MATB24 – Midterm –Fall 2019
Professors: Camelia Karimianpour, Christopher Kennedy
Solutions

MATB24 Midterm , Page 1 of 11
1. Finish each of following sentences by giving complete, precise definitions for, or precise
mathematical characterizations of, the italicized terms.
(a) (3 points) A subspace W of a real vector space V .
Solution:
W is a subspace of real vector space V i↵ W ✓ V and rw1+w2 2 W, 8w1, w2 2
W, r 2 R.
(b) (3 points) A linear transformation T : V ! W , where V and W are real vector
spaces.
Solution:
T : V ! W is a linear transformation i↵ T (V ) ✓ W and T (rv1 + v2) =
rT (v1) + T (v2), 8v1, v2 2 V, r 2 R.
(c) (3 points) A linearly dependent subset S (finite or infinite) of a vector space V .
Solution:
S is a linearly dependent subset of V i↵ there exist {s1, . . . , sn} ✓ S and
a1, . . . , an 2 R such that
Pn
j=1 ajsj = 0 and aj 6= 0 for some j 2 {1, . . . , n}.
(d) (3 points) An isomorphism between two vector spaces V and W .
Solution:
T is an isomorphism between vector spaces V and W i↵ T : V ! W is an
invertible (one-to-one and onto) linear transformation.
MATB24 Midterm , Page 2 of 11
2. State whether each statement is true or false and provide a short justification for your
claim (a short proof is you think the statement is true or a counter example if you think
it is false).
(a) (3 points) The orthogonal complement of Span
8<:
2412
3
359=; ⇢ R3 is the plane x+2y+
3z = 0 in R3.
Solution:
True, h(1, 2, 3)T , (x, y, z)T i = (x+2y+3z) = 0, 8(x, y, z) 2 {x+2y+3z = 0}.
(b) (3 points) Let U and W be non-trivial subspaces of V . Then U [W is a subspace
of V .
Solution:
False, for let U = {(x, 0)|x 2 R} and V = {(0, y)|y 2 R} be non-trivial sub-
spaces of R2. Then U [W is not closed under addition since U [W 3 (1, 1) =
(1, 0) + (0, 1) 2 U +W .
(c) (3 points) If V is not finitely generated then every non-trivial subspace of V is also
not finitely generated.
Solution:
False, for let P0 = span{1} be a non-trivial subspace of P = 1n=0Pn =
span{1, x, x2, . . . }.
MATB24 Midterm , Page 3 of 11
(d) (3 points) The set {ex, ex + e2x, ex + e2x + e3x} is a basis for Span{ex, e2x, e3x}.
Solution:
True, c1ex+c2e2x+c3e3x = (c1c2c3)ex+(c2c3)(ex+e2x)+c3(ex+e2x+e3x).
(e) (3 points) The space of all upper triangular 3⇥ 3 matrices is isomorphic to P5.
Solution:
True, with isomorphism
0@a b c0 d e
0 0 f
1A 7! a+ bx+ cx2 + dx3 + ex4 + fx5.
(f) (3 points) Let V and W be vector spaces, and suppose that T : V ! W is an
injective linear transformation. If there are vectors ~v1,~v2, . . . ,~vk in V such that the
vectors T (~v1), T (~v2), . . . , T (~vk) span W , then the vectors ~v1,~v2, . . . ,~vk span V .
Solution:
True, 8v 2 V, T (v) = w =Pkj=1 rjT (vj) = T (Pkj=1 rjvj) =) v =Pkj=1 rjvj.
MATB24 Midterm , Page 4 of 11
3. In each part, give an explicit example of the mathematical object described or explain
why such object does not exist.
(a) (3 points) A 4-dimensional subspace of F , the vector space of all functions from R
to R.
Solution:
Take P3 = span{1, x, x2, x3} ⇢ F .
(b) (3 points) A surjective and non-injective map from a 3-dimensional vector space V
other than R3 to a 3-dimensional vector space W other than R3.
Solution:
False, by the Rank-Nullity Theorem. Let T : V ! W be a surjective linear
map. Then dim(kerT ) = dim(V ) dim(ImT ) = dim(V ) dim(W ) = 3 3 =
0 =) kerT = {0V }, whence T is injective. It is true, however, if T need not
be linear.
(c) (3 points) Two di↵erent bases for M2⇥2(R), the vector space of all 2 ⇥ 2 matrices
with entries in R.
Solution:
Take B1 =
⇢✓
1 0
0 0
◆
,
✓
0 1
0 0
◆
,
✓
0 0
1 0
◆
,
✓
0 0
0 1
◆
and
B2 =
⇢✓
1 0
0 1
◆
,
✓
0 1
0 0
◆
,
✓
0 0
1 0
◆
,
✓
1 0
0 1
◆
.
MATB24 Midterm , Page 5 of 11
(d) (3 points) An isomorphism between M2⇥2(R), the vector space of all 2⇥2 matrices
with entries in R, and P3, the vector space of all cubic polynomials with coecients
in R.
Solution:
Take isomorphism T :
✓
a b
c d
◆
7! a+ bx+ cx2 + dx3.
(e) (3 points) Two di↵erent bases and ↵ for R2 and a change of basis matrix C that
changes -coordinate of a vector ~v 2 R2 into ↵-coordinates of ~v.
Solution:
Take = {(1, 1), (1,1)} and ↵ = {(1, 0), (0, 1)}. Then C!↵ =
✓
1 1
1 1
◆
.
(f) (3 points) An isomorphism between R3 and V = {A 2 M2⇥2(R)| tr(A) = 0}, that
is the vector space of all 2⇥ 2 matrices with entries in R that have zero trace1.
Solution:
Take isomorphism T : (a, b, c) 7!
✓
a b
c a
◆
.
1trace of a matrix is the sum ot its diagonal entries
MATB24 Midterm , Page 6 of 11
4. Let Pn be the vector space of all polynomials with degree less than or equal to n. Consider
the map D : Pn ! Pn, given by D(p(x)) = p0(x).
(a) (4 points) Show that D is a linear transformation.
Solution:
Check D(p(x) + q(x)) = (p+ q)0(x) = p0(x) + q0(x) = D(p(x)) +D(q(x)).
(b) (4 points) Prove that {a1x+ a2x2 + · · ·+ an1xn1 | ai 2 R} is a subspace of Pn.
Solution:
Let p(x) 2 S := {a1x+ a2x2 + · · ·+ an1xn1 | ai 2 R}. Then p(x) = 0 + a1x+
a2x2+· · ·+an1xn1+0xn 2 Pn. Moreover, rp(x)+q(x) =
Pn1
j=1 (raj+bj)x
n1 2
S, 8p(x) :=Pn1j=1 ajxj, q(x) :=Pn1j=1 bjxj 2 S, r 2 R, whence S is a subspace.
(c) (5 points) Suppose n = 2. Let A = {1, x, x2} and B = {1 + x+ x2, 1 x+ x2, 1 +
x x2} be bases for P2 (you don’t need to prove this). Suppose B is a matrix such
that B[p(x)]B = [p0(x)]B. Find B.
Solution:
Observe that 1 + 2x = (1 + x+ x2) 12(1 x+ x2) + 12(1 + x x2), 1 + 2x =
(1 + x + x2) 32(1 x + x2) 12(1 + x x2) and 1 2x = 1(1 + x + x2) +
3
2(1 x+ x2) + 12(1 + x x2) and conclude
B = [D]B =

[D(1 + x+ x2)]B | [D(1 x+ x2)]B | [D(1 + x x2)]B

=

[1 + 2x]B | [1 + 2x]B | [1 2x]B

=
0@ 1 1 11
2
3
2
3
2
1
2
1
2
1
2
1A .
MATB24 Midterm , Page 7 of 11
(d) (6 points) Let C be a matrix such that (C1BC)[p(x)]A = [p0(x)]A. Find C1.
Solution:
Recall that [D]A[p(x)]A = [p0(x)]A =) C1BC = [D]A =) C = CA!B and
conclude
C1 = CB!A =

[1 + x+ x2]A | [1 x+ x2]A | [1 + x x2]A

=
0@1 1 11 1 1
1 1 1
1A .
(e) (6 points) Suppose p(x) = a + bx + cx2 2 P2 is such that [p0(x)]B =
2423
5
35. Find b
and c.
Solution:
Calculate
0@ b2c
0
1A = [p0(x)]A = CB!A[p0(x)]B
=
0@1 1 11 1 1
1 1 1
1A0@23
5
1A
=
0@104
0
1A ,
whence (b, c) = (10, 2). Verify that 10 + 4x = 2(1 + x + x2) + 3(1 x + x2) +
5(1 + x x2).
MATB24 Midterm , Page 8 of 11
5. Let U = {
✓
a b
0 c
◆
|a, b, c 2 R} be the vector space of all upper triangular 2⇥2 matrices.
Let
B =
✓
1 0
0 0

,
1 1
0 0

,

0 1
0 1
◆
be a basis of U .
Let T : U ! R2 be the linear transformation defined by
T
✓
a b
0 c
◆
=
✓
a
b+ 2c
◆
.
(a) (5 points) Find a basis for the kernel of T .
Solution:
Check that
✓
a b
0 c
◆
2 kerT () T
✓
a b
0 c
◆
:=
✓
a
b+ 2c
◆
=
✓
0
0
◆
()
(a, b, c) = (0,2, 1), 2 R. Conclude kerT = span{(0,2, 1)T}.
(b) (5 points) Is T surjective? Justify.
Solution:
Yes, fix
✓
x
y
◆
2 R2. Then
✓
x y
0 0
◆
2 U satisfies T
✓
x y
0 0
◆
=
✓
x
y
◆
. Or, show by
Rank-Nullity that dim(ImT ) = dim(U) dim(kerT ) = 3 1 = 2.
(c) (5 points) Show that Null([T ]B,E) is 1-dimensional.
Solution:
First calculate matrix [T ]B,E
[T ]B,E =

[T (b1)]E | [T (b2)]E | [T (b3)]E

=
✓
1 1 0
0 1 1
◆
and deduce that kerT = {(1, 1,1) | 2 R}.
MATB24 Midterm , Page 9 of 11
6. Let V andW be n andm-dimensional vector spaces respectively. Let k be a fixed integer
between 1 and n. Suppose T : V ! W is a linear transformation. Let B = {~b1, · · · ,~bn}
be a basis for V and {~vk, · · · ,~vn} be a subset of V . Let [T ]B,E be the matrix of T with
respect to B. Assume the first k columns of [T ]B,E are zero and all other columns are
linearly independent.
(a) (6 points) Carefully prove that ker(T ) = Span{~b1, · · · ,~bk}.
Solution:
Recall that [T ]B,E =

[T (b1)]E | · · · | [T (bn)]E

and T = [·]1E [T ]B,E [·]B.
Next calculate
T (bj) =

[·]1E [T ]B,E [·]B

(bj)
= [·]1E ([T ]B,E([bj]B))
= [·]1E ([T ]B,E(ij)1i)
= [T (bj)]
1
E .
Deduce, for 1  j  k, T (bj) = [T (bj)]1E = [(0, . . . , 0)T ]1E = 0W and thus
conclude that span{b1, . . . bk} ✓ kerT .
Next, let b =
Pn
j=1 rjbj 2 kerT and T (bj) = wj, 8j 2 {k + 1, . . . , n} and
calculate
T (b) = T

nX
j=1
rjbj
!
=
nX
j=1
rjT (bj)
=
kX
j=1
rj[T (bj)]
1
E +
nX
j=k+1
rj[T (bj)]
1
E
=
kX
j=1
0W +
nX
j=k+1
rj[T (bj)]
1
E
=
nX
j=k+1
rj[T (bj)]
1
E = 0W .
Then deduce
Pn
j=k+1 rj[T (bj)]E =
hPn
j=k+1 rj[T (bj)]
1
E
i
E
= [0W ]E = (0, . . . , 0)T ,
whence rj = 0, 8j 2 {k + 1, . . . , n} by linear independence and thus conclude
that kerT ✓ span{b1, . . . , bk} to complete the proof.
MATB24 Midterm , Page 10 of 11
(b) (6 points) Suppose [~vj]B = ~ej + ~ek for k  j  n, where ~ej’s are standard vectors
in Rn. Prove that Img(T ) = Span{T (~vk), · · · , T (~vn)}.
Solution:
First let w 2 span{T (vk), . . . , T (vn)} and verify that w =
Pn
j=k+1 rjT (vj) =
T
⇣Pn
j=k+1 rjvj
⌘
2 ImgT =) span{T (vk), . . . , T (vn)} ✓ ImgT .
Next let w 2 ImgT . Then find u =Pnj=1 rjbj such that T (u) = w. Observing
[vj]B = ej + ek =) vj = bj + bk,
T (u) = T

nX
j=1
rjbj
!
= T

k1X
j=1
rjbj +

rk
nX
j=k+1
rj
!
bk +
nX
j=k+1
rj(bj + bk)
!
=
k1X
j=1
rjT (bj) +
1
2

rk
nX
j=k+1
rj
!
T (2bk) +
nX
j=k+1
rjT (bj + bk)
= 0W +
1
2

rk
nX
j=k+1
rj
!
T (vk) +
nX
j=k+1
rjT (vj)
=
1
2

rk
nX
j=k+1
rj
!
T (vk) +
nX
j=k+1
rjT (vj).
Then deduce w = T (u) 2 span{T (vk), . . . , T (vn)} and thus conclude ImgT ✓
span{T (vk), . . . , T (vn)} to complete the proof. Note that T (vk) = T (2bk) =
2T (bk) = 0W and so ImgT = span{T (vk+1), . . . , T (vn)} as well.
MATB24 Midterm , Page 11 of 11
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