Download Linear Algebra: Solving Systems of Linear Equations and Computing Subspace Bases and more Exams Algebra in PDF only on Docsity!
FINAL EXAM GUIDE SOLUTIONS
MATH 208A Exam Date: December 13, 2021
The exam will have 6 questions with each question being multipart. As with the midterm exam, the questions are not equally weighted. The exam covers all of the key concepts introduced in this course. Specifically, it covers the material in Sections 1.1, 1.2, 2.1, 2.1, 2.3, 3.1, 3.2, 3.3, 4.1, 4.2, 4.3, 4.4, 5.1, 5.2, 6.1, and 6.2 of the text. In particular, this means that the exam does not test on the orthogonality material covered in class. You are allowed one handwritten 8.5 by 11 sheet of notes is allowed (2-sided is OK), and you are allowed a nonprogrammable calculator (the Texas Instruments TI-30X IIS is the official Math Dept approved calculator). A loose description of the content of each question is given below along with sample questions for the purposes of illustration and practice. The rules for the exam are listed at the end of this guide.
Question 1: Samples Our general tool for solving systems of the form Ax = b, where A ∈ Rn×m^ and b ∈ Rn^ is to reduce the augmented system [A|b] to echelon form using the elementary row operations. Since the elementary row operations are equivalent to multiplying the augmented matrix [A|b] on the left by an invertible elementary matrix G 1 , we find after the first elementary row operation that we have the augmented matrix G 1 [A|b] = [G 1 A|G 1 b] and after the second, we have G 2 G 1 [A|b] = G 2 [G 1 A|G 1 b] = [G 2 G 1 A|G 2 G 1 b], and after k elementary row operations we have G[A|b] = Gk · · · G 2 G 1 [A|b] = [Gk · · · G 2 G 1 A|Gk · · · G 2 G 1 b] = [GA|Gb], where G = Gk · · · G 2 G 1. Now suppose we apply the same elementary row operations to the augmented matrix [A|I]. The we would get G[A|I] = Gk · · · G 2 G 1 [A|I] = [Gk · · · G 2 G 1 A|Gk · · · G 2 G 1 ] = [GA|G], that is, we would recover the matrix G that puts A into echelon form. (a) Consider the linear system x 1 + x 2 + 3 x 3 + 2 x 4 + 2 x 5 = 2 2 x 1 + x 2 + 5 x 3 + x 4 + 2 x 5 = − 2 4 x 1 + 3 x 2 + 11 x 3 + 3 x 4 + 4 x 5 = 2 3 x 1 + 2 x 2 + 8 x 3 + 3 x 4 + 4 x 5 = 0
(i) Write the augmented matrix A associated with this system. (ii) Compute an echelon form for this matrix. (iii) Compute a reduced echelon form for this augmented matrix.
Solution: 1 1 3 2 2 1 0 0 0 2 1 5 1 2 0 1 0 0 4 3 11 3 4 0 0 1 0 3 2 8 3 4 0 0 0 1 1 0 2 − 1 0 − 1 1 0 0 0 1 1 1 0 0 − 2 1 0 echelon 0 0 0 2 2 2 1 − 1 0 form 0 0 0 0 0 − 1 − 1 0 1 1 0 2 0 1 0 3 / 2 − 1 / 2 0 reduced 0 1 1 0 − 1 − 1 − 5 / 2 3 / 2 0 echelon 0 0 0 1 1 1 1 / 2 − 1 / 2 0 form 0 0 0 0 0 1 1 0 − 1 1
Then G = (^12)
with^ GA^ =^ B^ where
A =
and^ B^ =
(iv) Write a description of the set of solutions to this system in vector form. (v) If B is the reduced echelon form for the augmented matrix, compute an invertible matrix G ∈ R^4 ×^4 such that GA = B. (vi) Let c 0 = (2, − 2 , 2 , 0)T^ be the vector given by the right-hand side of the linear system given above. Compute the vector Gc 0 and explain the relationship of this vector to the solution set to this linear system.
Solution: Write ¯x = G
. Then the solution set is given by
solving the augmented system [B|x¯]. or
1 0 2 0 1 − 4 0 1 1 0 − 1 6 0 0 0 1 1 0 0 0 0 0 0 0
giving the solution set
: s, t ∈ R
(vii) Let c 1 = (1, 1 , 1 , 2)T^. Use the matrix G to describe the set of solutions to the linear system obtained from the system above by replacing the right-hand side vector c 0 by the the vector c 1.
Solution: Write G
so the associated augmented system is
Solution: Compute
[GA|G] =
G =
Therefore,
is a basis for Ran (A) (or just use the standard basis as A is onto),
is a basis for Nul (A),
is a basis for Ran
AT^
, and
A is onto so Nul
AT^
(ii) A =
(b) For the matrices A in part (a) chose a basis for the row space of A from the rows of A. (c) Answer the following questions for the matrices A = [S T ] and B = AT^ where S ∈ Rn×n^ is invertible and T ∈ Rn×t^ with t ≥ 1. Solution: To answer the following questions observe that rank(AT^ ) = rank(A) = rank(S) = n and nullity(A) = (n + t) − rank(A) = t. (i) Is A onto? Is A one to one? What is the rank of A? What is the nullity of A? Solution: A is onto since rank(A) = n. A is not one to one since nullity(A) = t ≥ 1. (ii) Is B onto? Is B one to one? What is the rank of B? What is the nullity of B? Solution: B is not onto since rank(B) = rank(AT^ ) = n < n + t. B is one to one since nullity(B) = n − rank(B) = 0.
(iii) Set L =
[
S T
0 K
]
where K ∈ Rt×t. Under what conditions is L both onto and one to one? Solution: K needs to be invertible since then L is invertible because det (L) = det (S)det (K) 6 = 0. Solution: For (d)-(j) see midterm guide 2.
(d) Let A ∈ R^2 ×^3 be such that Ran (A) = Span
[(
)]
. Give two different examples of such a matrix A.
(e) Suppose A ∈ R^3 ×^3 is such that Ran (A) = Span
= S.
(i) What is the nullity of A?
(ii) Give an example of a matrix A with Ran (A) = S and
(^) ∈ Nul (A).
(f) Let A, B ∈ Rn×m^ be equivalent matrices. Answer the following true or false questions. (i) Nul (A) = Nul (B) : © True © False (ii) Ran (A) = Ran (B) : © True © False (g) Let x ∈ Rn^ and y ∈ Rm^ be nonzero vectors and consider the matrix A = xyT^. (i) If A ∈ Rs×t, what are s and t? (ii) What are rank(A) and nullity(A)? (h) Compute a basis Ran (A), Nul (A), Ran
AT^
, and Nul
AT^
where
A =
(i) Compute a basis Ran (A), Nul (A), Ran
AT^
, and Nul
AT^
where
A =
(j) Let A ∈ Rn×k^ and B ∈ Rk×m. (i) If AB ∈ Rs×t^ what are s and t? (ii) If rank(A) = n and rank(B) = k, what can be said about rank(AB)? (iii) If nullity(A) = 0 and nullity(B) = 0, what can be said about nullity(AB)? (iv) If n = m, and k ≤ n, under what conditions is AB invertible?
Question 3: This question concerns linear transformations and their matrix representation, matrix algebra including the relationship between matrix multiplication and the composition of linear transformations, and the inverse of a matrix including its computation, and properties, their use in solving linear systems including the relationship to echelon form and the equivalence of linear systems.
(a) Let T : Rm^7 → Rn^ be a linear tranformation. Let {b 1 , b 2 ,... , bm} be a basis for Rm^ and suppose that T (bi) = yi for yi ∈ Rn, i = 1, 2 ,... , m. In terms of the matrices B = [b 1 b 2... bm] and Y = [y 1 y 2... ym] derive an expression for the matrix A such that T = TA, i.e. T (x) = Ax. Solution: If T = TA, then AB = Y so A = Y B−^1 since B is invertible. (b) Let T : R^3 7 → R^3 be such that
T (e 1 ) =
(^) , T (e 2 ) =
(^) , T (e 3 ) =
Find a matrix A such that T = TA. (Hint: Use the result from part (a) above.)
(a) Compute the coordinate transformation matrix for the basis
B =
that is, compute the matrix W such that [x]B = W y for any vector y ∈ R^3 where [x]B ∈ R^3 is the vector containing the coordinates of y in the basis B. Then give
the coordinates of the vector y =
(^) in the basis B.
Solution: W =
− 1
=
(^) so
[y]B = W y =
(b) Consider the following two bases for the subspace S ⊂ R^5 :
B 1 =
, B 2 =
Compute the coordinate transformation matrix C such that [y]B 2 = C[y]B 1 for all
y ∈ S. Then compute y and [y]B 2 for [y]B 1 =
. Next, compute the coordinate transformation matrix C such that [y]B 1 = C[y]B 2 for all y ∈ S. Then compute y
and [y]B 1 for [y]B 2 =
Solution: C = (^12)
[
]
and C−^1 =
[
]
so [y]B 1 = C[y]B 2 for all y ∈ S,
and [y]B 2 = C−^1 [y]B 1 for all y ∈ S.
(c) Consider the following two bases for the subspace S ⊂ R^5 :
B 1 =
, B 2 =
(i) Compute the coordinate transformation matrix C such that [y]B 1 = C[y]B 2 for all y ∈ S. (ii) Then compute y and [y]B 1 for [y]B 2 =
Solution: C =
[
]
and C−^1 =
[
]
so [y]B 1 = C[y]B 2 for all y ∈ S, and
[y]B 2 = C−^1 [y]B 1 for all y ∈ S.
(d) Do the vectors the sets B 1 and B 2 (given below) span the same subspace of R^5?
B 1 =
B 2 =
If they do span the same subspace, compute the coordinate transformation matrix C for which [y]B 1 = C[y]B 2 for all y ∈ R^5. (Hint: Start by trying to compute C and see what happens.)
Solution: C =
Question 5: This question concerns the determinant of a matrix including its properties, compu- tation, use, and relationship to linear systems.
(a) Use the properties of the deteminant to compute the determinant of the following matrices.
(i) A =
Solution: |A| =
∣ = 3^ ·^5 ·^ 1 = 15.
(ii) B =
Solution: |B| = 1 · 2 · (−5) · 6 = −60.
(iii) C =
Solution: Reduce 1 0 0 1 0 2 2 0 0 2 4 0 − 1 0 0 1 1 0 0 1 0 2 2 0 0 0 2 0 0 0 0 2
so |C| = 1 · 2 · 2 · 2 = 8. (b) Let M =
[
A B
C D
]
where A ∈ Rs×s^ is invertible, B ∈ Rs×t, C ∈ Rt×s, and
D ∈ Rt×t.
(b) Compute the eigenvalues and eigenvectors of the matrix A =
. The
give both the algebraic and geometric multiplicity of each of the eigenvalues of A.
Solution: The characteristic polynomial is p(λ) = (1 − λ)(2 − λ)(3 − λ) so the eigenvalues are 1,2, and 3 each with algebraic multiplicity 1 which implies that
their geometric multiplicity is also 1. The eigenvector for λ = 1 is
, the
eigenvector for λ = 2 is
, and the eigenvector for λ = 3 is
(c) Consider the two matrices
A =
B^ =
Compute the eigenvalues and eigenvectors of each and give both the algebraic and geometric multiplicity of each. Solution: By using the block diagonal structure it follows that he characteristic polynomial for both matrices is p(λ) = (λ − 2)^4. So λ = 2 is the only eigenvalue for both matrices. For the first matrix we get two eigenvectors
and
so the geometric multiplicity is 2. For the second matrix
is the only eigen-
vector so the geometric multiplicity is 1.
(d) Let A ∈ Rn×n^ and let P ∈ Rn×n^ be invertible. Show that if λ ∈ R is an eigen- value of A with associated eigenvector x, then λ is an eigenvalue of B = P AP −^1 associated with eigenvector y = P x. Solution: By = P AP −^1 P x = P Ax = λP x = λy. (e) Let u ∈ Rn^ be such that uT^ u = 1 and set A = In − uuT^. Compute the eigenvalues of A and their associated eigenvectors. In particular, show that A is diagonalizable.
Solution: First observe that A^2 = (I − uuT^ )(I − uuT^ ) = I − uuT^ − uuT^ + uuT^ uuT^ = (I − uuT^ ) = A. Since Au = 0, u is an eigenvector with eigenvalue 0. Observe that 0 = Aw = w − (uT^ w)u which implies that w is a multiple of u. Let {u, w 2 , w 3 ,... , wn} be a basis for Rn. Then Awi 6 = 0, i − 2 ,... , n since if Awi = 0 then wi is a multiple of u which contradicts the linear independence of the vectors {u, w 2 , w 3 ,... , wn}. Set wˆi = Awi = (I − uuT^ )wi, i = 2,... , n. Then A wˆi = AAwi = A^2 wi = Awi = ˆwi so that every ˆwi is an eigenvector of A with eigenvalue
- Moreover, the vectors ˆwi, i = 2,... , n are linearly independent since otherwise there exist μi such that 0 = μ 2 wˆ 2 + · · · + μn wˆn = (I − uuT^ )(μ 2 w 2 + · · · + μnwn) so that μ 2 w 2 + · · · + μnwn is a multiple of u which again contradicts the linear independence of the vectors {u, w 2 , w 3 ,... , wn}. Hence the geometric multiplicity of the eigenvalue 1 is (n − 1) and the geometric multiplicity of the eigenvalue 0 is
- Therefore, A is diagonalizable with A = P DP −^1 where P = [u, w 2 , w 3 ,... , wn] and
D =
