1. (10 points)
Fix positive integers m,n and consider the vector space V of all m×n matrices with entries
in the real numbers R.
(a) Find the dimension of V and prove your answer. Please carry out all the steps of your
proof.
(b) Let P be the subset of V consisting of m×n matrices each of whose row sum is 1. Prove
or disprove: P is a subspace of V.
(c) Assume m ≥ 2 and n ≥ 2. Find a subspace of V of dimension 2. Please explain your
answer, but you don’t have to give a proof.
Answer:
(a) V has dimension mn. To prove this statement, it is enough to exhibit a basis β of V
with mn elements. For 1 ≤ i ≤ m and 1 ≤ j ≤ n, let eij denote the matrix in V with a 1 in
the (i, j) position and zeros elsewhere. Note that there are mn such elements eij, and let β
be the set of these mn matrices. If β forms a basis of V, then V has dimension mn and we
are done.
First we show that the span of β is V. Indeed, if A ∈ V, then
A =
m∑
i=1
n∑
j=1
Aijeij
so β spans V.
Next we show that β is a linearly independent set, which would finish the proof that β is a
basis for V and hence the dimension of V is mn. Suppose that 0 = A =
∑m
i=1
∑n
j=1 aijeij for
scalars aij. By the definition of eij, we conclude that aij = Aij. But a matrix A = 0 if and
only if all of its entries Aij = aij are equal to 0. This shows that β is a linearly independent
set, and hence the dimension of V is mn.
1
(b) P cannot be a subspace of V. Indeed, let A ∈ P. By the definition of P, each of the
rows of A sums to 1. Now consider B = 2A, and note that each of the rows of B sums to 2.
Therefore, B 6∈ P, and so P is not closed under scalar multiplication. Therefore, P is not a
vector space, and hence not a subspace of V.
(c) It suffices to find a linearly independent subset of β V with 2 nonzero elements, since
then the span of β would be a two dimensional subspace of V. There are many such choices.
For example, using the notation from part (a) above, we could choose β = {e11, e12}.
2. (10 points)
P2(R) is the real vector space of real polynomials of degree at most 2. Let W be the following
subset of P2(R):
W = {f ∈ P2(R) | f(2) = f(1)}.
(a) Prove that W is a vector subspace of P2(R).
(b) Write down a basis for W . You do not need to prove that the set given is a basis, though
justification of how you found it must be given.
(c) W is isomorphic to Rd for what value of d Justify your answer.
Answer:
(a) W is a subspace because it satisfies the three standard properties.
1. The zero polynomial O(x) belongs in W : O(2) = 0 = O(1).
2. W is closed under addition: let f and g be two polynomials in W , their sum (f + g)
also belongs to W because
(f + g)(2) = f(2) + g(2) = f(1) + g(1) = (f + g)(1).
3. W is closed under scalar multiplication: let f be a polynomial in W and λ a real
number, their scalar product (λf) also belongs to W because
(λf)(2) = λf(2) = λf(1) = (λf)(1).
(b) Let f(t) = at2 + bt+ c ∈ W . Then
4a+ 2b+ c = f(2) = f(1) = a+ b+ c.
2
Therefore, 3a+ b = 0.
This means that c is a free variable and a, b are related by b = 3a. Therefore,
f(t) = a(t2 3t) + c
and a basis for W is {t2 3t, 1}.
(c) The dimension of W is 2 and so W is isomorphic to R2 because all 2-dimensional real
vector spaces are isomorphic to R2.
3. (10 points)
Let V denote the linear span of the following functions from R to R: e 2x, 1, e2x. Also
suppose that these functions form an ordered basis β for V. Let T : V → V be the linear
transformation defined by (Tf)(x) = f( x), and let D : V→ V be the linear transformation
defined by (Df)(x) = df(x)
dx
.
For the following questions, you must show your calculations, but you need not give a proof.
(a) Find the matrix [T ]β.
(b) Find the matrix [D]β.
(c) Find the matrix [TD]β.
Answer:
(a) We find that T maps e 2x, 1, e2x to e2x, 1, e 2x respectively. Therefore
[T ]β =
0 0 10 1 0
1 0 0
.
(b) We find that D maps e 2x, 1, e2x to 2e 2x, 0, 2e2x respectively. Therefore
[D]β =
2 0 00 0 0
0 0 2
.
(c) Using parts (a) and (b), we find
[TD]β = [T ]β[D]β =
0 0 10 1 0
1 0 0
