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National University of SingaporeDepartment of MathematicsSemester 2, 2017/ MA1101R Linear Algebra I May 2018 — Time allowed: 2 hoursStudent Number:Instructions to candidates: This examination paper consists of 6 questions, for a total of 80 points. Excluding the cover page, there are 12 printed page...

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National University of Singapore Department of Mathematics Semester 2, 2017/18 MA1101R Linear Algebra I May 2018 — Time allowed: 2 hours

Student Number:

Instructions to candidates: 1. This examination paper consists of 6 questions, for a total of 80 points. Excluding the cover page, there are 12 printed pages. 2. Answer all 6 questions. 3. This is a closed book examination, but you are allowed to bring one A4-sized double-sided helpsheet. 4. You are permitted to use any kind of calculator, except devices which can be used for communication and/or web-surfing. However various steps in the calculations should be laid out systematically. 5. Write down your student number in the space provided above. Do not write your name. 6. Write your answers in the space below each question. Only this booklet will be collected at the end of the examination. 7. The blank pages on the left can be used for rough work. Do not write below this box.

Question:

1

2

3

4

5

6

Total

Points:

10

15

10

15

15

15

80

Score:

1. Suppose v1 = (1, 1, 1, 1), v2 = (1, 2, 4, 5), and v3 = (10, −30, −40, −20). Let S = {v1 , v2 , v3 } and let U = span(S). (a) (6 points) Use the Gram-Schmidt process to find an orthonormal basis T for U. Solution: Set w1 = u1 = (1, 1, 1, 1). Next set w2 = v2 −

v2 · w1 2

kw1 k

w1 = (−2, −1, 1, 2).

Next set w3 = v3 −

v3 · w2 v3 · w1 w1 − 2 w2 = (16, −17, −13, 14). 2 kw2 k kw1 k

Now normalize these orthogonal vectors: 1 1 w2 w1 = (1, 1, 1, 1) = √ (−2, −1, 1, 2) u2 = kw1 k 2 kw2 k 10 1 w3 =√ (16, −17, −13, 14). u3 = kw3 k 910 u1 =

Thus T = {u1 , u2 , u3 } is an orthonormal basis for U .

(b) (4 points) Find the transition matrix from S to T . You may leave the entries of your answer in terms of square roots.        2 v1 · u1 6 v2 · u1 √ Solution: [v1 ]T =  v1 · u2  =  0 . [v2 ]T =  v2 · u2  =  10 . v1 · u3 0 v2 · u3 0     v3 · u1 −40 √ [v3 ]T =  v3 · u2  =  −7 10 . So the transition matrix from S to T √ v3 · u3 910 is:   2 6 −40 √ √  P = 0 10 −7 √ 10 . 0 0 910 

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 3 2 4 2. Let A be the matrix  2 0 2 . 4 2 3 

(a) (3 points) Find the characteristic polynomial of A and verify that the eigenvalues of A are λ = −1 and λ = 8.  λ − 3 −2 −4 Solution: λI − A =  −2 λ −2 . Hence det(λI − A) = λ3 − −4 −2 λ − 3 2 6λ − 15λ − 8. So the characteristic polynomial is λ3 − 6λ2 − 15λ − 8 = (λ − 8)(λ + 1)2 . Hence the eigenvalues are λ = −1 and λ = 8. 

(b) (3 points) Find a basis for the eigenspace E−1 of A. Solution: We find the general solution to the equation (−I − A) x = 0. We have     −4 −2 −4 1 12 1 Gaussian −I − A =  −2 −1 −2  −−−−−−→  0 0 0  −4 −2 −4

Elimination

0 0 0

Hence the general solution is   1      1 −1 −2 s − t −2 x y =   = s 1  + t 0 . s z t 1 0 Thus a basis for E−1

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    −1   − 12 is  1  ,  0  .   1 0

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Question 2 continues. . . (c) (3 points) Find a basis for the eigenspace E8 of A. Solution: We find the general We have  5 −2  8I − A = −2 8 −4 −2

solution to the equation (8I − A) x = 0.    1 0 −1 −4 Gaussian −2  −−−−−−→  0 1 − 12  . Elimination 0 0 0 5

Hence the general solution is       s x 1  y  =  1s  = s 1  . 2 2 z s 1    1  Thus a basis for E8 is  12  .   1

(d) (2 points) Find an invertible matrix P and a diagonal matrix D such that P−1 AP = D. Solution: Let  − 12 −1 1 P= 1 0 12  0 1 1 

 −1 0 0 D =  0 −1 0  . 0 0 8 

Then P−1 AP = D.

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Question 2 continues. . . (e) (4 points) Find a matrix B such that B3 = A. Show your steps.    (−1)3 0 0 −1 0 0 Solution: Let E =  0 −1 0 . Then E3 =  0 (−1)3 0  = 0 0 23 0 0 2   −1 0 0  0 −1 0  = D = P−1 AP. So A = PE3 P−1 = (PEP−1 )3 . To calcu0 0 8 −1 late P we do Gaussian elimination of the following augmented matrix:    1  8 −2 − 2 −1 1 1 0 0 1 0 0 − 29 9 9 Gaussian 5 .  1 0 12 0 1 0  −−−−−−→  0 1 0 − 49 − 29 9 Elimination 2 4 4 0 0 1 0 1 1 0 0 1 9 9 9 

1

Thus B = PEP−1 = 

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3 2 3 4 3

2 3 − 23 2 3

4 3 2 . 3 1 3

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 1 1 4    4 10 1   . 3. Let A =  7 17 3  2 4 3

(a) (4 points) Find the rank of A. Solution: Find a row echelon form of A.   1 1 4   Elimination  0 6 −15  A −−−−−−→  . Gaussian  0 0 0  0 0 0 The first and second columns are pivots while the third is not. So the rank is 2.

 1 4  (b) (6 points) Find the rank of B =  7 2 the value of λ.

 3 1 4 λ 10 1    . Your answer will depend on 1 17 3  2 4 3

     1  1          4   10  Solution: From part (a)   ,   is a basis for the column space  7   17       2 4           1 1  3      4   10          λ of A. We check when does   belong to span   ,   . We   7   17   1      4  2 2 have       1 1 3 1 1 3 1 1 3  4 10 λ  R2 −4R1  0 6 λ − 12  R3 − 10 R2  0 6 λ − 12      R3 −7R1   6  −−−−2−→   −−−−→   10λ  . R −2 R −20  R4 − 6 R2  0 0 − 6   7 17 1  4 1  0 10 0 2 −4 2 4 2 0 0 − 2λ 6         1  1 3             4   10   λ Therefore   belongs to span   ,   if and only if λ = 0.  1   7   17       2 4  2 So rank of B is 2 when λ = 0 and rank of B is 3 when λ 6= 0.

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  1 3 3 4. Let u =  1  and let T : R → R be defined by 0 T (x) =

 u · x u·u

u.

(a) (3 points) Verify that for any x, y ∈ R3 and for any c, d ∈ R T (cx + dy) = cT (x) + dT (y), thus proving that T is a linear transformation. Solution: We have T (cx + dy) =



   u · (cx + dy) cu · x + du · y u= u u·u u·u   cu · x du · y u = + u·u u·u cu · x du · y = u u+ u·u u·u = cT (x) + dT (y).

(b) (3 points) Find the standard matrix of the linear transformation T .       x x 1 3 Solution: Let x =  y  ∈ R . Note that u · x =  1  ·  y  = x + y z z 0     1 1    and u · u = 1 · 1  = 2. So 0 0    x+y  1 2 x + y    x+y  u= 1 = T (x) = 2 u·u 2 0 0   1    1 1 1 x 0 0 2 2 2 2 1 1 1 1       = x 2 + y 2 + z 0  = 2 2 0  y . z 0 0 0 0 0 0 u · x

1

1 2 1 2

 0 0 . So the standard matrix of T is  0 0 0 2 1 2

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Question 4 continues. . . (c) (3 points) Find a basis for the kernel of T , and write down the nullity of T . 1

1 2 1 2

 0 0 . Then the kernel of T is the nullspace of Solution: Let A =  0 0 0 A. We have   1 1 0 Gaussian A −−−−−−→  0 0 0  Elimination 0 0 0 2 1 2

Hence the general solution to Ax = 0 is         x 0 −1 −s y  =  s  = s  1  + t0  t z 0 1     0   −1    , So a basis for the kernel of T is 0  and hence the nullity 1   1 0 of T is 2.

(d) (3 points) Find a basis for the range of T , and write down the rank of T . 1 2 1 2

1 2 1 2

 0 0 . Then the range of T is the column space Solution: Let A =  0 0 0  1   2  of A. It is clear that  12  is a basis for the column space of A, and   0 hence for the range of T . Therefore the rank of T is 1. 

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Question 4 continues. . .       0  −1  1 (e) (3 points) Let B =  0  ,  1  ,  −1   . It is known that B is a basis 1 0 0   x 3  for R . For any x = y  ∈ R3 , find [T (x)]B . z Solution: We first find [u]B . Consider the augmented matrix    1 0 0 2 1 −1 0 1 Gaussian 0 1 −1 1  −−−−−−→  0 1 0 1  Elimination 0 0 1 0 0 0 1 0 

    2 x    Hence [u]B = 1 . For x = y , z 0   1 x+y  1 T (x) = 2 0 So    1 x + y  1  [T (x)]B =  2 0 B       1 2 x+y x + y x+y  1  =  1  =  x+y  . = 2 2 2 0 0 0 B

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5. In the question, all vectors are column vectors. Let A be an n × n matrix.   (a) (4 points) Show that for any u, w ∈ Rn , (Au) · w = u · AT w . Solution: Since u, w, Au, and AT w are all column vectors in Rn , we have       (Au) · w = (Au)T w = uT AT w = uT AT w = u · AT w .

(b) (3 points) Give an example of a 2 × 2 matrix A for which           1 1 1 1 . A · 6= · A 1 0 1 0 You should demonstrate that your example works. 

        0 1 1 0 1 1 1 Solution: Let A = . Then A = = . 0 1 1 0 1 1 1        1 0 1 1 0 And A = = . So 0 0 1 0 0          1 1 1 1 A · = · = 1 and 1 0 1 0          1 1 1 0 · A = · = 0. 1 0 1 0 Hence they are not equal.

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Question 5 continues. . . (c) (3 points) Let v1 , v2 , v3 be orthonormal vectors in Rn . Suppose w ∈ Rn . Define b1 = (Av1 ) · w, b2 = (Av2 ) · w, and b3 = (Av3 ) · w.

Define q = b1 v1 + b2 v2 + b3 v3 .

Calculate v1 · q, v2 · q, and v3 · q (hint: recall v1 · v1 = 1, v1 · v2 = 0 etc.). Solution: Since v1 , v2 , and v3 are orthonormal, we have v1 · q = v1 · (b1 v1 + b2 v2 + b3 v3 ) = b1 v1 · v1 + b2 v1 · v2 + b3 v1 · v3 = b1 .

v2 · q = v2 · (b1 v1 + b2 v2 + b3 v3 ) = b1 v2 · v1 + b2 v2 · v2 + b3 v2 · v3 = b2 .

v3 · q = v3 · (b1 v1 + b2 v2 + b3 v3 ) = b1 v3 · v1 + b2 v3 · v2 + b3 v3 · v3 = b3 .

(d) (5 points) Using the same definitions as in Part(c), show that for every v ∈ span{v1 , v2 , v3 }, (Av) · w = v · q. Solution: Write v = a1 v1 + a2 v2 + a3 v3 . Then Av = a1 Av1 + a2 Av2 + a3 Av3 . Therefore (Av) · w = (a1 Av1 + a2 Av2 + a3 Av3 ) · w

= a1 (Av1 ) · w + a2 (Av2 ) · w + a3 (Av3 ) · w

= a 1 b1 + a 2 b2 + a 3 b3

= a1 v1 · q + a2 v2 · q + a3 v3 · q

= (a1 v1 + a2 v2 + a3 v3 ) · q

= v · q.

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6. Let A be an n × n matrix. (a) (3 points) Suppose λ is an eigenvalue of A and suppose v is an eigenvector associated with λ. Show that for any m > 0, Am v = λm v. Hint: use induction. Solution: The base case is m = 1. In this case, since v is an eigenvector associated with λ, we have Am v = Av = λv = λm v. Now suppose that Am v = λm v for some m > 0. We check it for m + 1. We have Am+1 v = A(Am v) = A(λm v) = λm (Av) = λm (λv) = λm+1 v. So by induction Am v = λm v holds for all m > 0.

(b) (2 points) Let λ be a real number which is an eigenvalue of A. Show that if m > 0 and Am = I, then λ = ±1. Solution: Let v be an eigenvector associated with λ. By Part (a), v = Iv = Am v = λm v. So (λm − 1)v = 0. Since v is a non-zero vector, it follows that λm = 1. Since λ is a real number, λ = ±1.

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Question 6 continues. . . (c) (2 points) Suppose P and B are n × n matrices such that P is invertible and P−1 AP = B. Show that if B2 = I, then A2 = I. 2

Solution: Assume B2 = I. Then I = B2 = (P−1 AP) = P−1 A2 P. Thus P−1 A2 P = I, which implies A2 = PP−1 = I.

(d) (8 points) Assume that A is a symmetric matrix. Show that if m > 0 and Am = I, then A2 = I. Solution: Symmetric matrices are diagonalizable. There exist real numbers λ1 , . . . , λn and an invertible matrix P such that   λ1   .. P−1 AP =  .  = B. λn Note that each λi is an eigenvalue of A. Hence by Part (b), each λi is either 1 or −1. Therefore λi 2 = 1, for each 1 ≤ i ≤ n. So we have  2    λ1 1     .. .. B2 =  . .  = I. = λn2

1

So by Part (c), A2 = I.

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