Math304 Practice Exams PDF

Title Math304 Practice Exams
Course Linear Algebra
Institution Binghamton University
Pages 5
File Size 99.9 KB
File Type PDF
Total Downloads 72
Total Views 178

Summary

Practice exams for Linear Algebra course....


Description

Math 304-6

Linear Algebra

Spring 2020

Practice Exam 3

Feingold

SHOW ALL NECESSARY WORK. Note: AT means the transpose of A. (1) (20 Pts) Let L : R22 → R2 be given by 

a L c

   b a + 2b + 3c + 4d . = −a + b + 2c − 3d d

Let S = {v1 , v2 , v3 , v4 } be the standard basis of R22 and let T = {w1 , w2 } be the standard basis of R2 . Let other ordered bases be                    3 2 1 0 0 0 0 1 1 1 ′ ′ ′ ′ . , w2 = S = , , , and T = w1 = 0 1 1 1 1 0  2 1   0 ′1  ′ ′ ′ v1

v2

v3

v4

(a) (4 pts) Find the matrix T [L]S representing L from S to T . (b) (4 pts) Find the matrix T ′ [L]S ′ representing L from S ′ to T ′ without using transition matrices. (Do it directly.) (c) (12 pts) Find the transition matrices S PS ′ and T QT ′ and verify that your answers satisfy T ′ [L]S ′ = (T QT ′ )−1 T [L]S (S PS ′ ) .

(2) (15 Points, 3 points each) Answer each question separately. (a) If A ∈ Rnn and the homogeneous linear system AX = 0 has only the trivial solution, then what is the most you can say about det(A)? (b) Let L : V → V and suppose v ∈ V is an eigenvector for L with eigenvalue λ. Show that v is an eigenvector for L2 with eigenvalue λ2 . Do not assume L is diagonalizable. (c) Suppose A ∈ Rnn has characteristic polynomial det(λIn − A) = (λ − λ1 )k1 (λ − λ2 )k2 · · · (λ − λr )kr with r distinct real eigenvalues λ1 , λ2 , · · · , λr . For each 1 ≤ i ≤ r, what is the most you can say about the relationship between the algebraic multiplicity ki and the geometric multiplicity gi = dim(Aλi )? (d) In part (c), if you also know that A is diagonalizable, what else can you say about the relationship between ki and gi for each i? (e) Let E ∈ Rnn be an elementary matrix corresponding to an elementary row operation R, let A ∈ Rnn and let B be the matrix obtained by doing the row operation R to A. What is the relationship between det(A), det(B) and det(E)?

(3) (20 Points, 3 points each, (a) worth 8 pts) Answer each question separately.  2 3 4 1   1 1 −1 −1  (a) Find det  . 3 2 1 1  4 4 −2 −3 (b) If det(A) = 10, det(B) = 4 and det(C) = 3, find det(A−1 B 2 C T ). (c) Suppose S is a basis of V , T is a basis of W , dim(V ) = n, dim(W ) = m. Then for any linear L : V → W we defined a map T MS : Lin(V, W ) → Rnm by T MS (L) = m T [L]S . What property of the map T MS implies dim(Lin(V, W )) = dim(Rn )? n are similar, that is, B = P −1 AP for some invertible P ∈ Rn . (d) Suppose A, B ∈ Rn n What is the relationship between the characteristic polynomials CharA (λ) = det(λIn − A) and CharB (λ) = det(λIn − B)? (e) For A ∈ Rnn we know that CharA (λ) is a polynomial in the variable λ of degree n. What is the constant term of that polynomial?   7 −4 2 (4) (20 Points) Let A =  4 −1 2 . 4 −4 5 (a) (8 Pts) Find the characteristic polynomial of A, det(λI3 − A), find all eigenvalues, λi , of A and the corresponding algebraic multiplicities, ki . (b) (8 Pts) For each eigenvalue, λi , of A, find a basis for the eigenspace, Aλi , and the geometric multiplicity gi = dim(Aλi ). (c) (4 Pts)Determine whether or not A is diagonalizable. If it is, find D and P such that D = P −1 AP is diagonal. If not, explain why.

Math 304-6

Linear Algebra

Spring 2020

Practice Exam 3 Solutions

Feingold

1. (20 Points) 

       1 2 3 4 (a) (4 Pts) Find L(S): L(v1 ) = , L(v2 ) = , L(v3 ) = , L(v4 ) = . −3 −1 1 2         1 0  1 2 3 4  so [L] = 1 2 3 4 . Then [T | L(S)] =   S T 0 1  −1 1 2 −3 −1 1 2 −3  T L(S)

       7 5 7 5 ′ ′ ′ , L(v2 ) = , L(v3 ) = , L(v4 ) = . (b) (4 Pts) Find L(S ): = −3 −4  −1 3          5 7 5 22 17 1   1 0  23 2 3  7 Row reduce   to  0 1  −13 −13 −9 1  1 2  −3 −4 −1 3  I2  T′ L(S ′ ) T ′ [L]S ′   1 1 0 0   2 3 1 0 0 1   (c) (12 Pts) S PS ′ =  since S and T are the standard  and T QT ′ = 0 0 1 1 1 2 1 1 1 0 bases. To get T ′ QT = (T QT ′ )−1 , reduce ′



L(v1′)



2 1

(T QT ′ )−1 T [L]S (S PS ′ ) =

T′



  3  1 2  0

    0  1 0  2 −3  to   0 1  −1 2  1 I 2  T ′ QT T 

2 −3 −1 2

 



1 −1

2 3 4 1 2 −3



1 1 0 1



5 1 0 −3 0 1



2 −3 −1 2

 

1 −1



2 −3 −1 2

 

7 −3

17 −10



1 1 0

0 0 1 1 

0 0 1 1

0 1 = 1 0

0   23 22 17 1 1 = T ′ [L]S ′ checks. Also,  = 1 −13 −13 −9 1 1 0  1 1 0 0 2 3 4 1 0 0 1   = 1 2 −3 0 0 1 1   1 1 1 0 23 22 17 1 5 7 5 . = −13 −13 −9 1 −4 −1 3 

1 0 0 1

1 0 0 1



(2) (15 Points, 3 points each) Answer each question separately. (a) If A ∈ Rnn and the homogeneous linear system AX = 0 has only the trivial solution, then A has rank n and is invertible so det(A) 6= 0. (b) Let L : V → V and suppose v ∈ V is an eigenvector for L with eigenvalue λ. Then L(v) = λv so L2 (v) = L(L(v)) = L(λv ) = λL(v) = λλv = λ2 v so v is an eigenvector for L2 with eigenvalue λ2 . (c) Suppose A ∈ Rnn has characteristic polynomial det(λIn − A) = (λ − λ1 )k1 (λ − λ2 )k2 · · · (λ − λr )kr with r distinct real eigenvalues λ1 , λ2 , · · · , λr . In general, for each 1 ≤ i ≤ r, we know that 1 ≤ gi ≤ ki . (d) In part (c), if you also know that A is diagonalizable, then we know that gi = ki for each i. (e) Let E ∈ Rnn be an elementary matrix corresponding to an elementary row operation R, let A ∈ Rnn and let B be the matrix obtained by doing the row operation R to A. Then B = EA so det(B) = det(E) det(A). (3) (20 Points, 3 points each, (a) worth 8 pts) Answer each question separately.   1 1 −1 −1   1 1 −1 2 3 4 1  6 3   1 1 −1 −1  0 1 0 1 6 (a) det   = − det   = − det  3 2 1 1 0 −1 4 4 0 0 10 4 4 −2 −3 0 0 2 1 0 0 2     1 1 −1 −1 1 1 −1 −1 3  3  0 1 6 0 1 6 =4 − det   = det  0 0 2 1  0 0 0 2 0 0 0 2 0 0 2 1 (42 )(3) (b) If det(A) = 10, det(B) = 4 and det(C ) = 3, then det(A−1 B 2 C T ) = 10 =

−1  3  = 7  1

24 . 5

(c) Suppose S is a basis of V , T is a basis of W , dim(V ) = n, dim(W ) = m. Then for any linear L : V → W we defined a map T MS : Lin(V, W ) → Rm n by T MS (L) = T [L]S . What property of the map T MS implies dim(Lin(V, W )) = dim(Rnm)? Answer: The property that T MS is an isomorphism, that is, a bijective linear map. (d) Suppose A, B ∈ Rnn are similar, that is, B = P −1 AP for some invertible P ∈ Rnn . The relationship between the characteristic polynomials is that they are equal, CharA (λ) = det(λIn − A) = CharB (λ) = det(λIn − B). (e) For A ∈ Rnn we know that CharA (λ) is a polynomial in the variable λ of degree n. What is the constant term of that polynomial? Answer: The constant term is CharA (0) = det(−A) = (−1)n det(A).



 7 −4 2 (4) (20 Points) Let A =  4 −1 2 . 4 −4 5 (a) (8 Points) The characteristic polynomial is CharA (t) = det(λI3 − A) = − det(A−λI3 )     7−λ −4 2 3−λ 0 λ−3 −1 − λ 2  = − det  0 3 − λ λ − 3 − det  4 4 −4 5−λ 4 −4 5−λ     1 0 −1 1 0 −1 = −(3 − λ)2 det  0 1 −1  = −(3 − λ)2 det  0 1 −1  4 −4 5 − λ 0 0 5−λ = (λ − 3)2 (λ − 5)1 = λ3 − 11λ2 + 39λ − 45.

So the eigenvalues are λ1 = 3 and λ2 = 5, the roots of CharA (λ), with algebraic multiplicities k1 = 2 and k2 = 1. (b) (4 Points) For λ1 = 3, the eigenspace, Aλ1 , is found by row reducing [A − 3I3 |0]:       4 −4 2  0 x1 = r − 12 s 1 −1 21  0  4 −4 2  0  to  0 0 0  0  so x2 = r ∈ R   x3 = s ∈ R 0 0 0 0 4 −4 2  0 so the λ1 = 3 eigenspace    r − 12 s Aλ1 =  r  ∈ R3  s (4 Points) For λ2  2 4 4

      1 −1      r, s ∈ R has basis  1  ,  0  and g1 = 2 = k1 .     0 2

= 5, the eigenspace,    −4 2  0 1 −6 2  0  to  0 −4 0  0 0

so the λ1 = 5 eigenspace    r Aλ2 =  r  ∈ R3  r

Aλ2 , is found by row reducing [A − 5I3 |0]:   0 −1  0 x1 = r 1 −1  0  so x2 = r 0 0 0 x3 = r ∈ R

      1     r ∈ R has basis  1  and g2 = 1 = k2 .     1

3 (c) (4 Points) eigenbasis for  A is    since g1 + g2 = 3 and we found an   R ,  diagonalizable −1 3 0 0 1  1 −1 1  1 T =  1  ,  0  ,  1  . D =  0 3 0  and P = S PT =  1 0 1  is the   0 2 0 0 5 1 0 2 1 transition matrix such that D = P −1 AP is diagonal. As a numerical check:         1 −1 1 3 0 0 3 −3 5 7 −4 2 1 −1 1 P D =  1 0 1   0 3 0  =  3 0 5  =  4 −1 2   1 0 1  = AP. 0 2 1 0 0 5 0 6 5 4 −4 5 0 2 1...


Similar Free PDFs