Mathematics 110
Name:
GSI Name: (5pts)
Linear Algebra Midterm Exam, February 22, 2007
Write clearly, with complete sentences, explaining your work. You will be
graded on clarity, style, and brevity. If you add false statements to a correct
argument, you will lose points. Be sure to put your name and your GSI’s
name on every page.
1. Let V be a vector space over a field F .
(a) (5pts) What is the definition of a linear subspace of V ?
Answer: A linear subspace of V is a subset W of V containing
the zero vector and such that a1 w1 + a2 w2 belongs to W whenever
w1 and w2 belong to W and a1 and a2 belong to F .
(b) (10 pts) What is the definition of the span of a subset S of a vector
space V ? Prove that the span of S is the smallest linear subspace
of V containing S.
Answer: The span of S is the set of all vectors in V which can
be written as a linear combination of elements of S. Let W be
the span of S. First we claim that W is a linear subspace of V
containing S. By definition, (or convention), W contains the zero
P
vector. Suppose that w, w0 ∈ W and a, a0 ∈ F . Write w = ci si
P
and w0 = c0i s0i with si , s0i ∈ S and ci , c0i ∈ F . Then aw + a0 w0 =
P
aci si + a0 c0i s0i , which is a linear combination of vectors in S and
hence belongs to W . Finally, we note that S ⊆ W , because each
s ∈ S can be written as 1s, a linear combination of elements of
S. This proves the claim. To prove that W is the smallest linear
subspace of V containing S, we shall prove that W is contained in
very linear subspace W 0 of V containing S. Indeed, every w ∈ W
P
can be written as w = ci si with ci ∈ F and si ∈ S. Since each
si ∈ W 0 and W 0 is a linear subspace, if follows that w ∈ W 0 . This
proves the claim.
(c) (5 pts) State the replacement theorem.
Answer: Let S be a subset of V which spans V and let T be a
linearly independent subset of V containing exactly n elements.
There there is a subset S 0 of S containing exactly n elements such
that (S \ S 0 ) ∪ T spans V . In particular, S has at least n elements.
(d) (5pts) What is the definition of the dimension of a vector space?
Explain how the replacement theorem is used to show that this
definition make sense. What other ingredient is also used?
Answer: The dimension of a vector space V is the number of
elements in any basis for V . The replacement theorem is used
to show that this number is independent of the choice of basis.
Indeed, if B and B 0 are two bases of V , then B spans and B 0 is
linearly independent, so the cardinality of B 0 is less than or equal
to the cardinality of B. The same argument shows that the reverse
is also true, so the cardinalities are equal. Finally, we must also
use the fact that every vector space has at least one basis.
2. Let V and W be vector spaces over a field F .
(a) (5pts) What is the definition of a linear transformation from V to
W?
Answer: A linear transformation from V to W is a function
T from V to W such that T (a1 v1 + a2 v2 ) = a1 T (v1 ) + a2 T (v2 )
whenever a1 , a2 ∈ F and v1 , v2 ∈ V .
(b) (7 pts) If α := (v1 , v2 , . . . , vn ) is an ordered basis for V and β :=
(w1 , w2 , . . . wm ) is an ordered basis for W , what is the definition
of the matrix representation [T ]βα of a linear transformation from
V to W with respect to the bases α and β?
P
α
Answer: Write T (vj ) = m
i=1 aij wi . Then [T ]β is the mxn matrix
whose ijth entry is aij .
(c) (8 pts) Let V be the space of polynomials of degree at most 2 over
R and let α := (1, x, x2 ), an ordered basis for V . Let T : V → V be
the transformation sending p to p0 + 2p, where p0 is the derivative
of p. Find [T ]αα .
Answer: We compute T (1) = 2, T (x) = 1+2x, T (x2 ) = 2x+2x2 .
2
Thus


2 1 0


α
[T ]α =  0 2 2 
0 0 2
3. If V and W are vector spaces, let L(V, W ) denote the set of linear
transformations from V to W .
(a) (10 pts) Explain the definition of the sum S + T of two elements
S and T of L(V, W ), and in particular show why, with your definition, S + T ∈ L(V, W ).
Answer: We define S + T to be the function taking v ∈ V to
S(v) + T (v) ∈ W . To see that this is linear, we compute:
(S + T )(a1 v1 + a2 v2 ) =
=
=
=
S(a1 v1 + a2 v2 ) + T (a1 v1 + a2 v2 )
a1 S(v1 ) + a2 S(v2 ) + a1 T (v1 ) + a2 T (v2 )
a1 S(v1 ) + a1 T (v1 ) + a2 S(v2 ) + a2 T (v2 )
a1 (S + T )(v1 ) + a2 (S + T )(v2 ).
(b) (15 pts) Let V = W be the set of infinitely differentiable functions from R to R. Explain why the D: V → V sending f to its
derivative belongs to L(V, V ). Prove that (D, D2 , D3 ) is a linearly
independent sequence in L(V, V ).
Answer: The fact that D is linear is just a theorem in calculus:
D(a1 f1 + a2 f2 ) = a1 D(f1 ) + a2 D(f2 ). To prove that (D, D2 , D3 )
is linearly independent, suppose that a1 D + a2 D2 + a3 D3 = 0.
Evaluation at x, we see that a1 = 0. Evaluation at x2 , we see that
2a2 = 0, and hence that a2 = 0. Evaluation at x3 , we see that
3a3 = 0, hence a3 = 0.
4. Say whether each of the following is true or false. In each case, explain
your answer. In particular, if you claim that statement is true, give a
brief proof, and if you claim it is false, give a counterexample.
(a) (5 pts) No linear transformation from R2 to the space P2 of real
polynomials of degree less than or equal to 2 is surjective.
Answer: This is true, because R2 has dimension 2 but P2 has
dimension 3.
3
(b) (5 pts) No linear transformation from P2 to R2 is injective.
Answer: This is true. The first space has dimension 3 and the
second space has dimension 2, so by the rank nullity theorem, the
kernel must have dimension at least one.
(c) (5 pts) If A and B are square matrices and AB is the identity
matrix, then BA is also the identity matrix.
Answer: This is true. If AB is the identity matrix, then LA ◦ LB
is the identity transformation, and hence LB is injective. By the
rank-nullity theorem, LB is also surjective, hence bijective. Hence
B is invertible: there is a matrix B 0 such that BB 0 = B 0 B = I.
Then AB = I so ABB 0 = B 0 so A = B 0 so BA = I.
(d) (5 pts) If A and B are matrices and AB is the identity matrix,
then LA is injective.
Answer:This
is false. For example, let A := ( 1 0 ) ∈ M12 and
1
let B :=
∈ M2,1 . Then AB = ( 1 ) ∈ M1,1 , but LA : F 2 → F
0
0
sends
to zero, and hence is not injective.
1
(e) (5pts) If A and B are matrices and AB is the identity matrix,
then LA is surjective.
Answer: This is true, since LA ◦ LB = LAB = id.
4