficients to write the standard basis vectors in terms of the αi’s. We do this by row-reducing the augmented matrix 1 0 1 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 4 2 0 0 0 1 . The left side must reduce to the identity whlie the right side transforms to the inverse of P. Row reduction gives 1 0 1 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 4 2 0 0 0 1 . 40 Chapter 2: Vector Spaces → 1 0 1 0 1 0 0 0 0 0 −1 0 −1 1 0 0 0 1 0 0 0 0 1 0 0 1 4 2 0 0 0 1 . → 1 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 −1 0 −1 1 0 0 0 1 4 2 0 0 0 1 . → 1 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 −1 0 −1 1 0 0 0 0 4 2 0 0 −1 1 . → 1 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 −1 0 0 0 0 4 2 0 0 −1 1 . → 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 −1 0 0 0 0 0 2 −4 4 −1 1 . → 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 −1 0 0 0 0 0 1 −2 2 −1/2 1/2 . Thus {α1, . . . , α4} is a basis. Call this basis β. Thus (1, 0, 0, 0) = α3 − 2α4, (0, 1, 0, 0) = α1 − α3 + 2α4, (0, 0, 1, 0) = α2 − 12α4 and (0, 0, 0, 1) = 12α4. Thus [(1, 0, 0, 0)]β = (0, 0, 1,−2), [(0, 1, 0, 0)]β = (1, 0,−1, 2), [(0, 0, 1, 0)]β = (0, 1, 0,−1/2) and [(0, 0, 0, 1)]β = (0, 0, 0, 1/2). Exercise 2: Find the coordinate matrix of the vector (1, 0, 1) in the basis of C3 consisting of the vectors (2i, 1, 0), (2,−1, 0), (0, 1 + i, 1 − i), in that order. Solution: Using Theorem 7, page 52, the answer is P−1 101 where P = 2i 2 01 −1 1 + i0 0 1 − i . We find P−1 by row-reducing the augmented matrix 2i 2 0 1 0 01 −1 1 + i 0 1 00 0 1 − i 0 0 1 . The right side will transform into the P−1. Row reducing: 2i 2 0 1 0 01 −1 1 + i 0 1 00 0 1 − i 0 0 1 Section 2.4: Coordinates 43 → 1 1 1 1 + i 0 1 1+i2 −1+i 2 0 0 0 0 → 1 0 1−i2 3+i 2 0 1 1+i2 −1+i 2 0 0 0 0 Thus α1 = 1−i2 β1 + 1+i 2 β2 and α2 = 3+i 2 β1 + −1+i 2 β2. So finally, if B is the basis {β1, β2} then [α1]B = ( 1 − i 2 , 1 + i 2 ) [α2]B = ( 3 + i 2 , −1 + i 2 ) . Exercise 5: Let α = (x1, x2) and β = (y1, y2) be vectors in R2 such that x1y1 + x2y2 = 0, x21 + x 2 2 = y 2 1 + y 2 2 = 1. Prove thatB = {α, β} is a basis for R2. Find the coordinates of the vector (a, b) in the ordered basisB = {α, β}. (The conditions on α and β say, geometrically, that α and β are perpendicular and each has length 1.) Solution: It suffices by Corollary 1, page 46, to show α and β are linearly indepdenent, because then they generate a subspace of R2 of dimension two, which therefore must be all of R2. The second condition on x1, x2, y1, y2 implies that neither α nor β are the zero vector. To show two vectors are linearly independent we only need show neither is a non-zero scalar multiple of the other. Suppose WLOG that β = cα for some c ∈ R, and since neither vector is the zero vector, c , 0. Then y1 = cx1 and y2 = cx2. Thus the conditions on x1, x2, y1, y2 implies 0 = x1y1 + x2y2 = cx21 + cx 2 2 = c(x 2 1 + x 2 2) = c · 1 = c. Thus c = 0, a contradiction. It remains to find the coordinates of the arbitrary vector (a, b) in the ordered basis {α, β}. To find the coordinates of (a, b) we can row-reduce the augmented matrix [ x1 y1 a x2 y2 b ] . It cannot be that both x1 = x2 = 0 so assume WLOG that x1 , 0. Also it cannot be that both y1 = y2 = 0. Assume first that y1 , 0. Since order matters we cannot assume y1 , 0 WLOG, so we must consider both cases. Then note that x1y1 + x2y2 = 0 implies x2y2 x1y1 = −1 (18) Thus if x1y2 − x2y1 = 0 then x2x1 = y2 y1 from which (18) implies ( x2 x1 )2 = −1, a contradiction. Thus we can conclude that x1y2 − x2y1 , 0. We use this in the following row reduction to be sure we are not dividing by zero.[ x1 y1 a x2 y2 b ] → [ 1 y1/x1 a/x1 x2 y2 b ] → [ 1 y1/x1 a/x1 0 y2 − x2y1 x1 b − x2ax1 ] = [ 1 y1/x1 a/x1 0 x1y2−x2y1x1 bx1−ax2 x1 ] → [ 1 y1/x1 a/x1 0 1 bx1−ax2x1y2−x2y1 ] → 1 0 ay2−by1 x1y2−x2y1 0 1 bx1−ax2x1y2−x2y1 Now if we substitute y1 = −x2y2/x1 into the numerator and denominator of ay2−by1 x1y2−x2y1 and use x21 + x 2 2 = 1 it simplifies to ax1 + bx2. Similarly ay2−by1 x1y2−x2y1 simplifies to ay1 + by2. So we get[ 1 0 ax1 + bx2 0 1 ay1 + by2 ] . 44 Chapter 2: Vector Spaces Now assume y2 , 0 (and we continue to assume x1 , 0 since we assumed that WLOG). In this case y1 y2 = − x2 x1 (19) So if x1y2 − x2y1 = 0 then x2y1 x1y2 = 1. But then (19) implies ( x2 x1 )2 = −1 a contradition. So also in this case we can assume x1y2 − x2y1 , 0 and so we can do the same row-reduction as before. Thus in all cases (ax1 + bx2)α + (ay1 + by2)β = (a, b) or equivalently (ax1 + bx2)(x1, x2) + (ay1 + by2)(y1, y2) = (a, b). Exercise 6: Let V be the vector space over the complex numbers of all functions from R into C, i.e., the space of all complex- valued functions on the real line. Let f1(x) = 1, f2(x) = eix, f3(x) = e−ix. (a) Prove that f1, f2, and f3 are linearly independent. (b) Let g1(x) = 1, g2(x) = cos x, g3(x) = sin x. Find an invertible 3 × 3 matrix P such that g j = 3∑ i=1 Pi j fi. Solution: Suppose a + beix + ce−ix = 0 as functions of x ∈ R. In other words a + beix + ce−ix = 0 for all x ∈ R. Let y = eix. Then y , 0 and a + by + cy = 0 which implies ay + by 2 + c = 0. This is at most a quadratic polynomial in y thus can be zero for at most two values of y. But eix takes infinitely many different values as x varies in R, so ay + by2 + c cannot be zero for all y = eix, so this is a contradiction. We know that eix = cos(x) + i sin(x). Thus e−ix = cos(x) − i sin(x). Adding these gives 2 cos(x) = eix + e−ix. Thus cos(x) = 12 e ix + 12 e −ix. Subtracting instead of adding the equations gives eix − e−ix = 2i sin(x). Thus sin(x) = 12i e ix − 12i e −ix or equivalently sin(x) = − i2 e ix + i2 e −ix. Thus the requested matrix is P = 1 0 00 1/2 −i/20 1/2 i/2 . Exercise 7: Let V be the (real) vector space of all polynomial functions from R into R of degree 2 or less, i.e., the space of all functions f of the form f (x) = c0 + c1x + c2x2. Let t be a fixed real number and define g1(x) = 1, g2(x) = x + t, g3(x) = (x + t)2. Prove that B = {g1, g2, g3} is a basis for V . If f (x) = c0 + c1x + c2x2 what are the coordinates of f in this ordered basis B? Solution: We know V has dimension three (it follows from Example 16, page 43, that {1, x, x2} is a basis). Thus by Corollary 2 (b), page 45, it suffices to show {g1, g2, g3} span V . We need to solve for u, v,w the equation c2x2 + c1x + c0 = u + v(x + t) + w(x + t)2. Rearranging c2x2 + c1x + c0 = wx2 + (v + 2wt)x + (u + vt + wt2). Section 2.6: Computations Concerning Subspaces 45 It follows that w = c2 v = c1 − 2c2t u = c0 − c1t + c2t2. Thus {g1, g2, g3} do span V and the coordinates of f (x) = c2x2 + c1x + c0 are (c2, c1 − 2c2t, c0 − c1t + c2t2). Section 2.6: Computations Concerning Subspaces Exercise 1: Let s < n and A an s × n matrix with entries in the field F. Use Theorem 4 (not its proof) to show that there is a non-zero X in Fn×1 such that AX = 0. Solution: Let α1, α2, . . . , αn be the colunms of A. Then αi ∈ F s×1 ∀ i. Thus {α1, . . . , αn} are n vectors in F s×1. But F s×1 has dimension s < n thus by Theorem 4, page 44, α1, . . . , αn cannot be linearly independent. Thus ∃ x1, . . . , xn ∈ F such that x1α1 + · · · + xnαn = 0. Thus if X = x1 ... xn then AX = x1α1 + · · · + xnαn = 0. Exercise 2: Let α1 = (1, 1,−2, 1), α2 = (3, 0, 4,−1), α3 = (−1, 2, 5, 2). Let α = (4,−5, 9,−7), β = (3, 1,−4, 4), γ = (−1, 1, 0, 1). (a) Which of the vectors α, β, γ are in the subspace of R4 spanned by the αi? (b) Which of the vectors α, β, γ are in the subspace of C4 spanned by the αi? (c) Does this suggest a theorem? Solution: (a) We use the approach of row-reducing the matrix whose rows are given by the αi: 1 1 −2 13 0 4 −1 −1 2 5 2 → 1 1 −2 10 −3 10 −40 3 3 3 → 1 1 −2 10 0 13 −10 1 1 1 → 1 0 −3 00 1 1 10 0 13 −1 48 Chapter 2: Vector Spaces in R5 which are linear combinations of the vectors α1 = (1, 0, 2, 1,−1), α2 = (−1, 2,−4, 2, 0) α3 = (2,−1, 5, 2, 1), α4 = (2, 1, 3, 5, 2). Solution: We row-reduce the matrix whose rows are given by the αi’s. 1 0 2 1 −1 −1 2 −4 2 0 2 −1 5 2 1 2 1 3 5 2 → 1 0 2 1 −1 0 2 −2 3 −1 0 −1 1 0 3 0 1 −1 3 4 → 1 0 2 1 −1 0 1 −1 3 4 0 0 0 −3 −9 0 0 0 3 7 → 1 0 2 1 −1 0 1 −1 3 4 0 0 0 1 3 0 0 0 3 7 → 1 0 2 1 −4 0 1 −1 0 −5 0 0 0 1 3 0 0 0 0 −2 → 1 0 2 0 −4 0 1 −1 0 −5 0 0 0 1 3 0 0 0 0 1 → 1 0 2 0 0 0 1 −1 0 0 0 0 0 1 0 0 0 0 0 1 Let ρ1 = (1, 0, 2, 0, 0), ρ2 = (0, 1,−1, 0, 0), ρ3 = (0, 0, 0, 1, 0) and ρ4 = (0, 0, 0, 0, 1). Then the general element that is a linear combination of the αi’s is b1ρ1 + b2ρ2 + b3ρ3 + b4ρ4 = (b1, b2, 2b1 − b2, b3, b4). Exercise 6: Let V be the real vector space spanned by the rows of the matrix A = 3 21 0 9 0 1 7 −1 −2 −1 2 14 0 6 1 6 42 −1 13 0 . (a) Find a basis for V . (b) Tell which vectors (x1, x2, x3, x4, x5) are elements of V . Section 2.6: Computations Concerning Subspaces 49 (c) If (x1, x2, x3, x4, x5) is in V what are its coordinates in the basis chosen in part (a)? Solution: We row-reduce the matrix 3 21 0 9 0 1 7 −1 −2 −1 2 14 0 6 1 6 42 −1 13 0 → 1 7 −1 −2 −1 0 0 3 15 3 0 0 2 10 3 0 0 5 25 6 → 1 7 −1 −2 −1 0 0 1 5 1 0 0 2 10 3 0 0 5 25 6 → 1 7 0 3 0 0 0 1 5 1 0 0 0 0 1 0 0 0 0 1 → 1 7 0 3 0 0 0 1 5 0 0 0 0 0 1 0 0 0 0 0 (a) A basis for V is given by the non-zero rows of the reduced matrix ρ1 = (1, 7, 0, 3, 0), ρ2 = (0, 0, 1, 5, 0), ρ3 = (0, 0, 0, 0, 1). (b) Vectors of V are any of the form b1ρ1 + b2ρ2 + b3ρ3 = (b1, 7b1, b2, 3b1 + 5b2, b3) for arbitrary b1, b2, b3 ∈ R. (c) By the above, the element (x1, x2, x3, x4, x5) in V must be of the form x1ρ1 + x3ρ2 + x5ρ3. In other words if B = {ρ1, ρ2, ρ3} is the basis for V given in part (a), then the coordinate matrix of (x1, x2, x3, x4, x5) is [(x1, x2, x3, x4, x5)]B = x1x3x5 . Exercise 7: Let A be an m × n matrix over the field F, and consider the system of equations AX = Y . Prove that this system of equations has a solution if and only if the row rank of A is equal to the row rank of the augmented matrix of the system. Solution: To solve the system we row-reduce the augmented matrix [A |Y] resulting in an augmented matrix [R |Z] where R is in reduced echelon form and Z is an m× 1 matrix. If the last k rows of R are zero rows then the system has a solution if and only if the last k entries of Z are also zeros. Thus the only non-zero entries in Z are in the non-zero rows of R. These rows are already linearly independent, and they clearly remain independent regardless of the augmented values. Thus if there are solutions then the rank of the augmented matrix is the same as the rank of R. Conversely, if there are non-zero entries in Z in any of the last k rows then the system has no solutions. We want to show that those non-zero rows in the augmented matrix are linearly independent from the non-zero rows of R, so we can conclude that the rank of R is less than the rank of [R |Z]. Let S 50 Chapter 2: Vector Spaces be the set of rows of [R |Z] that contain all rows where R is non-zero, plus one additional row r where Z is non-zero. Suppose a linear combination of the elements of S equals zero. Since c · r = 0⇔ r = 0, at least one of the elements of S different from r must have a non-zero coefficient. Suppose row r′ ∈ S has non-zero coefficient c in the linear combination. Suppose the leading one in row r′ is in position i. Then the i-th coordinate of the linear combination is also c, because except for the one in the i-th position, all other entries in the i-th column of [R |Z] are zero. Thus there can be no non-zero coefficients. Thus the set S is linearly independent and |S | = |R| + 1. Thus the system has a solution if and only if the rank of R is the same as the rank of [R |Z]. Now A has the same rank as R and [R |Z] has the same rank as [A |Y] since they differ by elementary row operations. Thus the system has a solution if and only if the rank of A is the same as the rank of [A |Y]. Section 3.1: Linear Transformations 53 (c) What are the conditions on a, b, and c that (a, b, c) be in the null space of T? What is the nullity of T? Solution: (a) Let P = 1 −1 22 1 0 −1 −2 2 . Then T can be represented by x1x2x3 7→ P x1x2x3 . By Example 4, page 68, this is a linear transformation, where we’ve identified F3 with F3×1 and taken Q in Example 4 to be the identity matrix. (b) The range of T is the column space of P, or equivalently the row space of PT = 1 2 −1−1 1 −22 0 2 . We row reduce the matrix as follows → 1 2 −10 3 −30 −4 4 . → 1 2 −10 1 −10 0 0 . → 1 0 10 1 −10 0 0 . Let ρ1 = (1, 0, 1) and ρ2 = (0, 1,−1). Then elements of the row space are elements of the form b1ρ1 + b2ρ2 = (b1, b2, b1 − b2). Thus the rank of T is two and (a, b, c) is in the range of T as long as c = a − b. Alternatively, we can row reduce the augmented matrix 1 −1 2 a2 1 0 b −1 −2 2 c → 1 −1 2 a0 3 −4 b − 2a0 −3 4 a + c → 1 −1 2 a0 3 −4 b − 2a0 0 0 −a + b + c → 1 −1 2 a0 1 −4/3 (b − 2a)/40 0 0 −a + b + c 54 Chapter 3: Linear Transformations → 1 0 2/3 (b + 2a)/40 1 −4/3 (b − 2a)/40 0 0 −a + b + c from which we arrive at the condition −a + b + c = 0 or equivalently c = a − b. (c) We must find all X = abc such that PX = 0 where P is the matrix from part (a). We row reduce the matrix 1 −1 22 1 0 −1 −2 2 → 1 −1 20 3 −40 −3 4 → 1 −1 20 3 −40 0 0 → 1 −1 20 1 −4/30 0 0 → 1 0 2/30 1 −4/30 0 0 Therefore { a + 23 c = 0 b − 43 c = 0 So elements of the null space of T are of the form (− 23 c, 4 3 c, c) for arbitrary c ∈ F and the dimension of the null space (the nullity) equals one. Exercise 8: Describe explicitly a linear transformation from R3 to R3 which has as its range the subspace spanned by (1, 0,−1) and (1, 2, 2). Solution: By Theorem 1, page 69, (and its proof) there is a linear transformation T from R3 to R3 such that T (1, 0, 0) = (1, 0,−1), T (0, 1, 0) = (1, 0,−1) and T (0, 0, 1) = (1, 2, 2) and the range of T is exactly the subspace generated by {T (1, 0, 0),T (0, 1, 0),T (0, 0, 1)} = {(1, 0,−1), (1, 2, 2)}. Exercise 9: Let V be the vector space of all n × n matrices over the field F, and let B be a fixed n × n matrix. If T (A) = AB − BA verify that T is a linear transformation from V into V . Solution: T (cA1 + A2) = (cA1 + A2)B − B(cA1 + A2) = cA1B + A2B − cBA1 − BA2 = c(A1B − BA1) + (A2B − BA2) = cT (A1) + T (A2). Section 3.2: The Algebra of Linear Transformations 55 Exercise 10: Let V be the set of all complex numbers regarded as a vector space over the field of real numbers (usual oper- ations). Find a function from V into V which is a linear transformation on the above vector space, but which is not a linear transformation on C1, i.e., which is not complex linear. Solution: Let T : V → V be given by a + bi 7→ a. Let z = a + bi and w = a′ + b′i and c ∈ R. Then T (cz + w) = T ((ca + a′) + (cb + b′)i) = ca + a′ = cT (a + bi) + T (a′ + b′i) = aT (z) + T (w). Thus T is real linear. However, if T were complex linear then we must have 0 = T (i) = T (i · 1) = i · T (1) = i · 1 = i. But 0 , i so this is a contradiction. Thus T is not complex linear. Exercise 11: Let V be the space of n × 1 matrices over F and let W be the space of m × 1 matrices over F. Let A be a fixed m × n matrix over F and let T be the linear transformation from V into W defined by T (X) = AX. Prove that T is the zero transformation if and only if A is the zero matrix. Solution: If A is the zero matrix then clearly T is the zero transformation. Conversely, suppose A is not the zero matrix, suppose the k-th column Ak has a non-zero entry. Then T (k) = Ak , 0. Exercise 12: Let V be an n-dimensional vector space over the field F and let T be a linear transformation from V into V such that the range and null space of T are identical. Prove that n is even. (Can you give an example of such a linear transformation T?) Solution: From Theorem 2, page 71, we know rank(T ) + nullity(T ) = dim V . In this case we are assuming both terms on the left hand side are equal, say equal to m. Thus m + m = n or equivalently n = 2m which implies n is even. The simplest example is V = {0} the zero space. Then trivially the range and null space are equal. To give a less trivial example assume V = R2 and define T by T (1, 0) = (0, 0) and T (0, 1) = (1, 0). We can do this by Theorem 1, page 69 because {(1, 0), (0, 1)} is a basis for R2. Then clearly the range and null space are both equal to the subspace of R2 generated by (1, 0). Exercise 13: Let V be a vector space and T a linear transformation from V into V . Prove that the following two statements about T are equivalent. (a) The intersection of the range of T and the null space of T is the zero subspace of V . (b) If T (Tα) = 0, then Tα = 0. Solution: (a)⇒ (b): Statement (a) says that nothing in the range gets mapped to zero except for 0. In other words if x is in the range of T then T x = 0⇒ x = 0. Now Tα is in the range of T , thus T (Tα) = 0⇒ Tα = 0. (b)⇒ (a): Suppose x is in both the range and null space of T . Since x is in the range, x = Tα for some α. But then x in the null space of T implies T (x) = 0 which implies T (Tα) = 0. Thus statement (b) implies Tα = 0 or equivalently x = 0. Thus the only thing in both the range and null space of T is the zero vector 0. Section 3.2: The Algebra of Linear Transformations Page 76: Typo in line 1: It says Ai j, . . . , Am j, it should say A1 j, . . . , Am j. Exercise 1: Let T and U be the linear operators on R2 defined by T (x1, x2) = (x2, x1) and U(x1, x2) = (x1, 0). (a) How would you describe T and U geometrically? (b) Give rules like the ones defining T and U for each of the transformations (U + T ), UT , TU, T 2, U2. 58 Chapter 3: Linear Transformations = 0 0 00 0 00 0 0 . Exercise 5: Let C2×2 be the complex vector space of 2 × 2 matrices with complex entries. Let B = [ 1 −1 −4 4 ] and let T be the linear operator on C2×2 defined by T (A) = BA. What is the rank of T? Can you describe T 2? Solution: An (ordered) basis for C2×2 is given by A11 = [ 1 0 0 0 ] , A21 = [ 0 0 1 0 ] A12 = [ 0 1 0 0 ] , A22 = [ 0 0 0 1 ] . If we identify C2×2 with C4 by [ a b c d ] 7→ (a, b, c, d) then since A11 7→ A11 − 4A21 A21 7→ −A11 + 4A21 A12 7→ A12 − 4A22 A22 7→ −A12 + 4A22 the matrix of the transformation is given by 1 −4 0 0 −1 4 0 0 0 0 1 −4 0 0 −1 4 . To find the rank of T we row-reduce this matrix: → 1 −4 0 0 0 0 1 −4 0 0 0 0 0 0 0 0 . It has rank two so the rank, so the rank of T is 2. T 2(A) = T (T (A)) = T (BA) = B(BA) = B2A. Thus T 2 is given by multiplication by a matrix just as T is, but multiplication with B2 instead of B. Explicitly B2 = [ 1 −1 −4 4 ] [ 1 −1 −4 4 ] = [ 5 −5 −20 20 ] . Exercise 6: Let T be a linear transformation from R3 into R2, and let U be a linear transformation from R2 into R3. Prove that the transformation UT is not invertible. Generalize the theorem. Section 3.2: The Algebra of Linear Transformations 59 Solution: Let {α1, α2, α3} be a basis for R3. Then T (α1),T (α2),T (α3) must be linearly dependent in R2, because R2 has dimension 2. So suppose b1T (α1) + b2T (α2) + b3T (α3) = 0 and not all b1, b2, b3 are zero. Then b1α1 + b2α2 + b3α3 , 0 and UT (b1α1 + b2α2 + b3α3) = U(T (b1α1 + b2α2 + b3α3)) = U(b1T (α1) + b2T (α2) + b3T (α3) = U(0) = 0. Thus (by the definition at the bottom of page 79) UT is not non-singular and thus by Theorem 9, page 81, UT is not invertible. The obvious generalization is that if n > m and T : Rn → Rm and U : Rm → Rn are linear transformations, then UT is not invertible. The proof is an immediate generalization the proof of the special case above, just replace α3 with . . . , αn. Exercise 7: Find two linear operators T and U on R2 such that TU = 0 but UT , 0. Solution: Identify R2 with R2×1 and let T and U be given by the matrices A = [ 1 0 0 0 ] , B = [ 0 1 0 0 ] . More precisely, for X = [ x y ] . let T be given by X 7→ AX and let U be given by X 7→ BX. Thus TU is given by X 7→ ABX and UT is given by X 7→ BAX. But BA = 0 and AB , 0 so we have the desired example. Exercise 8: Let V be a vector space over the field F and T a linear operator on V . If T 2 = 0, what can you say about the relation of the range of T to the null space of T? Give an example of a linear operator T on R2 such that T 2 = 0 but T , 0. Solution: If T 2 = 0 then the range of T must be contained in the null space of T since if y is in the range of T then y = T x for some x so Ty = T (T x) = T 2x = 0. Thus y is in the null space of T . To give an example of an operator where T 2 = 0 but T , 0, let V = R2×1 and let T be given by the matrix A = [ 0 1 0 0 ] . Specifically, for X = [ x y ] . let T be given by X 7→ AX. Since A , 0, T , 0. Now T 2 is given by X 7→ A2X, but A2 = 0. Thus T 2 = 0. Exercise 9: Let T be a linear operator on the finite-dimensional space V . Suppose there is a linear operator U on V such that TU = I. Prove that T is invertible and U = T−1. Give an example which shows that this is false when V is not finite- dimensional. (Hint: Let T = D, be the differentiation operator on the space of polynomial functions.) Solution: By the comments in the Appendix on functions, at the bottom of page 389, we see that simply because TU = I as functions, then necessarily T is onto and U is one-to-one. It then follows immediately from Theorem 9, page 81, that T is invertible. Now TT−1 = I = TU and multiplying on the left by T−1 we get T−1TT−1 = T−1TU which implies (I)T−1 = (I)U and thus U = T−1. 60 Chapter 3: Linear Transformations Let V be the space of polynomial functions in one variable over R. Let D be the differentiation operator and let T be the operator “multiplication by x” (exactly as in Example 11, page 80). As shown in Example 11, UT = I while TU , I. Thus this example fulfills the requirement. Exercise 10: Let A be an m × n matrix with entries in F and let T be the linear transformation from Fn×1 into Fm×1 defined by T X = AX. Show that if m < n it may happen that T is onto without being non-singular. Similarly, show that if m > n we may have T non-singular but not onto. Solution: Let B = {α1, . . . , αn} be a basis for Fn×1 and let B′ = {β1, . . . , βm} be a basis for Fm×1. We can define a linear transformation from Fn×1 to Fm×1 uniquely by specifying where each member of B goes in Fm×1. If m < n then we can define a linear transformation that maps at least one member of B to each member of B′ and maps at least two members of B to the same member of B′. Any linear transformation so defined must necessarily be onto without being one-to-one. Similarly, if m > n then we can map each member of B to a unique member of B′ with at least one member of B′ not mapped to by any member of B. Any such transformation so defined will necessarily be one-to-one but not onto. Exercise 11: Let V be a finite-dimensional vector space and let T be a linear operator on V . Suppose that rank(T 2) = rank(T ). Prove that the range and null space of T are disjoint, i.e., have only the zero vector in common. Solution: Let {α1, . . . , αn} be a basis for V . Then the rank of T is the number of linearly independent vectors in the set {Tα1, . . . ,Tαn}. Suppose the rank of T equals k and suppose WLOG that {Tα1, . . . ,Tαk} is a linearly independent set (it might be that k = 1, pardon the notation). Then {Tα1, . . . ,Tαk} give a basis for the range of T . It follows that {T 2α1, . . . ,T 2αk} span the range of T 2 and since the dimension of the range of T 2 is also equal to k, {T 2α1, . . . ,T 2αk} must be a basis for the range of T 2. Now suppose v is in the range of T . Then v = c1Tα1 + · · · + ckTαk. Suppose v is also in the null space of T . Then 0 = T (v) = T (c1Tα1 + · · · + ckTαk) = c1T 2α1 + · · · + ckT 2αk. But {T 2α1, . . . ,T 2αk} is a basis, so T 2α1, . . . ,T 2αk are linearly independent, thus it must be that c1 = · · · = ck = 0, which implies v = 0. Thus we have shown that if v is in both the range of T and the null space of T then v = 0, as required. Exercise 12: Let p,m, and n be positive integers and F a field. Let V be the space of m × n matrices over F and W the space of p × n matrices over F. Let B be a fixed p × m matrix and let T be the linear transformation from V into W defined by T (A) = BA. Prove that T is invertible if and only if p = m and B is an invertible m × m matrix. Solution: We showed in Exercise 2.3.12, page 49, that the dimension of V is mn and the dimension of W is pn. By Theorem 9 page (iv) we know that an invertible linear transformation must take a basis to a basis. Thus if there’s an invertible linear transformation between V and W it must be that both spaces have the same dimension. Thus if T is inverible then pn = mn which implies p = m. The matrix B is then invertible because the assignment B 7→ BX is one-to-one (Theorem 9 (ii), page 81) and non-invertible matrices have non-trivial solutions to BX = 0 (Theorem 13, page 23). Conversely, if p = n and B is invertible, then we can define the inverse transformation T−1 by T−1(A) = B−1A and it follows that T is invertible. Section 3.3: Isomorphism Exercise 1: Let V be the set of complex numbers and let F be the field of real numbers. With the usual operations, V is a vector space over F. Describe explicitly an isomorphism of this space onto R2. Solution: The natural isomorphism from V to R2 is given by a + bi 7→ (a, b). Since i acts like a placeholder for addition in C, (a + bi) + (c + di) = (a + c) + (b + d)i 7→ (a + c, b + d) = (a, b) + (c, d). And c(a + bi) = ca + cbi 7→ (ca, cb) = c(a, b). Thus this is a linear transformation. The inverse is clearly (a, b) 7→ a + bi. Thus the two spaces are isomorphic as vector spaces over R. Exercise 2: Let V be a vector space over the field of complex numbers, and suppose there is an isomorphism T of V into C3. Let α1, α2, α3, α4 be vectors in V such that Tα1 = (1, 0, i), Tα2 = (−2, 1 + i, 0), Section 3.4: Representation of Transformations by Matrices 63 (c) How would you describe the range of T? Solution: (a) The four coordinates of T (z) are written as linear combinations of the coordinates of z (as a vector over R). Thus T is clearly a linear transformation. To see that T is one-to-one, let z = x + yi and w = a + bi and suppose T (z) = T (w). Then considering the top right entry of the matrix we see that 5y = 5b which implies b = y. It now follows from the top left entry of the matrix that x = a. Thus T (z) = T (w)⇒ z = w, thus T is one-to-one. (b) Let z1 = x + yi and z2 = a + bi. Then T (z1z2) = T ((ax − by) + (ay + bx)i) = [ (ax − by) + 7(ay + bx) 5(ay + bx) −10(ay + bx) (ax − by) − 7(ay + bx) ] . On the other hand, T (z1)T (z2) = [ x + 7y 5y −10y x − 7y ] [ a + 7b 5b −10b a − 7b ] = [ (ax − by) + 7(ay + bx) 5(ay + bx) −10(ay + bx) (ax − by) − 7(ay + bx) ] . Thus T (z1z2) = T (z1)T (z2). (c) The range of T has (real) dimension equal to two by part (a), and so the range of T is isomorphic to C as real vector spaces. But both spaces also have a natural multiplication and in part (b) we showed that T respects the multiplication. Thus the range of T is isomorphic to C as fields and we have essentially found an isomorphic copy of the field C in the algebra of 2 × 2 real matrices. Exercise 6: Let V and W be finite-dimensional vector spaces over the field F. Prove that V and W are isomorphic if and only if dim(V) = dim(W). Solution: Suppose dim(V) = dim(W) = n. By Theorem 10, page 84, both V and W are isomorphic to Fn, and consequently, since isomorphism is an equivalence relation, V and W are isomorphic to each other. Conversely, suppose T is an isomor- phism from V to W. Suppose dim(W) = n. Then by Theorem 10 again, there is an isomorphism S : W → Fn. Thus S T is an isomorphism from V to Fn implying also dim(V) = n. Exercise 7: Let V and W be vector spaces over the field F and let U be an isomorphism of V onto W. Prove that T → UTU−1 is an isomorphism of L(V,V) onto L(W,W). Solution: L(V,V) is defined on page 75 as the vector space of linear transformations from V to V , and likewise L(W,W) is the vector space of linear transformations from W to W. Call the function f . We know f (T ) is linear since it is a composition of three linear tranformations UTU−1. Thus indeed f is a function from L(V,V) to L(W,W). Now f (aT + T ′) = U(aT + T ′)U−1 = (aUT + UT ′)U−1 = aUTU−1 + UT ′U−1 = a f (T ) + f (T ′). Thus f is linear. We just must show f has an inverse. Let g be the function from L(W,W) to L(V,V) given by g(T ) = U−1TU. Then g f (T ) = U−1(UTU−1)U = T . Similarly f g = I. Thus f and g are inverses. Thus f is an isomorphism. Section 3.4: Representation of Transformations by Matrices Page 90: Typo. Four lines from the bottom it says “Example 12” where they probably meant Example 10 (page 78). Page 91: Just before (3-8) it says ”By definition”. I think it’s more than just by definition, see bottom of page 88. 64 Chapter 3: Linear Transformations Exercise 1: Let T be the linear operator on C2 defined by T (x1, x2) = (x1, 0). Let B be the standard ordered basis for C2 and let B′ = {α1, α2} be the ordered basis defined by α1 = (1, i), α2 = (−i, 2). (a) What is the matrix of T relative to the pair B, B′? (b) What is the matrix of T relative to the pair B′, B? (c) What is the matrix of T in the ordered basis B′? (d) What is the matrix of T in the ordered basis {α2, α1}? Solution: (a) According to the comments at the bottom of page 87, the i-th column of the matrix is given by [T i]B′ , where 1 = (1, 0) and 2 = (0, 1), the standard basis vectors of C2. Now T 1 = (1, 0) and T 2 = (0, 0). To write these in terms of α1 and α2 we use the approach of row-reducing the augmented matrix[ 1 −i 1 0 i 2 0 0 ] → [ 1 −i 1 0 0 1 −i 0 ] → [ 1 0 2 0 0 1 −i 0 ] . Thus T 1 = 2α1 − iα2 and T 2 = 0 · α1 + 0 · α2 and the matrix of T relative to B, B′ is[ 2 0 −i 0 ] . (b) In this case we have to write Tα1 and Tα2 as linear combinations of 1, 2. Tα1 = (1, 0) = 1 · 1 + 0 · 2 Tα2 = (−i, 0) = −i · 1 + 0 · 2. Thus the matrix of T relative to B′, B is [ 1 −i 0 0 ] . (c) In this case we need to write Tα1 and Tα2 as linear combinations of α1 and α2. Tα1 = (1, 0), Tα2 = (−i, 0). We row-reduce the augmented matrix: [ 1 −i 1 −i i 2 0 0 ] → [ 1 −i 1 −i 0 1 −i −1 ] → [ 1 0 2 −2i 0 1 −i −1 ] . Thus the matrix of T in the ordered basis B′ is [ 2 −2i −i −1 ] . (d) In this case we need to write Tα2 and Tα1 as linear combinations of α2 and α1. In this case the matrix we need to row-reduce is just the same as in (c) but with columns switched:[ −i 1 −i 1 2 i 0 0 ] → [ 1 i 1 i 2 i 0 0 ] → [ 1 i 1 i 0 −i −2 −2i ] → [ 1 i 1 i 0 1 −2i 2 ] → [ 1 0 −1 −i 0 1 −2i 2 ] Thus the matrix of T in the ordered basis {α2, α1} is [ −1 −i −2i 2 ] . Exercise 2: Let T be the linear transformation from R3 to R2 defined by T (x1, x2, x3) = (x1 + x2, 2x3 − x1). Section 3.4: Representation of Transformations by Matrices 65 (a) If B is the standard ordered basisfor R3 and B′ is the standard ordered basis for R2, what is the matrix of T relative to the pair B, B′? (b) If B = {α1, α2, α3} and B′ = {β1, β2}, where α1 = (1, 0,−1), α2 = (1, 1, 1), α3 = (1, 0, 0), β1 = (0, 1), β2 = (1, 0) what is the matrix of T relative to the pair B, B′? Solution: With respect to the standard bases, the matrix is simply[ 1 1 0 −1 0 2 ] . (b) We must write Tα1,Tα2,Tα3 in terms of β1, β2. Tα1 = (1,−3) Tα2 = (2, 1) Tα3 = (1, 0). We row-reduce the augmented matrix[ 0 1 1 2 1 1 0 −3 1 0 ] → [ 1 0 −3 1 0 0 1 1 2 1 ] . Thus the matrix of T with respect to B, B′ is [ −3 1 0 1 2 1 ] . Exercise 3: Let T be a linear operator on Fn, let A be the matrix of T in the standard ordered basis for Fn, and let W be the subspace of Fn spanned by the column vectors of A. What does W have to do with T? Solution: Since {α1, . . . , αn} is a basis of Fn, we know {T 1, . . . ,T n} generate the range of T . But T i equals the i-th column vector of A. Thus the column vectors of A generate the range of T (where we identify Fn with Fn×1). We can also conclude that a subset of the columns of A give a basis for the range of T . Exercise 4: Let V be a two-dimensional vector space over the field F, and let B be an ordered basis for V . If T is a linear operator on V and [T ]B = [ a b c d ] prove that T 2 − (a + d)T + (ad − bc)I = 0. Solution: The coordinate matrix of T 2 − (a + d)T + (ad − bc)I with respect to B is [T 2 − (a + d)T + (ad − bc)I]B = [ a b c d ]2 − (a + d) [ a b c d ] + (ad − bc) [ 1 0 0 1 ] Expanding gives = [ a2 + bc ab + bd ac + cd bc + d2 ] − [ a2 + ad ab + bd ac + cd ad + d2 ] + [ ad − bc 0 0 ad − bc ] = [ 0 0 0 0 ] . 68 Chapter 3: Linear Transformations → [ 1 c/a −b/a −d/a 0 ad−bca a2+b2 a ac+bd a ] . → [ 1 c/a −b/a −d/a 0 1 a 2+b2 ad−bc ac+bd ad−bc ] . → 1 0 ac+bdad−bc c 2+d2 ad−bc 0 1 a 2+b2 ad−bc ac+bd ad−bc . If b , 0 then a similar computation results in the same thing. Thus [T ]B = ac+bdad−bc c 2+d2 ad−bc a2+b2 ad−bc ac+bd ad−bc . Now ad− bc , 0 implies that at least one of a or b is non-zero and at least one of c or d is non-zero, it follows that a2 + b2 > 0 and c2 + d2 > 0. Thus (a2 + b2)(c2 + d2) , 0. Thus a2 + b2 ad − bc · c2 + d2 ad − bc , 0 Exercise 7: Let T be the linear operator on R3 defined by T (x1, x2, x3) = (3x1 + x3, −2x1 + x2, −x1 + 2x2 + 4x3). (a) What is the matrix of T in the standard ordered basis for R3. (b) What is the matrix of T in the ordered basis (α1, α2, α3) where α1 = (1, 0, 1), α2 = (−1, 2, 1), and α3 = (2, 1, 1)? (c) Prove that T is invertible and give a rule for T−1 like the one which defines T . Solution: (a) As usual we can read the matrix in the standard basis right off the definition of T : [T ]{1,2,3} = 3 0 1−2 1 0 −1 2 4 . (b) Tα1 = (4,−2, 3), Tα2 = (−2, 4, 9) and Tα3 = (7,−3, 4). We must write these in terms of α1, α2, α3. We do this by row-reducing the augmented matrix 1 −1 2 4 −2 70 2 1 −2 4 −31 1 1 3 9 4 → 1 −1 2 4 −2 70 2 1 −2 4 −30 2 −1 −1 11 −3 → 1 −1 2 4 −2 70 2 1 −2 4 −30 0 −2 1 7 0 → 1 −1 2 4 −2 70 1 1/2 −1 2 −3/20 0 1 −1/2 −7/2 0 Section 3.4: Representation of Transformations by Matrices 69 → 1 0 5/2 3 0 11/20 1 1/2 −1 2 −3/20 0 1 −1/2 −7/2 0 → 1 0 0 17/4 35/4 11/20 1 0 −3/4 15/4 −3/20 0 1 −1/2 −7/2 0 Thus the matrix of T in the basis {α1, α2, α3} is [T ]{α1,α2,α3} = 17/4 35/4 11/2−3/4 15/4 −3/2 −1/2 −7/2 0 . (c) We row reduce the augmented matrix (of T in the standard basis). If we achieve the identity matrix on the left of the dividing line then T is invertible and the matrix on the right will represent T−1 in the standard basis, from which we will be able read the rule for T−1 by inspection. 3 0 1 1 0 0−2 1 0 0 1 0 −1 2 4 0 0 1 → −1 2 4 0 0 13 0 1 1 0 0 −2 1 0 0 1 0 → 1 −2 −4 0 0 −13 0 1 1 0 0 −2 1 0 0 1 0 → 1 −2 −4 0 0 −10 6 13 1 0 30 −3 −8 0 1 −2 → 1 −2 −4 0 0 −10 0 −3 1 2 −10 −3 −8 0 1 −2 → 1 −2 −4 0 0 −10 −3 −8 0 1 −20 0 −3 1 2 −1 → 1 −2 −4 0 0 −10 1 8/3 0 −1/3 2/30 0 −3 1 2 −1 → 1 0 4/3 0 −2/3 1/30 1 8/3 0 −1/3 2/30 0 −3 1 2 −1 → 1 0 4/3 0 −2/3 1/30 1 8/3 0 −1/3 2/30 0 1 −1/3 −2/3 1/3 → 1 0 0 4/9 2/9 −1/90 1 0 8/9 13/9 −2/90 0 1 −1/3 −2/3 1/3 70 Chapter 3: Linear Transformations Thus T is invertible and the matrix for T−1 in the standard basis is 4/9 2/9 −1/98/9 13/9 −2/9 −1/3 −2/3 1/3 . Thus T−1(x1, x2, x3) = ( 4 9 x1 + 2 9 x2 − 1 9 x3, 8 9 x1 + 13 9 x2 − 2 9 x3, − 1 3 x1 − 2 3 x2 + 1 3 x3 ) . Exercise 8: Let θ be a real number. Prove that the following two matrices are similar over the field of complex numbers:[ cos θ − sin θ sin θ cos θ ] , [ eiθ 0 0 e−iθ ] (Hint: Let T be the linear operator on C2 which is represented by the first matrix in the standard ordered basis. Then find vectors α1 and α2 such that Tα1 = eiθα1, Tα2 = e−iθα2, and {α1, α2} is a basis.) Solution: Let B be the standard basis. Following the hint, let T be the linear operator on C2 which is represented by the first matrix in the standard ordered basis B. Thus [T ]B is the first matrix above. Let α1 = (i, 1), α2 = (i,−1). Then α1, α2 are clealry linearly independent so B′ = {α1, α2} is a basis for C2 (as a vector space over C). Since eiθ = cos θ + i sin θ, it follows that Tα1 = (i cos θ− sin θ, i sin θ+cos θ) = (cos θ+ i sin θ)(i, 1) = eiθα1 and similarly since and e−iθ = cos θ− i sin θ, it follows that Tα2 = e−iθα2. Thus the matrix of T with respect to B′ is [T ]B′ = [ eiθ 0 0 e−iθ ] . By Theorem 14, page 92, [T ]B and [T ]B′ are similar. Exercise 9: Let V be a finite-dimensional vector space over the field F and let S and T be linear operators on V . We ask: When do there exist ordered bases B and B′ for V such that [S ]B = [T ]B′? Prove that such bases exist if and only if there is an invertible linear operator U on V such that T = US U−1. (Outline of proof: If [S ]B = [T ]B′ , let U be the operator which carries B onto B′ and show that S = UTU−1. Conversely, if T = US U−1 for some invertible U, let B be any ordered basis for V and let B′ be its image under U. Then show that [S ]B = [T ]B′ .) Solution: We follow the hint. Suppose there exist bases B = {α1, . . . , αn} and B = {β1, . . . , βn} such that [S ]B = [T ]B′ . Let U be the operator which carries B onto B′. Then by Theorem 14, page 92, [US U−1]B′ = [U]−1B [US U −1]B[U]B and by the comments at the very bottom of page 90, this equals [U]−1 B [U]B[S ]B[U]−1B [U]B which equals [S ]B, which we’ve assumed equals [T ]B′ . Thus [US U−1]B′ = [T ]B′ . Thus US U−1 = T . Conversely, assume T = US U−1 for some invertible U. Let B be any ordered basis for V and let B′ be its image under U. Then [T ]B′ = [US U−1]B′ = [U]B′ [S ]B′ [U]−1B′ , which by Theorem 14, page 92, equals [S ]B (because U −1 carries B′ into B). Thus [T ]B′ = [S ]B. Exercise 10: We have seen that the linear operator T on R2 defined by T (x1, x2) = (x1, 0) is represented in the standard ordered basis by the matrix A = [ 1 0 0 0 ] . This operator satisfies T 2 = T . Prove that if S is a linear operator on R2 such that S 2 = S , then S = 0, or S = I, or there is an ordered basis B for R2 such that [S ]B = A (above). Solution: Suppose S 2 = S . Let 1, 2 be the standard basis vectors for R2. Consider {S 1, S 2}. If both S 1 = S 2 = 0 then S = 0. Thus suppose WLOG that S 1 , 0. Section 3.4: Representation of Transformations by Matrices 73 Now let α1 be any basis vector for S n−1(V) which we have shown has dimension one. Now S n−2(V) has dimension two and S takes this space onto a space S n−1(V) of dimension one. Thus there must be α2 ∈ S n−2(V) \ S n−1(V) such that S (α2) = α1. Since α2 is not in the space generated by α1 and {α1, α2} are in the space S n−2(V) of dimension two, it follows that {α1, α2} is a basis for S n−2(V). Now S n−3(V) has dimension three and S takes this space onto a space S n−2(V) of dimension two. Thus there must be α3 ∈ S n−3(V) \ S n−2(V) such that S (α3) = α2. Since α3 is not in the space generated by α1 and α2 and {α1, α2, α3} are in the space S n−3(V) of dimension three, it follows that {α1, α2, α3} is a basis for S n−3(V). Continuing in this way we produce a sequence of elements {α1, α2, . . . , αk} that is a basis for S n−k(V) and such that S (αi) = αi−1 for all i = 2, 3, . . . , k. In particular we have a basis {α1, α2, . . . , αn} for V and such that S (αi) = αi−1 for all i = 2, 3, . . . , n. Reverse the ordering of this bases to give B = {αn, αn−1, . . . , α1}. Then B therefore is the required basis for which the matrix of S with respect to this basis will be the matrix given in part (a). (d) Suppose S is the transformation of Fn×1 given by v 7→ Mv and similarly let T be the transformation v 7→ Nv. Then S n = T n = 0 and S n−1 , 0 , T n−1. Then we know from the previous parts of this problem that there is a basis B for which S is represented by the matrix from part (a). By Theorem 14, page 92, it follows that M is similar to the matrix in part (a). Likewise there’s a basis B′ for which T is represented by the matrix from part (a) and thus the matrix N is also similar to the matrix in part (a). Since similarity is an equivalence relation (see last paragraph page 94), it follows that since M and N are similar to the same matrix that they must be similar to each other. Exercise 13: Let V and W be finite-dimensional vector spaces over the field F and let T be a linear transformation from V into W. If B = {α1, . . . , αn} and B′ = {β1, . . . , βn} are ordered bases for V and W, respectively, define the linear transformations Ep,q as in the proof of Theorem 5: Ep,q(αi) = δi,qβp. Then the Ep,q, 1 ≤ p ≤ m, 1 ≤ q ≤ n, form a basis for L(V,W), and so T = m∑ p=1 n∑ q=1 ApqEp,q for certain scalars Apq (the coordinates of T in this basis for L(V,W)). Show that the matrix A with entries A(p, q) = Apq is precisely the matrix of T relative to the pair B, B′. Solution: Let Ep,qM be the matrix of the linear transformation E p,q with respect to the bases B and B′. Then by the formula for a matrix associated to a linear transformation as given in the proof of Theorem 11, page 87, Ep,qM is the matrix all of whose entries are zero except for the p, q-the entry which is one. Thus A = ∑ p,q Ap,qE p,q M . Since the association between linear transformations and matrices is an isomorphism, T 7→ A implies ∑ p,q ApqEp,q 7→ ∑ p,q ApqE p,q M . And thus A is exactly the matrix whose entries are the Apq’s. Section 3.5: Linear Functionals Page 100: Typo line 5 from the top. It says f (αi) = αi, should be f (αi) = ai. Page 100: In Example 22, it says the matrix 1 1 1t1 t2 t3t21 t22 t23 is invertible “as a short computation shows.” The way to see this is with what we know so far is to row reduce the matrix. As long as t1 , t2 we can get to 1 0 t2−t3t2−t1 0 1 t3−t1t2−t1 0 0 (t3−t1)(t3−t2)t22−t21 . 74 Chapter 3: Linear Transformations Now we can continue and obtain 1 0 t2−t3t2−t1 0 1 t3−t1t2−t1 0 0 1 as long as (t3 − t1)(t3 − t2) , 0. From there we can finish row-reducing to obtain the identity. Thus we can row-reduce the matrix to the identity if and only if t1, t2, t3 are distinct, that is no two of them are equal. Exercise 1: In R3 let α1 = (1, 0, 1), α2 = (0, 1,−2), α3 = (−1,−1, 0). (a) If f is a linear functional on R3 such that f (α1) = 1, f (α2) = −1, f (α3) = 3, and if α = (a, b, c), find f (α). (b) Describe explicitly a linear functional f on R3 such that f (α1) = f (α2) = 0 but f (α3) , 0. (c) Let f be any linear functional such that f (α1) = f (α2) = 0 and f (α3) , 0. If α = (2, 3,−1), show that f (α) , 0. Solution: (a) We need to write (a, b, c) in terms of α1, α2, α3. We can do this by row reducing the following augmented matrix whose colums are the αi’s. 1 0 −1 a0 1 −1 b1 −2 0 c → 1 0 −1 a0 1 −1 b0 −2 −1 c − a → 1 0 −1 a0 1 −1 b0 0 −1 c − a + 2b → 1 0 −1 a0 1 −1 b0 0 1 a − 2b − c → 1 0 0 2a − 2b − c0 1 0 a − b − c0 0 1 a − 2b − c Thus if (a, b, c) = x1α1 + x2α2 + x3α3 then x1 = 2a − 2b − c, x2 = a − b − c and x3 = a − 2b − c. Now f (a, b, c) = f (x1α1 + x2α2 + x3α3) = x1 f (α1) + x2 f (α2) + x3 f (α3) = (2a − 2b − c) · 1 + (a − b − c) · (−1) + (a − 2b − c) · 3 = (2a − 2b − c) − (a − b − c) + (3a − 6b − 3c) = 4a − 7b − 3c. In summary f (α) = 4a − 7b − 3c. (b) Let f (x, y, z) = x − 2y − z. The f (1, 0, 1) = 0, f (0, 1,−2) = 0, and f (−1,−1, 0) = 1. Section 3.4: Representation of Transformations by Matrices 75 (c) Using part (a) we know that α = (2, 3,−1) = −α1 − 3α3 (plug in a = 2, b = 3, c = −1 for the formulas for x1, x2, x3). Thus f (α) = − f (α1) − 3 f (α3) = 0 − 3 f (α3) and since f (α3) , 0, −3 f (α3) , 0 and thus f (α) , 0. Exercise 2: Let B = {α1, α2, α3} be the basis for C3 defined by α1 = (1, 0,−1), α2 = (1, 1, 1), α3 = (2, 2, 0). Find the dual basis of B. Solution: The dual basis { f1, f2, f3} are given by fi(x1, x2, x3) = ∑3 j=1 Ai jx j where (A1,1, A1,2, A1,3) is the solution to the system 1 0 −1 11 1 1 02 2 0 0 , (A2,1, A2,2, A2,3) is the solution to the system 1 0 −1 01 1 1 12 2 0 0 , and (A3,1, A3,2, A3,3) is the solution to the system 1 0 −1 01 1 1 02 2 0 1 , We row reduce the generic matrix 1 0 −1 a1 1 1 b2 2 0 c → 1 0 0 a + b − 1 2 c 0 1 0 c − b − a 0 0 1 b − 12 c . a = 1, b = 0, c = 0⇒ f1(x1, x2, x3) = x1 − x2 a = 0, b = 1, c = 0⇒ f2(x1, x2, x3) = x1 − x2 + x3 a = 0, b = 0, c = 1⇒ f3(x1, x2, x3) = − 12 x1 + x2 − 1 2 x3. Then { f1, f2, f3} is the dual basis to {α1, α2, α3}. Exercise 3: If A and B are n × n matrices over the field F, show that trace(AB) = trace(BA). Now show that similar matrices have the same trace. Solution: (AB)i j = ∑n k=1 AikBk j and (BA)i j = ∑n k=1 BikAk j. Thus trace(AB) = n∑ i=1 (AB)ii = n∑ i=1 n∑ k=1 AikBki = n∑ i=1 n∑ k=1 BkiAik = n∑ k=1 n∑ i=1 BkiAik = n∑ k=1 (BA)kk = trace(BA). Suppose A and B are similar. Then ∃ an invertible n × n matrix P such that A = PBP−1. Thus trace(A) = trace(PBP−1) = trace((P)(BP−1)) = trace((BP−1)(P)) = trace(B). Exercise 4: Let V be the vector space of all polynomial functions p from R into R which have degree 2 or less: p(x) = c0 + c1x + c2x2. Define three linear functionals on V by f1(p) = ∫ 1 0 p(x)dx, f2(x) = ∫ 2 0 p(x)dx, f3(x) = ∫ 3 0 p(x)dx. 78 Chapter 3: Linear Transformations The general element of W0 is therefore f (x1, x2, x3, x4) = (c3 − 2c4)x1 + (c3 + c4)x2 + c3x3 + c4x4, for arbitrary elements c3 and c4. Thus W0 has dimension 2 as expected. Exercise 8: Let W be the subspace of R5 which is spanned by the vectors α1 = 1 + 22 + 3, α2 = 2 + 33 + 34 + 5 α3 = 1 + 42 + 63 + 44 + 5. Find a basis for W0. Solution: The vectors α1, α2, α3 are linearly independent as can be seen by row reducing the matrix 1 2 1 0 00 1 3 3 11 4 6 4 1 → 1 2 1 0 00 1 3 3 10 2 5 4 1 → 1 0 −5 −6 −20 1 3 3 10 0 −1 −2 −1 → 1 0 −5 −6 −20 1 3 3 10 0 1 2 1 → 1 0 0 4 30 1 0 −3 −20 0 1 2 1 . Thus W has dimension 3 and {α1, α2, α3} is a basis for W. We know every functional is given by f (x1, x2, x3, x4, x5) = c1x2 + c2x2 + c3x3 + c4x4 + c5x5 for some c1, . . . , c5. From the row reduced matrix we see that the general solution for an element of W0 is f (x1, x2, x3, x4, x5) = (−4c4 − 3c5)x1 + (3c4 + 2c5)x2 − (2c4 + c5)x3 + c4x4 + c5x5. Exercise 9: Let V be the vector space of all 2 × 2 matrices over the field of real numbers, and let B = [ 2 −2 −1 1 ] . Let W be the subspace of V consisting of all A such that AB = 0. Let f be a linear functional on V which is in the annihilator of W. Suppose that f (I) = 0 and f (C) = 3, where I is the 2 × 2 identity matrix and C = [ 0 0 0 1 ] . Find f (B). Section 3.4: Representation of Transformations by Matrices 79 Solution: The general linear functional on V is of the form f (A) = aA11 +bA12 +cA21 +dA22 for some a, b, c, d ∈ R. If A ∈ W then [ x y z w ] [ 2 −2 −1 1 ] = [ 0 0 0 0 ] implies y = 2x and w = 2y. So W consists of all matrices of the form[ x 2x y 2y ] Now f ∈ W0 ⇒ f ([ x 2x y 2y ]) = 0 ∀ x, y ∈ R⇒ ax + 2bx + cy + 2dy = 0 ∀ x, y ∈ R⇒ (a + 2b)x + (c + 2d)y = 0 ∀ x, y ∈ R ⇒ b = − 12 a and d = − 1 2 c. So the general f ∈ W 0 is of the form f (A) = aA11 − 1 2 aA12 + cA21 − 1 2 cA22. Now f (C) = 3⇒ d = 3⇒ − 12 c = 3⇒ c = −6. And f (I) = 0⇒ a − 1 2 c = 0⇒ c = 2a⇒ a = −3. Thus f (A) = −3A11 + 3 2 A12 − 6A21 + 3A22. Thus f (B) = −3 · 2 + 3 2 · (−2) − 6 · (−1) + 3 · 1 = 0. Exercise 10: Let F be a subfield of the complex numbers. We define n linear functionals on Fn (n ≥ 2) by fk(x1, . . . , xn) = n∑ j=1 (k − j)x j, 1 ≤ k ≤ n. What is the dimension of the subspace annihilated by f1, . . . , fn? Solution: N fk is the subspace annihilated by fk. By the comments on page 101, N fk has dimension n − 1. Now the standard basis vector 2 is in N f2 but is not in N f1 . Thus N f1 and N f2 are distinct hyperspaces. Thus their intersection has dimension n− 2. Now 3 is in N f3 but is not in N f1 ∪N f2 . Thus N f1 ∩N f2 ∩N f3 is the intersection of three distinct hyperspaces and so has dimension n − 3. Continuing in this way, i < ∪i−1j=1N fi . Thus ∪ i j=1N fi is the intersection of i distinct hyperspaces and so has dimension n − i. Thus when i = n we have ∪nj=1N fi has dimension 0. Exercise 11: Let W1 and W2 be subspace of a finite-dimensional vector space V . (a) Prove that (W1 + W2)0 = W01 ∩W 0 2 . (b) Prove that (W1 ∩W2)0 = W01 + W 0 2 . Solution: (a) f ∈ (W1 +W2)0⇒ f (v) = 0 ∀ v ∈ W1 +W2⇒ f (w1 +w2) = 0 ∀ w1 ∈ W1, w2 ∈ W2⇒ f (w1) = 0 ∀ w1 ∈ W1 (take w2 = 0) and f (w2) = 0 ∀ w2 ∈ W2 (take w1 = 0). Thus f ∈ W01 and f ∈ W 0 2 . Thus f ∈ W 0 1 ∩W 0 2 . Thus (W1 + W2) 0 ⊆ W01 ∩W 0 2 . Conversely, let f ∈ W01∩W 0 2 . Let v ∈ W1+W2. Then v = w1+w2 where wi ∈ Wi. Thus f (v) = f (w1+w2) = f (w1)+ f (w2) = 0+0 (since f ∈ W01 and f ∈ W 0 2 ). Thus f (v) = 0 ∀ v ∈ W1 + W2. Thus f ∈ (W1 + W2) 0. Thus W01 ∩W 0 2 ⊆ (W1 + W2) 0. Since (W1 + W2)0 ⊆ W01 ∩W 0 2 and W 0 1 ∩W 0 2 ⊆ (W1 + W2) 0 it follows that W01 ∩W 0 2 = (W1 + W2) 0 (b) f ∈ W01 + W 0 2 ⇒ f = f1 + f2, for some fi ∈ W 0 i . Now let v ∈ W1 ∩W2. Then f (v) = ( f1 + f2)(v) = f1(v) + f2(v) = 0 + 0. Thus f ∈ (W1 ∩W2)0. Thus W01 + W 0 2 ⊆ (W1 ∩W2) 0. 80 Chapter 3: Linear Transformations Now let f ∈ (W1 ∩W2)0. In the proof of Theorem 6 on page 46 it was shown that we can choose a basis for W1 + W2 {α1, . . . , αk, β1, . . . , βm, γ1, . . . , γn} where {α1, . . . , αk} is a basis for W1 ∩W2, {α1, . . . , αk, β1, . . . , βm} is a basis for W1 and {α1, . . . , αk, γ1, . . . , γn} is a basis for W2. We expand this to a basis for all of V {α1, . . . , αk, β1, . . . , βm, γ1, . . . , γn, λ1, . . . , λ`}. Now the general element v ∈ V can be written as v = k∑ i=1 xiαi + m∑ i=1 yiβi + n∑ i=1 ziγi + ∑̀ i=1 wiλi (20) and f is given by f (v) = k∑ i=1 aixi + m∑ i=1 biyi + n∑ i=1 cizi + ∑̀ i=1 diwi for some constants ai, bi, ci, di. Since f (v) = 0 for all v ∈ W1 ∩W2, it follows that a1 = · · · = ak = 0. So f (v) = ∑ biyi + ∑ cizi + ∑ diwi. Define f1(v) = ∑ cizi + ∑ diwi and f2(v) = ∑ biyi. Then f = f1 + f2. Now if v ∈ W1 then v = k∑ i=1 xiαi + m∑ i=1 yiβi so that the coefficients zi and wi in (20) are all zero. Thus f1(v) = 0. Thus f1 ∈ W01 . Similarly if v ∈ W2 then the coefficients yi and wi in (20) are all zero and thus f2(v) = 0. So f2 ∈ W2. Thus f = f1 + f2 where f1 ∈ W01 and f2 ∈ W 0 2 . Thus f ∈ W 0 1 + W 0 2 . Thus (W1 ∩W2)0 ⊆ W01 + W 0 2 . Thus (W1 ∩W2)0 ⊆ W01 + W 0 2 . Exercise 12: Let V be a finite-dimensional vector space over the field F and let W be a subspace of V . If f is a linear functional on W, prove that there is a linear functional g on V suvch that g(α) = f (α) for each α in the subspace W. Solution: Let B be a basis for W and let B′ be a basis for V such that B ⊆ B′. A linear function on a vector space is uniquely determined by its values on a basis, and conversely any function on the basis can be extended to a linear function on the space. Thus we define g on B by g(β) = f (β) ∀ β ∈ B. Then define g(β) = 0 for all β ∈ B′ \ B. Since we have defined g on B′ it defines a linear functional on V and since it agrees with f on a basis for W it agrees with f on all of W. Exercise 13: Let F be a subfield of the field of complex numbers and let V be any vector space over F. Suppose that f and g are linear functionals on V such that the function h defined by h(α) = f (α)g(α) is also a linear functional on V . Prove that either f = 0 or g = 0. Solution: Suppose neither f nor g is the zero function. We will derive a contradiction. Let v ∈ V . Then h(2v) = f (2v)g(2v) = 4 f (v)g(v). But also h(2v) = 2h(v) = 2 f (v)g(v). Therefore f (v)g(v) = 2 f (v)g(v) ∀ v ∈ V . Thus f (v)g(v) = 0 ∀ v ∈ V . Let B be a basis for V . Let B1 = {β ∈ B | f (β) = 0} and B2 = {β ∈ B | g(β) = 0}. Since f (β)g(β) = 0 ∀ β ∈ B, we have B = B1 ∪ B2. Suppose B1 ⊆ B2. Then B2 = B and consequently g is the zero function. Thus B1 * B2. And Section 3.4: Representation of Transformations by Matrices 83 If f (I) = n then c = 1 and we have the trace function. Exercise 17: Let W be the space of n × n matrices over the field F, and let W0 be the subspace spanned by the matrices C of the form C = AB − BA. Prove that W0 is exactly the subspace of matrices which have trace zero. (Hint: What is the dimension of the space of matrices of trace zero? Use the matrix ’units,’ i.e., matrices with exactly one non-zero entry, to construct enough linearly independent matrices of the form AB − BA.) Solution: Let W ′ = {w ∈ W | trace(w) = 0}. We want to show W ′ = W0. We know from Exercise 3 that trace(AB − BA) = 0 for all matrices A, B. Since matrices of the form AB−BA span W0, it follows that trace(M) = 0 for all M ∈ W0. Thus W0 ⊆ W ′. Since the trace function is a linear functional, the dimension of W ′ is dim(W)−1 = n2−1. Thus if we show the dimension of W0 is also n2−1 then we will be done. We do this by exhibiting n2−1 linearly independent elements of W0. Denote by Ei j the ma- trix with a one in the i, j position and zeros in all other positions. Let Hi j = Eii − E j j. Let B = {Ei j | i , j} ∪ {H1,i | 2 ≤ i ≤ n}. We will show that B ⊆ W0 and that B is a linearly independent set. First, it clear that they are linearly independent be- cause Ei j is the only vector in B with a non-zero value in the i, j position and H1,i is the only vector in B with a non-zero value in the i, i position. Now 2Ei j = Hi jEi j − Ei jHi j and Hi j = Ei jE ji − E jiEi j. Thus Ei j ∈ W0 and Hi j ∈ W0. Now |B| = |{Ei j | i , j}| + |{H1,i | 2 ≤ i ≤ n}| = (n2 − n) + (n − 1) = n2 − 1 Thus we are done. Section 3.6: The Double Dual Exercise 1: Let n be a positive integer and F a field. Let W be the set of all vectors (x1, . . . , xn) in Fn such that x1 +· · ·+xn = 0. (a) Prove that W0 consists of all linear functionals f of the form f (x1, . . . , xn) = c n∑ j=1 x j. (b) Show that the dual space W∗ of W can be ‘naturally’ identified with the linear functionals f (x1, . . . , xn) = c1x1 + · · · + cnxn on Fn which satisfy c1 + · · · + cn = 0. Solution: (a) Let g be the functional g(x1, . . . , xn) = x1 + · · · + xn. Then W is exactly the kernel of g. Thus dim(W) = n − 1. Let αi = 1 − i+1 for i = 1, . . . , n − 1. Then {α1, . . . , αn−1} are linearly independent and are all in W so they must be a basis for W. Let f (x1, . . . , xn) = c1x1 + · · · + cnxn be a linear functional. Then f ∈ W0 ⇒ f (α1) = · · · = f (αn) = 0⇒ c1 − ci = 0 ∀ i = 2, . . . , n⇒ ∃ c such that ci = c ∀ i. Thus f (x1, . . . , xn) = c(x1 + · · · + xn). (b) Consider the sequence of functions W → (Fn)∗ → W∗ where the first function is (c1, . . . , cn) 7→ fc1,...,cn where fc1,...,cn (x1, . . . , xn) = c1x1 + · · · cnxn and the second function is restriction from F n to W. We know both W and W∗ have the same dimension. Thus if we show the composition of these two functions is one-to-one then it must be an isomorphism. Suppose (c1, . . . , cn) ∈ W 7→ fc1,...,cn = 0 ∈ W ∗. Then ∑ ci = 0 and ∑ cixi = 0 for all (x1, . . . , xn) ∈ W. In other words ∑ ci = 0 and ∑ cixi = 0 for all (x1, . . . , xn) such that ∑ xi = 0. 84 Chapter 3: Linear Transformations Let {α1, . . . , αn−1} be the basis for W from part (a). Then fc1,...,cn (αi) = 0 ∀ i = 1, . . . , n−1; which implies c1 = ci ∀ i = 2, . . . , n. Thus ∑ ci = (n − 1)c1. But ∑ ci = 0, thus c1 = 0. Thus fc1,...,cn is the zero function. Thus the mapping W → W∗ is a natural isomorphism. We therefore naturally identify each element in W∗ with a linear functional f (x1, . . . , xn) = c1x1 + · · · cnxn where ∑ ci = 0. Exercise 2: Use Theorem 20 to prove the following. If W is a subspace of a finite-dimensional vector space V and if {g1, . . . , gr} is any basis for W0, then W = r⋂ i=1 Ngi . Solution: {g1, . . . , gr} a basis for W0 ⇒ gi ∈ W0 ∀ i ⇒ gi(W) = 0 ∀ i ⇒ W ⊆ Ngi ∀ i ⇒ W ⊆ ⋂r i=1 Ngi . Let n = dim(V). By Theorem 2, page 71, we know dim(Ng1 ) = n − 1. Since the gi’s are linearly independent, g2 is not a multiple of g1, thus by Theorem 20, Ng1 * Ng2 . Thus dim(Ng1 ∩ Ng2 ) ≤ n − 2. By Theorem 20 again, Ng3 * Ng1 ∩ Ng2 since g3 is not a linear combination of g1 and g2. Thus dim(Ng1 ∩ Ng2 ∩ Ng3 ) ≤ n − 3. By induction dim( ⋂r i=1 Ngi ) ≤ n − r. Now by Theorem 16, dim(W) = n − r. Thus since W ⊆ ⋂r i=1 Ngi , it follows that dim( ⋂r i=1 Ngi ) ≥ n − r. Thus it must be that dim( ⋂r i=1 Ngi ) = n − r and it must be that W = ⋂r i=1 Ngi since we have shown the left hand side is contained in the right hand side and both sides have the same dimension. Exercise 3: Let S be a set, F a field, and V(S ; F) the space of all functions from S into F: ( f + g)(x) = f (x) + g(x) (c f )(x) = c f (x). Let W be any n-dimensional subspace of V(S ; F). Show that there exist points x1, . . . , xn in S and functions f1, . . . , fn in W such that fi(x j) = δi j. Solution: I’m not sure using the double dual is really the easiest way to prove this. It can be done rather easily directly by induction on n (in fact see question 121704 on math.stackexchange.com). However, since H&K clearly want this done with the double dual. At first glance you might try to think of W as a dual on S and W∗ as the double dual somehow. But that doesn’t work since S is just a set. Instead I think you have to consider the double dual of W, W∗∗ to make it work. I came up with the following solution. Let s ∈ S . We first show that the function φs :W → F w 7→ w(s) is a linear functional on W (in other words for each s, we have φs ∈ W∗). Let w1,w2 ∈ W, c ∈ F. Then φs(cw1 + w2) = (cw1 + w2)(s) which by definition equals cw1(s) + w2(s) which equals cφs(w1) + φs(w2). Thus φs is a linear functional on W. Suppose φs(w) = 0 for all s ∈ S , w ∈ W. Then w(s) = 0 ∀ s ∈ S , w ∈ W, which implies dim(W) = 0. So as long as n > 0, ∃ s1 ∈ S such that φs1 (w) , 0 for some w ∈ W. Equivalently there is an s1 ∈ S and a w1 ∈ W such that w1(s1) , 0. This means φs1 , 0 as elements of W ∗. It follows that 〈φs1〉, the subspace of W ∗ generated by φs1 , has dimension one. By scaling if necessary, we can further assume w1(s1) = 1. Now suppose ∀ s ∈ S that we have φs ∈ 〈φs1〉, the subspace of W ∗ generated by φs1 . Then for each s ∈ S there is a c(s) ∈ F such that φs = c(s)φs1 in W ∗. Then for each s ∈ S , w(s) = c(s)w(s1) for all w ∈ W. In particular w1(s) = c(s) (recall w1(s1) = 1). Let w ∈ W. Let b = w(s1). Then w(s) = c(s)w(s1) = bw1(s) ∀ s ∈ S . Notice that b depends on w but does not depend on s. Thus w = bw1 as functions on S where b ∈ F is a fixed constant. Thus w ∈ 〈w1〉, the subspace of W generated Section 3.7: The Transpose of a Linear Transformation 85 by w1. Since w was arbitrary, it follows that dim(W) = 1. Thus as long as dim(W) ≥ 2 we can find w2 ∈ W and s2 ∈ S such that 〈w1,w2〉 (the subspace of W generated by w1,w2) and 〈φs1 , φs2〉 (the subspace of W ∗ generated by {φs1 , φs2 }) both have dimension two. Let W0 = 〈w1,w2〉. Then we’ve shown that {φs1 , φs2 } is a basis for W ∗ 0 . Therefore there’s a dual basis {F1, F2} ⊆ W∗∗0 ; so that Fi(φs j ) = δi j, i, j ∈ {1, 2}. By Theorem 17, ∃ corresponding w1,w2 ∈ W so that Fi = Lwi (in the notation of Theorem 17). Therefore, δi j = Fi(φs j ) = Lwi (φs j ) = φs j (wi) = wi(s j), for i, j ∈ {1, 2}. Now suppose ∀ s ∈ S that we have φs ∈ 〈φs1 , φs2〉 ⊆ W ∗. Then ∀ s ∈ S , there are constants c1(s), c2(s) ∈ F and we have w(s) = c1(s)w(s1) + c2(s)w(s2) for all w ∈ W. Similar to the argument in the previous paragraph, this implies dim(W) ≤ 2 (for w ∈ W let b1 = w(s1) and b2 = w(s2) and argue as before). Therefore, as long as dim(W) ≥ 3 we can find s3 so that 〈φs1 , φs2 , φs3〉 ⊆ W ∗, the subspace of W∗ generated by φs1 , φs2 , φs3 , has dimension three. And as before we can find w3 ∈ W such that wi(s j) = δi j, for i, j ∈ {1, 2, 3}. Continuing in this way we can find n elements s1, . . . , sn ∈ S such that φs1 , . . . , φsn are linearly independent in W ∗ and corre- sponding elements w1, . . . ,wn ∈ W such that wi(s j) = δi j. Let fi = wi and we are done. Section 3.7: The Transpose of a Linear Transformation Exercise 1: Let F be a field and let f be the linear functional on F2 defined by f (x1, x2) = ax1 +bx2. For each of the following linear operators T , let g = f t f , and find g(x1, x2). (a) T (x1, x2) = (x1, 0); (b) T (x1, x2) = (−x2, x1); (c) T (x1, x2) = (x1 − x2, x1 + x2). Solution: (a) g(x1, x2) = T t f (x1, x2) = f (T (x1, x2)) = f (x1, 0) = ax1. (b) g(x1, x2) = T t f (x1, x2) = f (T (x1, x2)) = f (−x2, x1) = −axs + bx1. (c) g(x1, x2) = T t f (x1, x2) = f (T (x1, x2)) = f (x1 − x2, x1 + x2) = a(x1 − x2) + b(x1 + x2) = (a + b)x1 + (b − a)x2. Exercise 2: Let V be the vector space of all polynomial functions over the field of real numbers. Let a and b be fixed real numbers and let f be the linear functional on V defined by f (p) = ∫ b a p(x)dx. If D is the differentiation operator on V , what is Dt f ? Solution: Let p(x) = c0 + c1x + · · · + cnxn. Then Dt f (p) = f (D(p)) = f (c1 + 2c2x + 3c3x2 + · · · + ncnxn−1) =c1 + c2x2 + · · · cnxn |b0 =p(b) − p(a) Exercise 3: Let V be the space of all n × n matrices over a field F and let B be a fixed n × n matrix. If T is the linear operator on V defined by T (A) = AB − BA, and if f is the trace function, what is T t f ? 88 Chapter 3: Linear Transformation Chapter 4: Polynomials Section 4.2: The Algebra of Polynomials Exercise 1: Let F be a subfield of the complex numbers and let A be the following 2 × 2 matrix over F A = [ 2 1 −1 3 ] . For each of the following polynomials f over F, compute f (A). (a) f = x2 − x + 2; (b) f = x3 − 1; (c) f = x2 − 5x + 7; Solution: (a) A2 = [ 2 1 −1 3 ] · [ 2 1 −1 3 ] = [ 3 5 −5 8 ] Therefore f (A) = A2 − A + 2 = [ 3 5 −5 8 ] − [ 2 1 −1 3 ] + [ 2 0 0 2 ] = [ 3 4 −4 7 ] (b) A2 = [ 2 1 −1 3 ] · [ 2 1 −1 3 ] · [ 2 1 −1 3 ] = [ 3 5 −5 8 ] · [ 2 1 −1 3 ] = [ 1 18 −18 19 ] Therefore f (A) = A3 − 1 89 90 Chapter 4: Polynomials = [ 1 18 −18 19 ] − [ 1 0 0 1 ] = [ 0 18 −18 18 ] (c) f (A) = A2 − 5A + 7 = [ 3 5 −5 8 ] − 5 [ 2 1 −1 3 ] + [ 7 0 0 7 ] = [ 0 0 0 0 ] Exercise 2: Let T be the linear operator on R3 defined by T (x1, x2, x3) = (x1, x3,−2x2 − x3). Let f be the polynomial over R defined by f = −x3 + 2. Find f (T ). Solution: The matrix of T with respect to the standard basis is A = 1 0 00 0 10 −2 −1 Thus A2 = 1 0 00 0 10 −2 −1 · 1 0 00 0 10 −2 −1 = 1 0 00 −2 −10 2 −1 So A3 = 1 0 00 −2 −10 2 −1 · 1 0 00 0 10 −2 −1 = 1 0 00 2 −10 2 3 Thus −A3 + 2 = − 1 0 00 2 −10 2 3 + 2 0 00 2 00 0 2 = −1 0 00 0 10 −2 −1 . This corresponds to the transformation f (T )(x1, x2, x3) = (−x1, x3,−x2 − x3). Section 4.2: The Algebra of Polynomials 93 Since F has characterisitc zero we can find a, b ∈ F, such that a , b. Consider a and b as constant polynomials in F. Then T (a) = T (b) = 0. Thus T is not one-to-one. Thus T is not invertible. (b) Clearly D is a function from F[x] to F[x]. We must show D is linear. D n∑ i=0 cixi + n∑ i=0 c′i x i = D n∑ i=0 (ci + c′i)x i = n∑ i=1 i(ci + c′i)x i−1 = n∑ i=1 (icixi−1 + ic′i x i−1) = D n∑ i=0 cixi + D n∑ i=0 c′i x i . and D r n∑ i=0 cixi = D n∑ i=0 rcixi = n∑ i=1 rcixi−1 = r n∑ i=1 cixi−1 = r · D n∑ i=0 cixi . Thus D is linear. Suppose f (x) = ∑n i=0 cix i is in the null space of D. Then D( f ) = ∑n i=1 icix i−1 = 0. A polynomial is zero if and only if every coefficient is zero. Thus it must be that 0 = c1 = c2 = c3 = · · · . So it must be that f (x) = c0 a constant polynomial. Thus the null space of D contains the constant polynomials. Since D( f ) = 0 for all constant polynomials, the null space of D consists of exatly the constant polynomials. (c) D T n∑ i=0 cixi = D n∑ i=0 ci 1 + i xi+1 . The first non-zero term of this sum is the linear term c0x. Thus when we apply D the sum still starts at zero: = n∑ i=0 (i + 1) ci 1 + i xi+1−1 = n∑ i=0 cixi. Thus D T n∑ i=0 cixi = n∑ i=0 cixi. Thus DT = I. Let f (x) = 1. Then T D( f ) = T (D( f )) = T (0) = 0. Thus T D(1) , 1. Thus T D , I. (d) This follows rather easily from part (e). And likewise (e) follows rather easily from (d). Thus one can derive (d) straight from the definition of T and then derive (e) from it, or one can derive (e) straight from the definition of D and then derive (d) 94 Chapter 4: Polynomials from it. I’ve chosen to do the latter. In part (e) below the product formula is proven straight from the definition. So we will use it here to prove this part. In particular, we apply the product formula from part (e) to (T f )(Tg) D[(T f )(Tg)] = (Tg)D(T f ) + (T f )D(Tg). By part (c) DT = I so this is equivalent to D[(T f )(Tg)] = f (Tg) + (T f )g. Thus D[(T f )(Tg)] − f (Tg) = (T f )g. Now apply T to both sides T ( D[(T f )(Tg)] − f (Tg) ) = T ((T f )g). Since T is a linear transformation this is equivalent to T ( D[(T f )(Tg)] ) − T [ f (Tg)] = T ((T f )g). We showed in part (c) that T D , I, however if f has constant term equal to zero then in fact T (D( f )) does equal f . Now T f and Tg have constant term equal to zero, so (T f )(Tg) has constant term zero, thus T ( D[(T f )(Tg)] ) = (T f )(Tg). Thus (T f )(Tg) − T [ f (Tg)] = T ((T f )g). (e) I believe they are after the product formula here: D( f g) = f D(g) + gD( f ). (25) We prove this by brute force appealing just to the definition and to the product formula for polynomials. Let f (x) = ∑n i=0 cix i and g(x) = ∑m i=0 dix i. Then using the product formula (4-8) on page 121 we have D( f g) = D n∑ i=0 cixi m∑ i=0 dixi = D n+m∑ i=0 i∑ j=0 c jdi− j xi And using the linearity of the differentiation operator D this equals n+m∑ i=0 i i∑ j=0 c jdi− j xi−1. (26) Now we write down the sum for the right hand side of (25): f D(g) + gD( f ) = n∑ i=0 cixi m∑ i=1 idixi−1 + n∑ i=1 icixi−1 m∑ i=0 dixi. (27) Consider x · n∑ i=0 cixi m∑ i=1 idixi−1. Section 4.3: Lagrange Interpolation 95 This equals n∑ i=0 cixi m∑ i=1 idixi and since 0d0 = 0 we can write this as n∑ i=0 cixi m∑ i=0 idixi. It’s straightforward to apply (4-8) page 121 to this product. In (4-8) we let fi = ci and gi = idi and it equals m+n∑ i=0 i∑ j=0 (i − j)c jdi− j xi. The constant terms is zero thus we can write it as m+n∑ i=1 i∑ j=0 (i − j)c jdi− j xi. And thus the sum n∑ i=0 cixi m∑ i=1 idixi−1 equals m+n∑ i=1 i∑ j=0 (i − j)c jdi− j xi−1. Similary the second sum is n+m∑ i=1 i∑ j=0 jc jdi− j xi. Thus (27) does equal (26). (f) Suppose V is finite dimensional. Let {b1, · · · , bn} be a basis for V . Let d = maxi=1,···n deg(bi). It follows from Theo- rem 1(v) and induction that the degree of a linear combination of polynomials is no larger than the max of the degress of the individual polynomials involved in the linear combination. Thus no element of V has degree greater than d. Now let f ∈ V be any non-zero element. Let d′ = deg( f ). Then T f has degree d′ + 1, T 2 f has degree d′ + 2, etc. Thus for some n, deg(T n f ) > d. If T n f ∈ V then this is a contradiction. Thus if T n f ∈ V for all f ∈ V it must be that V is not finite dimensional. (g) Let {b1, · · · , bn} be a basis for V . Let d = maxi=1,···n deg(bi). For any f ∈ F[x], we know deg(D f ) < deg( f ). Thus Dd+1bi = 0 for all i = 1, . . . , n. Since Dd+1(bi) = 0 for all elements of the basis {b1, . . . , bn} it follows that Dd+1( f ) = 0 for all f ∈ V . Section 4.3: Lagrange Interpolation Exercise 1: Use the Lagrange interpolation formula to find a polynomial f with real coefficients such that f has degree ≤ 3 and f (−1) = −6, f (0) = 2, f (1) = −2, f (2) = 6.

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