Advanced Linear Algebra

Midterm 1 Math 4377 / 6308 (Spring 2015) March 5, 2015

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20 points 1. Mark each statement True or False. Justify each answer. (If true, cite appropriate
facts or theorems. If false, explain why or give a counterexample that shows why the
statement is not true in every case).

(1) If S is a linearly dependent set, then each vector in S is a linear combination of

other vectors in S.
(2) Any set containing the zero vector is linearly dependent.
(3) Subsets of linearly dependent sets are linearly dependent.
(4) Subsets of linearly independent sets are linearly independent.
(5) Every vector space that is generated by a finite set has a basis.
(6) Every vector space has a finite basis.
(7) If a vector space has a finite basis, then the number of vectors in every basis is the
same.
(7’) The dimension of Mm×n (F ) is m + n.
(8) Suppose that V is a finite-dimensional vector space, that S1 is a linearly independent
subset of V , and that S2 is a subset of V that generates V . Then S1 cannot contain
more vectors than S2 .
(9) If V is a vector space having dimension n, and if S is a subset of V with n vectors,
then S is linearly independent if and only if S spans V .
(10) If T : V → W is linear, then nullity(T ) + rank(T ) = dim(W ).

20 points 2. The first four Chebyshev polynomials are 1, x, 2x2 − 1, and 4x3 − 3x. These polynomials
arise naturally in the study of certain important differential equations. Show that the
first four Chebyshev polynomials form a basis of P3 (R).

20 points 3. Let T : R3 → R3 be given by

T (a, b, c) = (3a, −2a + c, b).

Prove that T is an isomorphism and find T −1 .

20 points 4. Let T : R3 → R2 be given by T (a, b, c) = (a − b, 2c).

(a) Show that T is a linear transformation.

(b) Find bases for the null space and the range of T .
(c) Compute the nullity and rank of T , and verify the dimension theorem.

20 points 5. Let W1 and W2 be subspaces of a vector space V .

(a) Prove that W1 + W2 is a subspace of V that contains both W1 and W2 .

(b) Prove that any subspace of V that contains both W1 and W2 must also contain
W1 + W2 .

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 Midterm 1 (cont.) Math 4377 (15549) / 6308 (14674) (2015 Spring) March 5, 2015

(c) Suppose
W1 = span{u1 , · · · , up }, W2 = span{v1 , · · · , vq }.
where u1 , · · · , up and v1 , · · · , vq are vectors in V . Show that

W1 + W2 = span{u1 , · · · , up , v1 , · · · , vq }.

20 points 6. (BONUS PROBLEM) Let V and W be finite-dimensional vector spaces and T : V →

W be an isomorphism. Let V0 be a subspace of V .

(a) Prove that T (V0 ) is a subspace of W .

(b) Prove that dim(V0 ) = dim(T (V0 )).

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Midterm 1 (cont.) Math 4377 (15549) / 6308 (14674) (2015 Spring) March 5, 2015

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Problem 1.

(1) False.

(2) True.

(3) , False.

(4) True.

(5) True.

(6) False.

(7) True.

(7’) False.

(8) True.

(9) True.

(10) False.

Problem 2.
Let β = {1, x, 2x2 − 1, 4x3 − 3x} and let γ = {1, x, x2 , x3 } be the standard ordered basis for
P3 (R). We have the coordinate vectors of β in γ as:
       
1 0 −1 0
 , [x]γ =   , [−1 + 2x2 ]γ =  0  [−3x + 4x3 ]γ = −3
0 1    
[1]γ = 
0 0 2 0
0 0 0 4

Note that the matrix with the coordinate vectors as columns have four pivots
 
1 0 −1 0
0 1 0 −3
Q= 0 0 2

0
0 0 0 4

Then {[1]γ , [x]γ , [−1 + 2x2 ]γ , [−3x + 4x3 ]γ } is linearly independent. By Thorem 2.21, β is
linearly independent. Combined with the fact that |β| = dim(P3 (R)) = 4, β is a basis for
P3 (R).

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Midterm 1 (cont.) Math 4377 (15549) / 6308 (14674) (2015 Spring) March 5, 2015

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Problem 3.
Let β = {e1 , e2 , e3 } be the standard ordered basis for R3 . The matrix representation of T in
β is  
3 0 0
[T ]β = −2 0 1
0 1 0
Note that the augmented matrix
   
3 0 0 1 0 0 1 0 0 1/3 0 0
0 1 0 1 0 ∼ 0 1 0 0 0 1 = I3 | ([T ]β )−1
 
[[T ]β | I3 ] = −2
0 1 0 0 0 1 0 0 1 2/3 1 0
So we have  
1/3 0 0
[T ]−1
β =  0 0 1
2/3 1 0
By Theorem 2.18, T is invertible and [T −1 ]β = ([T ]β )−1 . We have
T −1 (a, b, c) = (a/3, c, 2a/3 + b).

Problem 4.
(a) Note that, in the matrix and column vector notation, we have
 
3 1 −1 0
x ∈ R 7→ T (x) = Ax, A = .
0 0 2
Then, T is linear.
(b) To find the null space of T , row reduce the augmented matrix corresponding to Ax = 0
   
1 −1 0 0 1 −1 0 0
∼
0 0 2 0 0 0 1 0
We have      
x1 x2 1
x2  = x2  = x2 1
x3 0 0
Then   
 1 
the null space of T = span 1
0
 

The range of T is the column space of A and we have

   
1 0
the range of T = span {pivot columns of A} = span ,
0 1

(c) The nullity of T is 1 and the rank of T is 2 We have

nullity(T ) + rank(T ) = 1 + 2 = 3 = dim(R3 ).
Then, the dimension theorem is verified.

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Midterm 1 (cont.) Math 4377 (15549) / 6308 (14674) (2015 Spring) March 5, 2015

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Problem 5.
(a) W1 + W2 is a subspace of W : Closed under vector addition, because if u, v ∈ W1 + W2 ,
then there exist u1 , v1 ∈ W1 and u2 , v2 ∈ W2 such that u = u1 + u2 and v = v1 + v2 ,
and then u + v = u1 + u2 + v1 + v2 = (u1 + v1 ) + (u2 + v2 ) ∈ W1 + W2 . For scalar
multiplication, au = a(u1 + u2 ) = au1 + au2 ∈ W1 + W2 . Finally, W1 + W2 contains 0
since both W1 , W2 are subspaces and therefore contain 0. W1 + W2 contains both W1
and W2 : Every vector in W1 + W2 has the form x + y with x ∈ W1 , y ∈ W2 . Set y = 0
to obtain all vectors in W1 and x = 0 to obtain all vectors in W2 . That is, any vector
x ∈ W1 or y ∈ W2 is also present in W1 + W2 .
(b) A subspace W of V that contains both W1 and W2 must also contain all vectors of
the form x + y with x ∈ W1 , y ∈ W2 , since it is closed under addition. Therefore it
contains W1 + W2 .
(c) (⇒) W1 + W2 ⊆ span{u1 , · · · , up , v1 , · · · , vq }: For any u ∈ W1 = span{u1 , · · · , up }
and v ∈ W2 = span{v1 , · · · , vq }, there exist c1 , · · · , cp and d1 , · · · , dq such that
u = c1 u1 + · · · + cp up , v = d1 v1 + · · · + dq vq .
Then
u + v = c1 u1 + · · · + cp up + d1 v1 + · · · + dq vq ∈ span{u1 , · · · , up , v1 , · · · , vq }
(⇐) span{u1 , · · · , up , v1 , · · · , vq } ⊆ W1 +W2 : For any w ∈ span{u1 , · · · , up , v1 , · · · , vq },
there exist c1 , · · · , cp and d1 , · · · , dq such that
w = (c1 u1 + · · · + cp up ) + (d1 v1 + · · · + dq vq ) ∈ W1 + W2

Problem 6. (BONUS PROBLEM)

(a) (1) T (V0 ) contains 0W , since 0V ∈ V0 and T (0V ) = 0W . (2) Let u1 , u2 ∈ T (V0 ), then
there exist v1 , v2 ∈ V0 such that T (v1 ) = u1 and T (v2 ) = u2 . Then v1 + v2 ∈ V0 , and
T (V0 ) 3 T (v1 + v2 ) = T (v1 ) + T (v2 ) = u1 + u2 . (3) Similarly for scalar multiplication,
let u ∈ T (V0 ), then there exists v ∈ V0 such that T (v) = u. Then av ∈ V0 , and
T (V0 ) 3 T (av) = aT (v) = au. Combining (1)-(3) shows that T (V0 ) is a subspace of
W.
(a’) Let β = {u1 , · · · , un } be a basis for V0 . T being linear, we have
T (V0 ) = {T (v), ∀v ∈ V0 } = {T (a1 u1 + · · · + an un ), ∀a1 , · · · , an ∈ F }
= {a1 T (u1 ) + · · · + an T (un ), ∀a1 , · · · , an ∈ F } = span{T (u1 ), · · · , T (un )}
Then T (V0 ) is a subspace of W .
(b) Let β = {u1 , · · · , un } be a basis for V0 . T (β) is then a basis for T (V0 ) (from (a’)),
since it spans T (V0 ) and its vectors are linearly independent:
a1 T (u1 ) + · · · + an T (un ) = T (a1 u1 + · · · + an un ) = 0
gives a1 u1 + · · · + an un = 0 since T is an isomorphism, and a1 = · · · = an = 0 since β
is a basis for V0 . Thus, n = dim(V0 ) = dim(T (V0 )).

When you finish this exam, you should go back and reexamine your work for
any errors that you may have made.

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