Solutions to Practice Midterm 1
By ChuWee Lim
Author’s note: I’ve attempted to be as complete as possible in my solution, hence the wordiness.
In the midterm, you can probably miss one or two small steps.
(Question 1): True or False (with justification): If
A
and
B
are
n
by
n
matrices with entries
from
F
, then
AB
= 0 if and only
BA
= 0.
Solution.
False. Counterexample: if
A
= (
0 0
0 1
) and
B
= (
0 1
0 0
), then
AB
= (
0 0
0 0
) but
BA
=
(
0 1
0 0
).
(Question 2): True or False (with justification): If
A
and
B
are
n
by
n
matrices with entries
from
R
, then
AB
= 7
I
n
if and only
BA
= 7
I
n
.
Solution.
True. First, a general statement. We claim that a linear map
T
:
R
n
→
R
n
is an
isomorphism iff it is onetoone. Now,
⇒
is obvious. For the converse, suppose
T
is onetoone.
Then
n
= dim(
R
n
) = dim
N
(
T
) + dim
R
(
T
).
Since
T
is onetoone, dim
N
(
T
) = 0, and so
dim
R
(
T
) =
n
. Since
R
(
T
)
⊆
R
n
, this implies
R
(
T
) =
R
n
.
Now look at
L
A
and
L
B
as linear maps
R
n
→
R
n
. By symmetry, it suﬃces to prove =
⇒
.
Since
L
A
L
B
= 7
·
1
V
is onetoone,
L
A
is also onetoone. By the above paragraph,
L
A
is an
isomorphism and thus has an inverse
U
, i.e.
UL
A
=
L
A
U
= 1
V
. So,
UL
A
L
B
=
L
B
=
⇒
7
U
=
L
B
=
⇒
U
=
1
7
L
B
.
Thus,
1
7
L
B
is the inverse of
L
A
and
1
7
L
B
L
A
= 1
V
=
⇒
L
B
L
A
= 7
·
1
V
.
(Question 3): True or False (with justification): If
x, y
∈
V
and
a, b
∈
F
, then
ax
+
by
= 0 if
and only if
x
is a scalar multiple of
y
, or
y
is a scalar multiple of
x
.
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 Summer '08
 GUREVITCH
 Linear Algebra, Algebra, Vector Space, linear transformation

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