MA1101R AY19/20 Sem 1 Finals PDF

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MA1101RStudent Number:Seat Number:National University of SingaporeMA1101R Linear Algebra ISemester I (2019 – 2020)Time allowed: 2 hoursINSTRUCTIONS TO CANDIDATES Write down your student number and seat number clearly in the space provided at the top of this page. Do not writeyour name. This booklet ...

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MA1101R

Student Number:

Seat Number: National University of Singapore MA1101R

Linear Algebra I

Semester I (2019 – 2020) Time allowed: 2 hours

INSTRUCTIONS TO CANDIDATES

1. Write down your student number and seat number clearly in the space provided at the top of this page. Do not write your name. 2. This booklet (and only this booklet) will be collected at the end of the examination. 3. This examination paper contains SIX (6) questions and

Examiner’s Use Only Questions 1 2

comprises FIFTEEN (15) printed pages. 3 4. Answer ALL questions. 5. This is a CLOSED BOOK (with helpsheet) examination. 6. You are allowed to use one A4-size helpsheet. 7. You may use scientific calculators. However, you should lay

4 5 6

out systematically the various steps in the calculations. Total

Marks

Page 2

Question 1 

1

 0 Let A =    −1 0

MA1101R

[10 marks] −1 0

1 0





 1 −1 0 0 −1      2 −2 0   with reduced row echelon form R =  0 0 1 0 1  .    1 −1 1 0 0 0 1 1  −1 1 0 0 0 0 0 0 0

2

1

(i) Use R to find a basis for the column space V of A.       1 1 −12        0 0  0             (ii) Let u1 =   0, u2 = 2 , u3 =  9 .        0 0  11 

0 0 0 Show that S = {Au1 , Au2 , Au3 } is an orthogonal basis for V .   1   0  (iii) Find the coordinate vector [w]S of w =  1 ∈ V with respect to the basis S in part (ii).   0

(iv) Is it possible to find a one-dimensional subspace of V that does not contain any column of A? Justify your answer.

Show your working below.         1 0 2              0   2  −2      (i) Basis for V :  −1 ,  1  , −1.              0 −1 1  (Note that any three linearly independent columns of A also form a basis.) (ii) 

1





1





10



      0  4   −4      Au1 =   , Au2 =   , Au3 =   10  −1 1       0 −2 2

Check the dot products: Au1 · Au2 = 0, Au1 · Au3 = 0, Au2 · Au3 = 0. This implies S is an orthogonal set, and hence is linearly independent. Since dim V = 3 (from part (i)) and Au1 , Au2 , Au3 belongs to the column space V of A, so S is an orthogonal basis for V . Continue on next page if you need more writing space.

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More working space for Question 1. (iii) Since S is orthogonal, w · Au1 w · Au3 w · Au2 Au3 Au2 + Au1 + ∥Au3 ∥2 ∥Au2 ∥2 ∥Au1 ∥2 1+1 10 + 10 1−1 Au3 Au1 + 2 Au2 + 2 = 2 2 2 2 2 2 2 10 + 42 + 102 + 22 1 +0 +1 +0 1 +4 +1 +2 ) ( 1 1 . So [w]S = 0, , 11 11

w =

(iv) Yes. We can take linear of a linearcombination the span     of the columns of A:   1 0  1                    0 2   2        e.g. span   +   = span   .       −1  1   0           0   −1 −1  This is a one dimensional subspace of V which does not contain any column of A.

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Page 4

Question 2

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[10 marks]



          −1 2 1 0 1 3 −2 1             Let A =  0 4 0 and v1 = 0  , v2 =  0  , v3 =  1 , v4 = 0 , v5 =  1  . 1 1 2 1 −1 1 2 3 (i) Determine which of the five vectors v1 to v5 are eigenvectors of A. (ii) Write down all the eigenvalues of A. Justify your answers. (iii) Write down a basis for each of the eigenspaces of A. (iv) Find an invertible matrix P and a diagonal matrix D such that A3 = P DP −1 . (v) Is AAT orthogonally diagonalizable? Why? Show your working below. (i)       4 2 0   = 4v , Av =   = 2v , Av =   = 4v Av1 =  0 0 4 3 1 2 2 3 4 −2 8     7 −4     Av4 =  0 , Av5 =  4  = 4v5 5

4

So all except v4 are eigenvectors of A.

(ii) From (i), we have two eigenvalues 2 and 4. Since both v1 and v3 are linearly independent eigenvectors associated to 4, so the multiplicity of eigenvalue 4 is at least 2. As A is a 3 × 3 matrix, we conclude that 2 and 4 are the only eigenvalues of A. (iii) We deduce from (i) that the eigenspace E2 associated to 2 is one dimensional, and a   1   basis is given by v2 =  0 .

−1 Similarly, we deduce that the E to 4 is two dimensional, and a 4 associated  eigenspace   1 0    basis can be given by v1 =  0 and v3 = 1. 1 2 (Other possible bases for E4 : the pair v1 , v5 or the pair v3 , v5 .)

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More working space for Question 2. (iv) The eigenvalues of A3 are 23 and 43 (repeated) with corresponding eigenvectors same as those of A .

 1 0   Hence P =  0 0 1 (depending on the choices of eigenvectors in (iii)), −1 1 2   8 0 0   and D =  0 64 0 . 0

1

0

64

(v) Yes. AAT is a symmetric matrix, and hence is orthogonally diagonalizable.

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Page 6

Question 3 

[10 marks]

 0    0 1 1   and b = Let A =   1 0 0   0 −1 1 1

MA1101R

0

  1   1   .   0  0

(i) Show that the linear system Ax = b is inconsistent. (ii) Find the least squares solution of the system in (i). (iii) Find the projection p of b onto the column space of A. (iv) Find the smallest possible value of ∥Av − b∥ among all vectors v ∈ R3 . (v) Note that the three columns of A form an orthogonal set. Extend this set to an orthogonal basis for R4 . Show your working below. (i) 

1

0

0 1



   0 1 1 1  G.E.  (A | b) =   1 0 0 0  −→   0 −1 1 0

So Ax =b is inconsistent.    2 0 0 1     T T (ii) A A =  0 2 0  and A b = 1. 0 0 2



1 0 0

1

 0 1 1 1  0 0 2 1  0 0 0 −1

     

1

1  2

  So the solution of AT Ax = AT b is x =  12  1 2

which gives the least squares solution for Ax = b.   1 1  2  2    1  (iii) The projection is given by p = A  12  =  1 . 2  1 2 0 (iv) The samllest possible value of ∥Av − b∥ is given by           1 1 1    −2  √  2 √         1  1    0  2 1 1           . ∥p − b∥ =  1  −   =   1  = + = 2 4 4  2   0    2       0 0   0 

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More working space for Question 3. (v) We need to add one more vector to the set. This vector can be given by   − 21    0   p−b =  1   2  0

which is orthogonal to the column space of A, and hence to the three columns of A.

Continue on pages 14–15 if you need more writing space.

Page 8

Question 4

MA1101R

[10 marks]

Let T : R3 → R3 be a linear transformation such that      0 1       T  1  = v1 , T  1   = v2 , 1

1

  0    T  0  = v3 1

where v1 , v2 and v3 are non-zero vectors.     0 1       (i) Find T  0  and T  1  as linear combinations of v1 , v2 and v3 . 0

0

(ii) Find the standard matrix A for T in terms of v1 , v2 and v3 . (iii) Suppose v1 , v2 and v3 are linearly independent. Show that ker(T ) = {0}. (iv) Suppose T (v1 ) = 2v1 , T (v2 ) = 3v2 , T (v3 ) = 5v3 . Find v1 , v2 and v3 . Show your working below. (i)            0 1 0 1 1            T  0  = T  1 − 1 = T  1  − T   1  = v1 − v2 . 1 1 1 1 0            0 0 0 0 0             T  1  = T  1 − 0 = T  1  − T   0 = v2 − v3 . 1 1 1 1 0

(ii) From (i) we have:     1 0     A  0 = v1 − v2 and A  1 = v2 − v3 . 0 0 And  from  the given condition, we have: 0   A  0 = v3 . 1 These are the three columns of A.

Hence A = (v1 − v2 | v2 − v3 | v3 ).

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More working space for Question 4. (iii) We shall show S = {v1 − v2 , v2 − v3 , v3 } is linearly independent. Set up the vector equation

c1 (v1 − v2 ) + c2 (v2 − v3 ) + c3 v3 = 0 Rearranging the terms gives: c1 v1 + (c2 − c1 )v2 + (c3 − c2 )v3 = 0.

Since v1 , v2 and v3 are linearly independent, this implies: c1 = 0, c2 − c1 = 0, c3 − c2 = 0,

which will further give

c1 = c2 = c3 = 0. So S is linearly independent. Since the standard matrix A of T has three linearly independent columns, it is invertible. This implies the nullspace of A = Ker(T ) = {0}. (iv) From the given information, we have Av1 = 2v1 , Av2 = 3v2 , Av3 = 5v3 (∗) Since v1 , v2 , v3 are non-zero vectors, they are eigenvectors of A with eigenvalues 2, 3, 5 respectively. Since all the eigenvalues are non-zero, A is invertible. Hence the linear systems Ax = v1 , Ax = v2 , Ax = v3 all have unique solutions. From the given conditions of T , we have    1 0    A  1 = v1 , A  1 = v2 , 1 1

  0   A  0 = v3 1

On the other hand, by (∗), we have A( 21 v1 ) = v1 , A( 31 v2 ) = v2 , A51v3 ) = v3 . By comparison, we have:     1 2     1 v = 1 ⇒ v1 =  2; 2 1 1 2     0 0    1 v = 1 ⇒ v2 =  3 ; 3 2

3 1     0 0     1 v = 0 ⇒ v3 =  0. 5 3 1 5

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Question 5

MA1101R

[10 marks]

Suppose A is a 3 × 5 matrix with row space given by span{(1, 2, 3, 4, 5)}. (i) What are the rank and nullity of A? (ii) Write down the reduced row echelon form of A. (iii) Find a basis for the nullspace of A. (iv) Find the general solution of the non-homogeneous system Ax = b where b is the first column of A. 

1



  (v) Suppose the first column of A is  0 . Do we have enough information to determine −2 the matrix A? Why?

Show your working below. (i) rank(A) = 1 (since the row space is spanned by one non-zero vector). nullity(A) = 5 − 1 = 4 (by Dimension Theorem)   1 2 3 4 5   (ii) rref = 0 0 0 0 0

0 0 0 0 0 (iii) Let the variables of Ax = 0 be x1 , x2 , x3 , x4 , x5 .

Then set x2 = s, x3 = t, x4 = u, x5 = v where s, t, u, v are parameters. Then x1 = −2s − 3t − 4u − 5v .

So a general solution of the system is:             x1 −5 −4 −3 −2 −2s − 3t − 4u − 5v             s 0   0 0  1    x2                        x3 =  t  = s 0  + t  1  + u  0  + v  0                 0   1 0  0    x4  u x5

v

0

0

0

1

So a basis for the nullspace of A is given by:          −2 −3 −4 −5                  1 0 0 0                     0  , 1  , 0  , 0  .                      0   0   1   0         0 0 0 1 Continue on next page if you need more writing space.

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More working space for Question 5.   1   0     (iv) Since b is the first column of A, a solution of Ax = b is x =  0 .   0  0

  1   0    So a general solution of Ax = b = (general solution of Ax = 0) +  0:   0 0

  x1   x2    x3  =     x4  x5

       

1 − 2s − 3t − 4u − 5v s t u v



   .   

(v) Yes. Since rank(A) = 1, all columns of A are scalar multiples of the first column. Also, since the row space of A is span{(1, 2, 3, 4, 5)}, all the rows of A are scalar multiples of (1, 2, 3, 4, 5).

Hence we must have 

1

2

3

4

5



  A= 0 0 0 0 0  −2 −4 −6 −8 −10

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Question 6

MA1101R

[10 marks]

Prove the following statements. (a) If A is an n × n matrix such that A2 = I, then rank(I + A) + rank(I − A) = n. (Hint: rank(M + N ) ≤ rank(M ) + rank(N ))

(b) There are no orthogonal matrices A and B (of the same order) such that A2 − B 2 = AB . (Hint: Prove by contradiction. Recall that the product of two orthogonal matrices is an orthogonal matrix.)

Show your working below. (a) A2 = I ⇒ I − A2 = 0

⇒ (I − A)(I + A) = 0

⇒ column space of I + A ⊆ nullspace of I − A

⇒ rank(I + A) ≤ nullity(I − A) = n − rank(I − A) ⇒ rank(I + A) + rank(I − A) ≤ n − − − − − (1)

On the other hand, rank(I + A) + rank(I − A) ≥ rank[(I + A) + (I − A)] = rank(2I) = n − − − − − (2). By (1) and (2), we have rank(I + A) + rank(I − A) = n.

(b) Suppose A and B are orthogonal matrices such that A2 − B 2 = AB . A2 − AB = B 2 ⇒ A(A − B) = B 2 ⇒ A − B = A−1 B 2 = AT B 2 . AB + B 2 = A2 ⇒ (A + B )B = A2 ⇒ A + B = A2 B −1 = A2 B T . Since product of orthogonal matrices is orthogonal, A − B and A + B are both orthogonal. So (A − B )−1 = (A − B )T = AT − B T and (A + B )−1 = (A + B )T = AT + B T Then I = (AT − B T )(A − B ) = 2I − AT B − B T A − − − (1) I = (AT + B T )(A + B) = 2I + AT B + B T A − − − (2) Adding (1) and (2): 2I = 4I, which is a contradiction. Hence such orthogonal matrices A and B do not exist.

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More working space for Question 6.

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More working spaces. Please indicate the question numbers clearly.

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More working spaces. Please indicate the question numbers clearly.

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