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Solutions to exam 1 of linear algebra 2, covering topics such as isomorphisms, inverse transformations, and finding matrices. It includes examples and explanations for various linear transformations and their inverses.
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Tuesday, October 2, 2007
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Linear Algebra 2 Exam 1 Solution (Continued)
10 points 1. Let T : V → W be a transformation where V and W are vector spaces. Write down the property(-ies) (or condition(s)) that T should satisfy in order to be a linear transforma- tion. Answer. For any scalar a and any vectors v, w ∈ V, T (v + w) = T (v) + T (w), T (av) = aT (v). The following is an equivalent statement: for any scalars a, b and any vectors v, w ∈ V, T (av + bw) = aT (v) + bT (w).
10 points 2. Which of the following functions are linear transformations T : R^2 → R^2 and which are not? If it is a linear transformation, prove it by using the definition. If it is not, justify your answer (i.e., give the reason why you think it is not a linear transformation). 05 points (a)T (x, y) = (3x − 2 y, 2 x + 3y).
Answer. We observe T (x, y) = (3x − 2 y, 2 x + 3y) = x(3, 2) + y(− 2 , 3) = xA + yB, where A = (3, 2) and B = (− 2 , 3). For any scalars a, b and any vectors v 1 = (x 1 , y 1 ), v 2 = (x 2 , y 2 ) ∈ R^2 , we have av 1 + bv 2 = (ax 1 + bx 2 , ay 1 + by 2 ), T (av 1 + bv 2 ) = T (ax 1 + bx 2 , ay 1 + by 2 ) = (ax 1 + bx 2 )A + (ay 1 + by 2 )B = a (x 1 A + y 1 B) + b (x 2 A + y 2 B) = aT (v 1 ) + bT (v 2 ). Thus, T is a linear transformation.
05 points (b)T (x, y) =
( (^) x π ,^ |y|
Answer. We observe T (x, y) =
( (^) x π ,^ |y|
= x(1/π, 0) + |y|(0, 1) = xA + |y|B, where A = (1/|π|, 0) and B = (0, 1). For any scalars a, b and any vectors v 1 = (x 1 , y 1 ), v 2 = (x 2 , y 2 ) ∈ R^2 , we have av 1 + bv 2 = (ax 1 + bx 2 , ay 1 + by 2 ), T (av 1 + bv 2 ) = T (ax 1 + bx 2 , ay 1 + by 2 ) = (ax 1 + bx 2 )A + |ay 1 + by 2 |B 6 = a (x 1 A + |y 1 |B) + b (x 2 A + |y 2 |B) = aT (v 1 ) + bT (v 2 ). Thus, T is not a linear transformation. In general, |y 1 + y 2 | ≤ |y 1 | + |y 2 | (called the “Triangle Inequality”) and the equality holds when y 1 = y 2 , one of y 1 , y 2 is zero, or both y 1 and y 2 are zeros.
Linear Algebra 2 Exam 1 Solution (Continued)
20 points 5. Let T : R^2 → R^2 be a linear transformation defined by
T (x, y) = √^12 (x − y, x + y).
10 points (a)Is there a natural number n which makes T n^ = I? If so, find it. If not, justify your answer (i.e., give the reason why you think there is no such a number). Here I is the identity operator, i.e., I(x, y) = (x, y) and T n(x, y) = T (T (· · · T (x, y))) (n times). Answer 1. Let M be a matrix defined by
M = √^1 2
Then we observe T (x, y) = (x, y)M. So we have the relation T n(x, y) = (x, y)Mn and it implies that T n^ = I (identity operator) if and only if Mn^ = I (identity matrix). A simple computation shows
M^2 =
Therefore T 8 = I. (Since T 8 = I, so T 16 = I = T 24 = · · · = T 8 k, where k = 1, 2 ,.... Usually we are interested in the smallest number in this kind of problem.) Answer 2. Recall the rotation operator S : R^2 → R^2 defined by S(x, y) = (x cos θ − y sin θ, x sin θ + y cos θ). When putting θ = π/4, S becomes T. So T rotates a point by the angle π/4 in the counterclockwise direction and T n^ rotates a point by the angle n × (π/4). Hence T 8 rotates a point by the angle 2π, i.e., sends a point to its first place. For this reason, we have T 8 = I.
10 points (b)Find the inverse of the matrix corresponding to T.
Answer. Originally the problem was as follows: “Is T an isomorphism? If so, find the inverse of T. If not, justify your answer.” However, since I gave you the
class, I changed the problem.
Below, I will give you the full correct explanation with other methods for the original problem.
If T has the inverse, then by the definition, T is an isomorphism. So let us try to find the inverse. We discuss three methods to find the inverse of T : (1) using the standard basis {e 1 = (1, 0), e 2 = (0, 1)} for R^2 and (2) using the matrix M defined in (a) above and (3) using the rotation (geometrical viewpoint).
Linear Algebra 2 Exam 1 Solution (Continued)
(1) Standard Basis {e 1 = (1, 0), e 2 = (0, 1)} for R^2 : from the definition of T , i.e.,
T (x, y) = √^1 2 (x − y, x + y) ,
it is easy to see
T (e 1 ) = T (1, 0) = √^1 2
(e 1 + e 2 ) ,
T (e 2 ) = T (0, 1) = √^1 2
(−e 1 + e 2 ).
Solving the linear system for e 1 and e 2 , we get
e 1 = √^12 (T (e 1 ) − T (e 2 )) , e 2 = √^12 (T (e 1 ) + T (e 2 )).
Now let a linear transformation S : R^2 → R^2 be the inverse of T , then S should satisfy S · T = I and so applying S to e 1 and e 2 above, we have
S(e 1 ) = √^12 ((S · T )(e 1 ) − (S · T )(e 2 )) = √^12 (e 1 − e 2 ) , S(e 2 ) = √^1 2 ((S · T )(e 1 ) + (S · T )(e 2 )) = √^1 2 (e 1 + e 2 ).
Hence, if we define S : R^2 → R^2 by S(x, y) = S(xe 1 + ye 2 ) = xS(e 1 ) + yS(e 2 ) = √x 2 (e 1 − e 2 ) + √y 2 (e 1 + e 2 )
= √x 2 (1, −1) + √y 2 (1, 1) = √^12 (x + y, −x + y) ,
i.e., S(x, y) = √^12 (x + y, −x + y) ,
then S becomes the inverse of T. One can check that S is really the inverse of T by showing S · T = I and T · S = I.
(2) Matrix M: Let S correspond to the inverse M−^1 of the matrix M (M is found in (a) above), i.e., S(x, y) = (x, y)M−^1. Then since MM−^1 = I = M−^1 M, so we have T · S = I and S · T = I. Thus in this way, we can find the inverse of T. Let us find the inverse S of T explicitly. Since M has the inverse M−^1 ,
M−^1 = √^12
Exam on Chapter 11 Representing Linear Transformations by Matrices Math 341 Linear Algebra 2 Section 51
Answer. We express T (e 1 ) and T (e 2 ) in terms of f 1 and f 2 :
T (e 1 ) = T (1; 2) = 1 + 2x = f 1 + 2f 2 ; T (e 2 ) = T (3; 4) = 3 + 4x = 3f 1 + 4f 2 :
Hence, by collecting the coecients of f 1 and f 2 , we form the matrix M :
2 4 1 3 2 4
3 (^5) :
Exam on Chapter 11 (Continued)
2 4 2 3 4 5
3 (^5) :
Find T (a; b).
Answer. Step 1. Compute M et 1 and Met 2 : (eti means the column vector of ei.)
Met 1 =
2 4 2 3 4 5
3 5
2 4 1 0
3 (^5) =
2 4 2 4
3 (^5) ; (1)
Met 2 =
2 4 2 3 4 5
3 5
2 4 0 1
3 (^5) =
2 4 3 5
3 (^5) : (2)
Step 2. Express T (e 1 ) and T (e 2 ) in terms of f 1 and f 2 : By the formula given in the textbook, we have from equations (1), (2) above
T (e 1 ) = 2f 1 + 4f 2 = 2 1 + 4 x = 2 + 4x; T (e 2 ) = 3f 1 + 5f 2 = 3 1 + 5 x = 3 + 5x:
Therefore, we have T (e 1 ) = 2 + 4x; T (e 2 ) = 3 + 5x:
Step 3. Use results to nd T (a; b):
(a; b) = (a; 0) + (0; b) = a(1; 0) + b(0; 1) = ae 1 + be 2 ; T (a; b) = T (ae 1 + be 2 ) = aT (e 1 ) + bT (e 2 ) = a(2 + 4x) + b(3 + 5x) = 2a + 4ax + 3b + 5bx = 2a + 3b + (4a + 5b)x:
Exam on Chapter 11 (Continued)
=
(^3) x 2 ;^
x 2
(^3) x 2 ^2 y;^
x 2 +^ y
:
Thus, the inverse of T is S : R^2! R^2 de ned by
S(x; y) =
(^3) x 2 ^2 y;^
x 2 +^ y
:
Let us check. If S is really the inverse of T , we should have S T = I = T S, i.e., S(T (x; y)) = I(x; y) = (x; y) and T (S(x; y)) = I(x; y) = (x; y). It is not dicult to get
T (x; y) = (2x + 4y; x + 3y):
With this and the de nition of S, we have
S(T (x; y)) = S(2x + 4y; x + 3y) = 3(2x^2 + 4 y) 2(x + 3y); ^2 x^ + 4 2 y+ x + 3y
!
= (3x + 6y 2 x 6 y; x 2 y + x + 3y) = (x; y) = I(x; y): T (S(x; y)) = T
(^3) x 2 ^2 y;^
x 2 +^ y
2
(^3) x 2 ^2 y
x 2 + y
;
(^3) x 2 ^2 y
x 2 + y
3 x 4 y 2 x + 4y; 32 x 2 y 32 x + 3y
= (x; y) = I(x; y):
Yes, S is really the inverse of T.
Exam on Chapter 11 (Continued)
2 (^66) (^66) 4
3 (^77) (^77) 5 :
Suppose T has the following matrix [ T ]A (^0) B 0 relative to the ordered basis A^0 = f e^01 = (1; 1 ; 1); e^02 = (1; 1 ; 0); e^03 = (1; 1 ; 2) g = B^0 for the domain and the range R^3 :
2 (^66) (^66) 4
3 (^77) (^77) 5 :
Find the matrices P and Q such that [ T ]AB = P [ T ]A (^0) B 0 Q. Answer. By the formula, we have P = [ I ]B (^0) B and Q = [ I ]AA 0 :
2 (^66) (^66) 4
3 (^77) (^77) 5 ;^ [^ I^ ]AA^0 =
2 (^66) (^66) 4
3 (^77) (^77) 5 :
Therefore, we have [ T ]AB = [ I ]B (^0) B [ T ]A (^0) B 0 [ I ]AA 0 , i.e., 2 (^66) (^66) 4
3 (^77) (^77) 5
2 (^66) (^66) 4
3 (^77) (^77) 5
2 (^66) (^66) 4
3 (^77) (^77) 5
2 (^66) (^66) 4
3 (^77) (^77) 5
Exam on Chapter 11 (Continued)
One can easily check: 2 4 A^ B O D
3 5
2 4 A
3 (^5) = I =
2 4 A
3 5
2 4 A^ B O D
3 (^5) :
II. Let us solve the problem by using the concepts or techniques that we have studied. For a given (m + k) (m + k) matrix M , we can nd a linear transformation T : Rm+k^! Rm+k^ having M as its matrix relative to the standard basis. (We have studied this subject in Section 11.2 Basic Theorems.) The linear transformation T is obtained as follows:
T (x; y) = (Ax + By; Dy) ;
where x = (x 1 ; x 2 ; : : : ; xm) and y = (y 1 ; y 2 ; : : : ; yk). We recall that M has the inverse if and only if T has the inverse. So let us nd the inverse S of T. We assume that S : Rm+k^! Rm+k^ is de ned by
S(x; y) = (U x + V y; W x + Zy) ;
(why?) and determine U , V , W and Z. In order for S to be the inverse of T , S should satisfy
(S T )(x; y) = S (T (x; y)) = (x; y); (T S)(x; y) = T (S(x; y)) = (x; y):
We look for U , V , W and Z satisfying the two equations above. (i) (S T )(x; y) = (x; y): A simple computation shows
(x; y) = S (T (x; y)) = S (Ax + By; Dy) = (U (Ax + By) + V (Dy) ; W (Ax + By) + Z (Dy)) = (UAx + (UB + V D) y; W Ax + (W B + ZD) y) :
So we have four equations
UA = I; U B + V D = O (zero matrix); W A = O; W B + ZD = I:
From the rst and the second equation, we get
U = A ^1 ; and V = A ^1 BD ^1 ; and W = O:
(ii) (T S)(x; y) = (x; y): A simple computation shows
(x; y) = T (S(x; y)) = T (U x + V y; W x + Zy)
Exam on Chapter 11 (Continued)
= (A (U x + V y) + B (W x + Zy) ; D (W x + Zy)) = ((AU + BW ) x + (AV + BZ) y; DW x + DZy) :
So we have four equations
AU + BW = I; AV + BZ = O (zero matrix); DW = O; DZ = I:
From the last equation, we get Z = D ^1 :
Therefore, nally we deduce that S : Rm+k^! Rm+k^ de ned by
S(x; y) =
A ^1 x A ^1 BD ^1 y; D ^1 y
is the inverse of T. In other words, M is invertible if and only if T has the inverse S, which is equivalent that both A and D are invertible. Hence it proves the rst statement in the given problem. S has the matrix MS relative to the standard basis
2 4 A
3 (^5) :
Since S is the inverse of T , the matrix MS should be the inverse of M , i.e., MS = M ^1 , i.e.,
2 4 A
3 (^5) ;
which gives the answer to the second problem.
