Midterm: Abstract Linear Algebra

April 24, 2025

Problem 1. Determine whether the following statements are True or False: (No need to
verify your answer, just write T/F)
(1) The set of degree n polynomials over R form a vector space (under natural addition
and scaler).
(2) Every finite dimensional, non-zero vector space over R has more than one basis.
(3) A vector space over F of dimension n ≥ 1 is isomorphic to Fn .
(4) AB = BA if both A and B are square matrices.
(5) A linear equation always has a solution if the number of equations is less than the
number of variables.
(6) Let A, B be two matrices, then (AB)T = B T AT .
(7) If A is left invertible and A is a square, then A is invertible.
(8) Let A be left invertible and B be right invertible, then AB is invertible.

Solution:
(1) F: 0 is not in the set. I wrote degree n instead of degree ≤ n.
(2) T: There is a basis since finite generating set contains a basis, a rescaling gives a
different basis.
(3) T: Find a basis B, the map [ · ]B is an isomorphism to Fn .
(4) F: many counter-examples.
(5) F: many counter-examples.
(6) T: write down the matrix multiplication to prove this, you may easily see (AB)T ̸=
T T
A B since the rows/columns may not match.
(7) T: A is left invertible so there’s a pivot in each row of the reduced echelon form, but
since A is a square, this means the echelon form is In , so A is invertible.
(8) F: e.g. take A = In , B right invertible but not a square.

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Problem 2. Write down the matrix A acting on R2 that turns a square with sides (1, 0)T , (0, 1)T
into a parallelogram with sides (1, 0)T , (1, 1)T . Compute A2 and A4 .

Solution:
     
1 1 2 1 2 4 1 4
A= ,A = ,A =
0 1 0 1 0 1

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  
1 0 1
Problem 3. Compute the inversion of A = 0 2 5 and find all a, b ∈ R such that the
3 1 8
   
1 0 1   1
0 x1
2 5   2
 
equation 
3 x2 =   has a solution.
1 8 3
x3
−1 −2 a b
Solution:
Do rowreduction, we get 
11/5 1/5 −2/5
A−1 =  3 1 −1 
−6/5 −1/5 2/5
Thus A⃗x = (1, 2, 3)T if and only if ⃗x = A−1 (1, 2, 3)T = (7/5, 2, −2/5). So the equation
has a solution if and only if −7/5 − 4 − 2/5a = b, or equivalently 2a + 5b + 27 = 0.

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Problem 4. (9 pts)
Let A ∈ Mn×n (F) and C ∈Mm×m(F) be invertible matrices, let B ∈ Mn×m (F). Find the
A B
inverse of the blocked matrix .
0 C
Solution:      
A B E F In 0
Suppose · = Im+n = , then we have
0 C G H 0 Im

AE + BG = In ,

CG = 0,
AF + BH = 0,
CH = Im .
This implies
G = C −1 0 = 0,
AE = In ,
H = C −1 ,
AF + BH = 0.
So we get
G = 0,
E = A−1 , H = C −1 ,
F = −A−1 BH = −A−1 BC −1 .
Plugging back, we may check that
  −1
−A−1 BC −1
 −1
−A−1 BC −1
    
A B A A A B
· = Im+n = · ,
0 C 0 C −1 0 C −1 0 C

so this is indeed the inverse matrix.

Remark: I made a typo and wrote B ∈ Mm×n (F). If you spotted it and wrote in the
exam that this is not a blocked matrix, you will also get full credit.
Lots of you tried to use row reduction to find the inverse, which is also correct and doable,
but you have to be very careful about which side you’re multiplying: row reductions always
act on the left! So row reduction should be like
I A−1 B A−1 0 I A−1 B − A−1 B · I A−1 0
     
A B I 0
→ →
0 C 0 I 0 I 0 C −1 0 I 0 C −1

I 0 A−1 −A−1 BC −1
 
=
0 I 0 C −1
To avoid mistakes when doing row reduction, always check your result by multiplying it
to the original matrix!

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Problem 5. Let V be a vector space. Suppose W1 and W2 are subspaces of V , such that
W1 ∩ W2 = {⃗0}. Show that dim(W1 + W2 ) = dim(W1 ) + dim(W2 ).
Here W1 + W2 = {⃗v + w⃗ : ⃗v ∈ W1 , w
⃗ ∈ W2 }. This is (and you do not need to prove) a
subspace of V .

Solution:
If dim W1 or dim W2 are 0 or +∞ this is automatically true.
Let A = {⃗v1 , ..., ⃗vn } be a basis of W1 and B = {w ⃗ m } be a basis of W2 , then A ∪ B
⃗ 1 , ..., w
is a generating set of W1 + W2 since any v + w ∈ W1 + W2 is a linear Pn combination.
We claim that A ∪ B is also linearly independent. Suppose i=1 ai⃗vi + m
P
j=1 bj w
⃗ j = 0,
then we have n m
X X
ai⃗vi = − bj w ⃗j
i=1 j=1

where LHS is a vector in W1 and RHS is a vector in W2 , therefore since W1 ∩ W2 = ⃗0, both
sides must be ⃗0. Since A and B are basis in W1 and W2 , this implies ai = 0, bj = 0, ∀i, j.
This shows that A ∪ B is also linearly independent.
We’ve shown A ∪ B is a generating set of W1 + W2 and it is linearly independent, so it
is a basis.
So we have dim(W1 + W2 ) = |A ∪ B| = |A| + |B| = dim(W1 ) + dim(W2 )
Remark:
Be careful when proving linear independence: it’s not enough that the basis A and B are
⃗ j ∈ B is linearly independent from A and w
distinct; it is also not enough to say each w ⃗j ∈ B
is linearly independent other vectors in B, you need to show wj is linearly independent with
the union A ∪ (B − {wj }) which is not implied by being linearly independent with both.

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Problem 6. 1
(1) Let W be a subspace of V , show that dim W ≤ dim V .
(2) Let A ∈ Mm×n (R), B ∈ Mn×k (R), prove that

rank(AB) ≤ max(rank(A), rank(B)).

(Hint: one way to prove this is to use the rank theorem: rank(C) = rank(C T ).)

Solution:
(1) If dim V = ∞ or dim W = 0, this is automatically true.
Now let A be a basis in W then it is linearly independent in V , so we have dim W =
|A| ≤ dim V .
(2) We have Im(AB) ⊂ Im(A) since AB⃗x = A(B⃗x). Thus by (1) we have rank(AB) =
dim(Im(AB)) ≤ dim(Im(A)) = rank(A).
Applying the same reasoning to the conjugates, we also have rank(B T AT ) ≤ rank(B T ).
So the rank theorem shows rank(AB) = rank(B T AT ) ≤ rank(B T ) = rank(B).
So we get rank(AB) ≤ min(rank(A), rank(B)).
Remark:
I made a typo in the problem: putting max instead of min. So I gave full credit to anyone
who proved rank(AB) ≤ rank(A) or rank(AB) ≤ rank(B).
There are many alternative ways of proving this.
1. You may realize columns of AB are linear combination of columns of A, which shows
Im(AB) ⊂ Im(A). Also, row space of AB are linear combination of rows of B, so you get
Im((AB)T ) ⊂ Im(B T ).
2. You can also do row reduction. Let P A be echelon form of A, then it has m − rankA
many zero rows, so P AB must also have at least m − rankA zero rows. So the number of
pivots of AB is ≤ rankA, which shows rank(AB) ≤ rank(A).
3. You can apply the rank formula and analyze kernels, using ker(B) ⊂ ker(AB) and
ker(AT ) ⊂ ker(B T AT ) to get the inequality.
Analyzing m, n, k is not enough, as rank(A), rank(B) can be much smaller than those.