2020/21 Finals Answer Key PDF

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MA1101R

Question 1  1  0 Let A =  0  a

0 1 0 b

[20 marks]  0 1  1 0  be a 4 × 4 matrix where a, b, c, d are some real numbers. 1 1  c d

(i) (4 marks) Find det A and write down the condition in terms of a, b, c, d such that the homogeneous system Ax = 0 has non-trivial solutions.

(ii) (4 marks) Let S = {(a, b, c, d) | Ax = 0 has only the trivial solution}. Is S a subspace of R4 ? Why?

(iii) (4 marks) Given rankA = 3, find the general solution of Ax = 0. Show your working.  1   1  (iv) (4 marks) Given that  1 is an eigenvector of A, find the condition satisfied by a, b, c, d.   1 (v) (4 marks) If a, b, c, d are all equal, find a basis for the column space of A in terms of a. Explain how you derive your answer.   1 0 0 1           1 1 0 0 1 1   0 1 1 0    = 0 1 1 − 0 0 1  (i) det A =      0 0 1 1     b c d a b c  a b c d   ) (    ) (  1 1 0 1  0 1  0 0          =  −  − − +   = d − c + b − a.  c d b d  a c  a b  So the system Ax = 0 has non-trivial solutions if and only if d − c + b − a = 0.

(ii) Ax = 0 has only the trivial solution if and only if det A = 0.

From (i), we can rewrite the set notation of S as {(a, b, c, d) | d − c + b − a = 0}.

Let u = (1, 0, 0, 0) and v = (0, 1, 0, 0). Both vectors belong to S . Then u + v = (1, 1, 0, 0) ∈ S .

So S does not satisfied the closure property and hence is not a subspace of R4 .

(iii) Since the first three rows of A are linearly independent, in order that rankA = 3, the   1 0 0 1   0 1 1 0  . last row (a, b, c, d) is “redundant” and hence a row echelon form of A is   0 0 1 1   0 0 0 0 Denote the four variables of Ax = 0 by x, y, z, w, by back substitution, we get the general solution w = t, z = −t, y = t, x = −t for t ∈ R.

2



 x −1       y  1     . z      In matrix form, this is given by = t  −1 w 1     2 1       1  2  .   (iv) A   =   2 1     1 a+b+c+d   1    1  If   1 is an eigenvector of A, we must have a + b + c + d = 2.   1   1 0 0 1    0 1 1 0 .  (v) In this case, A =   0 0 1 1   a a a a Note that the last row (a, a, a, a) = a(1, 0, 0, 1) + a(0, 1, 1, 0) is a linear combination of first and third row.

  1 0 0 1    0 1 1 0  . So a row echelon form of A is   0 0 1 1   0 0 0 0 Hence we can take any three columns of A   asthe  basis  the column space.  for  1 0 0              0 1 1        In particular, we can take the basis   ,   ,   .  0 0 1         a a a 

3

Question 2a

[12 marks]

Let S = {(1, 1, 2, 0), (2, 2, 4, 0), (0, 0, 1, 3), (1, 1, 3, 3), (1, 1, 1, −3)} and V = span(S). (i) (4 marks) Find a basis S ′ for V such that S ′ ⊆ S and write down dim V .

(ii) (4 marks) Is V = span{(2, 2, 5, 3), (2, 2, 3, −3), (1, 1, 0, −6)}? Justify your answer.

(iii) (4 marks) Let W = {(x, y, z, w) | x − y + z − w = 0}. Find W ∩ V . Give your answer as a linear span.  1 2 0  1 2 0 (i) Let B =  2 4 1  0 0 3 So columns 1 and 3

1 1 3 3 of



 1   1 E  0  GJ −→ 0 1   0 −3 B are linearly 1

2 0 1

1



 0 1 1 −1  . 0 0 0 0   0 0 0 0 independent.

Hence S ′ = {(1, 1, 2, 0), (0, 0, 1, 3)} form a basis for V and dim V = 2. (ii) Stack the two spanning set of  1 0 2 2   1 0 2 2   2 1 5 3  0 3 3 −3

vectors column-wise   1 1 0   1  E  0 1  GJ −→    0 0 0   −6 0 0

This represents a consistent system, hence

as an augmented matrix as follow:  2 2 1  1 −1 −2   0 0 0   0 0 0

span{(2, 2, 5, 3), (2, 2, 3, −3), (1, 1, 0, −6)} ⊆ V − − − − − (1) Flipping the two sets of vectors around:    2 2 1 1 0    2 2 1 1 0  E    GJ  −→ 5 3  0 2 1     3 −3 −6 0 3

1 0 −3/4 1/4

1/2



 0 1 −5/4 1/4 −1/2   0 0 0 0 0   0 0 0 0 0

This again represents a consistent system, hence

V ⊆ span{(2, 2, 5, 3), (2, 2, 3, −3), (1, 1, 0, −6)} − − − − − (2) By (1) and (2), we get V = span{(2, 2, 5, 3), (2, 2, 3, −3), (1, 1, 0, −6)}. (iii) Observe that the vector (1, 1, 3, 3) ∈ V satisfies the equation x − y + z − w = 0 and hence it belongs to W .

Hence W ∩ V is non-trivial and this implies dim W ∩ V ≥ 1.

On the other hand, the vector (1, 1, 2, 0) ∈ V does not satisfy the equation x−y+z−w =

0 and hence it does not belong to W .

Hence W ∩ V is a proper subset of V and this implies dim W ∩ V < dim V = 2. So we conclude that dim W ∩ V = 1.

Since (1, 1, 3, 3) ∈ W ∩ V , we have W ∩ V = span{(1, 1, 3, 3)}.

4

Question 2b

[8 marks]

Let W be a subspace of Rn and W ⊥ = {w ∈ Rn | w · v = 0 for all v ∈ W }. (i) (2 marks) Show that W ∩ W ⊥ = {0}.

(ii) (6 marks) Show that every vector v ∈ Rn can be written uniquely as v = v1 + v2 where v1 ∈ W and v2 ∈ W ⊥ .

(You may assume in part (ii) that W and W ⊥ are associated to the row space and

nullspace of certain matrix.) (i) Let x ∈ W ∩ W ⊥ .

Then x ∈ W and x ∈ W ⊥ .

So x · x = 0 ⇒ x = 0. Hence we conclude that W ∩ W ⊥ = {0}.   r1    r2   (ii) Let {r1 , r2 , . . . , rk } be a basis for W and let A =   ..  be a k × m matrix with i-th .  rk row equal to ri . Then W is the row space of A and W ⊥ is the nullspace of A. Let {s1 , s2 , . . . , sh } be a basis for W ⊥ .

Then by dimension theorem, k + h = rank(A) + nullity(A) = n. To show that {r1 , r2 , . . . , rk , s1 , s2 , . . . , sh } is a basis for Rn , we just need to show the set is linearly independent:

a 1 r 1 + · · · + a k r k + b1 s 1 + · · · + bh s h = 0. This can be rewritten as a1 r1 + · · · + ak rk = −b1 s1 − · · · − bh sh (∗) Since LHS of (∗) belongs to W and RHS of (∗) belongs to W ⊥ , both sides belong to W ∩ W ⊥ = {0 }. Hence a1 r1 + · · · + ak rk = 0 implies a1 = · · · = ak = 0

and b1 s1 + · · · + bh sh = 0 implies b1 = · · · = bh = 0.

Therefore we conclude that {r1 , r2 , . . . , rk , s1 , s2 , . . . , sh } is linearly independent. For any v ∈ Rn , v = c1 r1 + · · · + ck rk + d1 s1 + · · · + dh sh = v1 + v2 where v1 = c1 r1 + · · · + ck rk ∈ W and v2 = d1 s1 + · · · + dh sh ∈ W ⊥ . Furthermore, the decomposition v = v1 + v2 is unique: Suppose v = u1 + u2 where u1 ∈ W and u2 ∈ W ⊥ . Then v1 + v2 = u1 + u2 ⇒ v1 − u1 = u2 − v2 (∗∗).

Like before, LHS of (∗∗) belongs to W and RHS of (∗∗) belongs to W ⊥ . So v1 − u1 = 0 ⇒ v1 = u1 and u2 − v2 = 0 ⇒ v2 = u2 .

5

Question 3a

[14 marks]   1 1 0 0 1      1 2 0 0 −1     .    Let A =   1 2 3 4 and b =  1  1 2 3 4

−1

      1 −3 0       1 1  −2       (i) (4 marks) Let S = {u1 , u2 , u3 } where u1 =  1 , u2 =  1  , u3 =  1 .       1 1 1 Show that S is an orthogonal basis for the column space V of A. (ii) (2 marks) Normalise S to get an orthonormal basis T = {v1 , v2 , v3 } for V .

(iii) (4 marks) Find the least squares solutions of Ax = b.

(iv) (4 marks) Extend the basis T in part (ii) to an orthonormal basis T ′ = {v1 , v2 , v3 , v4 } for R4 without using Gram-Schmidt. (i) Direct checking: u1 · u2 = 0, u1 · u3 = 0, u2 · u3 = 0.

So S is an orthogonal set and hence it is linearly independent. Denote the four columns of A by c1 , c2 , c3 , c4 . Then we have u1 = c1 , u2 = −3c1 + 2c2 , u3 = −c2 + c3 . Hence S ⊆ V (the column space of A).

Check that rank(A) = 3. So dim V = 3. Hence Sisan orthogonal basisfor V .   1 −3 0           1  1  , v2 = √1  1  , v3 = √1 −2 (ii) v1 =       2  1 12  1  6 1 1 1 1 T T (iii) We solveA Ax = A b.   4 6 6 9 0      6 12 12 16   −2 T    AT A =   6 12 18 24  and A b =  0 .     9 16 24 33 1 (



1 0 0

1

1



   0 1 0 −1/2 −1    A A | A b −→   0 0 1 4/3 1/3   0 0 0 0 0 T

T

)

GJ E

By back substitution, we get the general solution: w = t, z =

1 1 4 − t, y = −1 + t, x = 1 − t. 3 3 2

6

So the least squares solutions of Ax = b are:     1−t x      y  −1 + 1 t 2      z  =  1 4t  .    3 −3  w t   1    −1  (iv) Let v =   1  be one of the least squares solutions in (iii).  3 0   1    −1  Then p = Av =   0  is the projection of b onto V .   0   0    0  Hence p − b =   1  is orthogonal to V , and hence to v1 , v2 , v3 .   −1   0   0   So we can take v4 = ± √12   1 .   −1

7

Question 3b

[6 marks]

Let S = {u1 , u2 , . . . , um } and T = {v1 , v2 , . . . , vm } be two orthonormal bases for a proper subspace V of Rn . ) ) ( ( Let C = u1 u2 · · · um and D = v1 v2 · · · vm be matrices formed using the basis vectors of S and T as their columns respectively. Determine whether the following are true or false. Justify your answers. (i) (2 marks) C and D are orthogonal matrices. (ii) (2 marks) If the reduced row echelon form of (D | C) is given by (I | P ), then P is

the transition matrix from S to T . (iii) (2 marks) C T D is the transition matrix from T to S .

(i) False. The sizes of C and D are n × m, so they are non-square matrices, and hence cannot be orthogonal matrices.

(ii) False. The size of P is n × m, So it is a non-square matrix, and hence cannot be a transition matrix.

(iii) True.   T u1 u1 · v1   T ( ) u ·v  u2   v1 v2 · · · vm =  2 1 CT D =   ..  ..   .  .  T um

u1 · v2

u2 · v2

· · · u1 · vm

 · · · u2 · vm   ...  

um · v1 um · v2 · · · um · vm

which is the transition matrix from T to S .



8

Question 4a 1 0  0 2  Let C =  0 2 1 0

[12 marks]  0 1  2 0 . 2 0  0 1

(i) (4 marks) Find the characteristic polynomial and all the eigenvalues of C. Show your working. (ii) (4 marks) Find a basis for each eigenspace of C. Show your working. (iii) (4 marks) Find a matrix P that orthogonally diagonalizes C and write down the corresponding diagonal matrix D. Explain how your answers are derived.   x − 1      0 0 −1   x − 2 −2  0   0 x − 2 −2     0 x − 2 −2 0      = (x−1)  −2 x − 2 (i) det(xI−C) =  0 + 0 −2 x − 2       −2 x − 2 0   0  0   0 x − 1   −1 0 0   −1 0 0 x − 1 2 2 2 = (x − 1) [(x − 2) − 4] − [(x − 2) − 4] = (x2 − 2x)(x2 − 4x) = x2 (x − 2)(x − 4) So the eigenvalues of C: 0, 2 and 4.

(ii) For λ = 0: 

−1

0

0

−1 0





1 0 0 1 0



     0 −2 −2 0 0  GJ E  0 1 1 0 0       0 −2 −2 0 0  −→  0 0 0 0 0      −1 0 0 −1 0 0 0 0 0 0 By back substitution, we get the general solution: w = t, z = s, y = −s, x = −t.

So the eigenspace E0 for λ = 0 is {(−t, −s, s, t)T | s, t ∈ R} and a basis for E0 is       0 −1           −1   0    ,   .  1   0             0 1  For λ = 2: 

1

0

0

  0

0

−2

0

−1   

−2 0

−1 0 0

0

0

0

1



 0  0

0   

GJ E

−→



1 0 0 −1 0

 0 1 0

0

0 0 1

0

0 0 0   

0



 0  0

0   

9

By back substitution, we get the general solution: w = t, z = 0, y = 0, x = t. So the eigenspace E2 for λ = 2 is {(t, 0, 0, t)T | t ∈ R} and a basis for E2 is     1          0    .  0          1   For λ = 4:



3

0

0

  0 2 −2   0 −2 2  −1 0 0

−1 0 0

0 3





1 0

0

0 0



     0  E  0 1 −1 0 0   GJ −→ 0 0 0 1 0  0     0 0 0 0 0 0

By back substitution, we get the general solution: w = 0, z = t, y = t, x = 0. So the eigenspace E4 for λ = 4 is {(0, s, s, 0)T | s ∈ R} and a basis for E4 is     0          1    .  1           0 

(iii) The four eigenvectors in the bases for the eigenspaces:           −1 1 0   0             −1  0   0  1   , , ,              1   0   0  1       0 1 1 0 are orthogonal.

We normalise these vectors to get an orthogonal  0 −1 1  1 −1 0 0 P = √  0 0 2 1 0 1 1

matrix  0  1  1  0

 0  0 that orthogonally diagonalise C to give the diagonal matrix D =  0  0

0 0 0



 0 0 0 . 0 2 0  0 0 4

10

Question 4b

[8 marks]

Let M is an n × n matrix such that M 2 = M and both 0 and 1 are eigenvalues of M . (i) (4 marks) Show that the column space of M is the eigenspace E1 associated to eigenvalue 1. (ii) (4 marks) Show that M is diagonalizable (i) Let v ∈ E1 (the eigenspace associated to eigenvalue 1).

Then M v = v which implies v belongs to the column space of M . Hence E1 ⊆ column space of M (1).

Let v ∈ column space of M . Then v = M w for some w ∈ Rn .

So v = M 2 w = M (M w) = M v. This implies v ∈ E1 .

Hence column space of M ⊆ E1 (2).

By (1) and (2), we conclude that column space of M = E1 (ii) From (i), dim E1 = dim(column space of M ) = rank M . On the other hand, the eigenspace E0 associated to eigenvalue 0 is the nullspace of M . So dim E0 = dim(nullspace of M ) = nullity M . By Dimension Theorem, dim E1 + dim E0 = rank M + nullity M = n. This implies there are n linearly independent eigenvectors of M , and hence M is diagonalizable.

11

Question 5a

[12 marks]

Let T : R3 → R3 be a linear transformation such that             0 0 1 2 1 0            T 1 = 1 , T  0 =  0 , T 2 = 2 . 2 2 2 4 0 4 (i) (3 marks) Find the standard matrix of T . Show how your answer is derived. (ii) (3 marks) Find the kernel of T . Give your answer as a linear span. (iii) (3 marks) Find the largest possible subspace V of R3 such that every vector v ∈ V

maps to itself under T . Explain how your answer is derived.  1   3 (iv) (3 marks) Are there any vector v ∈ R such that T (v) = 2? Justify your answer. 0

(i) Let A be the standard matrix of T .            0 1 2 1 0 0            A  1 = 1  , A 0 =  0 , A 2 =  2 . 4 0 4 2 2 2    0 0 1 1    By stacking, we have: A 1 0 2 = 1 2 2 2 0   −1  4 0 1 1 0 2 0  = 1   ⇒ A =  1 0 2 1 0 2  0 3 8 2 2 0 2 4 4 (ii) ker T = nullspace A: 

4 −2 1 0

1 0 3 8

3 2



2 0



 0 2 4 4  −2 1  3 0. 2 2 

1 0 1/4 0

 GJ E  0 0  −→  0 1 0 0 2 0

0 0



 0  0

1 By back substitution, we get the general solutions: z = t, y = 0, x = − t. 4 So ker T = span{(−1/4, 0, 1)}.

(iii) The largest possible subspace V of R3 such that T (v) = v for all v ∈ V is E1 , the eigenspace of A associated to eigenvalue 1.

From the given conditions as well as part (ii), we know A has three distinct eigenvalues 1, 2, 0. Soeach has dimension 1.   eigenspace 0 0 From A 1 1 2   

=

2    

  0  1

 2       

, we know that E1 = span



.

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1   (iv) No. This is the same as saying  2 ∈ column space of A. 0     4 −2 1 3 1 0 1/4 0 1  GJ E    0 3 0 6  −→  0 1 0 0  3 8 2 2 0 0 0 0 1

  1   Since the above system is inconsistent, we conclude that T (v) = 2  for any vector 0 v ∈ R3 .

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Question 5b

[8 marks]

Let T : Rn → Rn be a linear transformation and R(T ) is the range of T . Denote T 1 = T and T k+1 = T ◦ T k for all (integers) k ≥ 1.

(i) (2 marks) Show that R(T k+1) ⊆ R(T k ) for all k ≥ 1.

(ii) (6 marks) Suppose T m is the zero transformation for some m > n. Show that T n must be the zero transformation. (Note that T itself need not be the zero transformation.) Hint: Show that if R(T k ) = R(T k+1 ) for some k ≥ 1, then R(T k ) = R(T h ) for all h ≥ k . (i) For any k ≥ 1, let v ∈ R(T k+1 ). So v = T k+1 (w) = T k (T (w)) for some w ∈ Rn . Hence v ∈ R(T k ).

This implies R(T k+1) ⊆ R(T k ). (ii) First of all, we show that if R(T k ) = R(T k+1), then R(T k ) = R(T k+2 ). Let v ∈ R(T k ). Then v ∈ R(T k+1 ). So v = T k+1 (w) = T (T k (w)) for some w ∈ Rn . Let u = T k (w) ∈ R(T k ) = R(T k+1). So u = T k+1 (x) for some x ∈ Rn . So v = T (u) = T (T k+1(x)) = T k+2 (x) ∈ R(T k+2). Hence we have R(T k ) ⊆ R(T k+2 ).

On the other hand, by (i), R(T k+2) ⊆ R(T k+1 ) = R(T k ).

So we conclude R(T k ) = R(T k+2 ).

Inductively, if we have R(T k ) = R(T k+1), then R(T k ) = R(T h ) for all h ≥ k .

Now let p be the smallest integer such that R(T P ) = R(T p+1). Then

R(T ) ) R(T 2 ) ) · · · ) R(T p ) = R(T p+1 ) = · · · = R(T m ) = {0}. This implies n ≥ rank (T ) > rank (T 2 ) > · · · > rank (T p ) = rank (T p+1 ) = · · · = rank(T m ) = 0 since T m is the zero transformation. Since we have a strictly decreasing sequence from rank(T ) to rank(T p ), and rank(T p ) = 0, this implies p ≤ n.

Consequently, rank(T n ) = 0 ⇒ R(T n ) = {0} ⇒ T n is the zero transformation....