MATH 110  SOLUTIONS TO MIDTERM
LECTURE 1, SUMMER 2009
GSI: SANTIAGO CA
˜
NEZ
1.
Give two equivalent definitions (or characterizations) of each of the following:
(a) A list (
v
1
, . . . , v
n
) is linearly dependent
(b) An operator
T
∈ L
(
V
) can be represented by an uppertriangular matrix
(c) A scalar
λ
∈
F
is an eigenvalue of an operator
T
∈ L
(
V
)
(d) A vector space
V
is the direct sum of subspaces
U
and
W
Solution.
Here are some possible answers.
(a) One of the
v
i
can be written as a linear combination of the rest; or, there exist scalars
a
1
, . . . , a
n
∈
F
, not all zero, such that
a
1
v
1
+
· · ·
+
a
n
v
n
= 0
(b) There exists a basis (
v
1
, . . . , v
n
) of
V
with respect to which
M
(
T
) is uppertriangular;
or, there exists a basis (
v
1
, . . . , v
n
) of
V
such that for each
k
, span(
v
1
, . . . , v
k
) is
T
invariant
(c) There exists a nonzero
v
∈
V
such that
Tv
=
λv
; or, the operator
T

λI
is not injective
(d)
V
=
U
+
W
and
U
∩
W
=
{
0
}
; or, for each
v
∈
V
, there is a unique way to write
v
as
v
=
u
+
w
with
u
∈
U
and
w
∈
W
2.
Determine whether the following statements are true or false. You do not need to provide
justifications for your answers. There will be 1 point subtracted for each incorrect response,
so do not guess just for the sake of guessing.
(a) Any operator on a real vector space has a 1dimensional invariant subspace.
(b) If
V
=
U
⊕
W
and
p
(
z
) =
z
2
, then
p
(
P
U,W
) =
P
U,W
.
(c) If
v
1
, . . . , v
n
∈
V
, then span(
v
1
, . . . , v
n
) is
n
dimensional.
(d) Any operator on a 10dimensional complex vector space has a 4dimensional invariant
subspace.
Solution.
Justifications were not required, but we will provide some below.
(a) False.
This is the same as saying that any operator on a real vector space has an
eigenvector, which is not true; for example, take the rotation by 90
◦
operator on
R
2
.
(b) True. The projection operator
P
U,W
satisfies
P
2
U,W
=
P
U,W
.
(c) False. This is true if and only if (
v
1
, . . . , v
n
) is linearly independent; for example, take
each
v
i
to be zero — then their span is 0dimensional.
(d) True. This follows from the last homework; explicitly, let (
v
1
, . . . , v
10
) be a basis of
V
with respect to which the operator is uppertriangular — then span(
v
1
, . . . , v
4
) is 4dimensional
and invariant.
3.
Let
n
be a positive integer and let
U
be the subset of
P
n
(
R
) defined by
U
=
{
p
(
x
)
∈ P
n
(
R
)

p
00
(1)

p
0
(2) +
p
(3) = 0
}
.
(a) Prove that
U
is a subspace of
P
n
(
R
).
(b) Find a linear map
T
:
P
n
(
R
)
→
R
such that null
T
=
U
, and use this to compute dim
U
.
(You do not need to prove that the map you write down is in fact linear)
Date
: July 16, 2009.
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GSI: SANTIAGO CA
˜
NEZ
Solution.
(a) First, the zero polynomial clearly satisfies the defining properties of
U
, so 0
∈
U
.
Now, let
p
(
x
)
, q
(
x
)
∈
U
. Then
p
00
(1)

p
0
(2) +
p
(3) = 0 and
q
00
(1)

q
0
(2) +
q
(3) = 0
,
so
(
p
+
q
)
00
(1)

(
p
+
q
)
0
(2)+(
p
+
q
)(3) = (
p
00
(1)

p
0
(2)+
p
(3))+(
q
00
(1)

q
0
(2)+
q
(3)) = 0+0 = 0
.
Thus (
p
+
q
)(
x
)
∈
U
so
U
is closed under addition.
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 Math, Linear Algebra, Vector Space, 0w, real vector space, 0 W

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