47.
Hard. Basically, follow the hint, and be very careful to think clearly.
*62.
I guess the most direct way to see this is to let
v
1
, v
2
, . . ., v
m
be the columns
of
A
and
w
1
, w
2
, . . ., w
m
be the columns of
B
. Then
Im
(
A
) =
Span
(
v
1
, v
2
, . . ., v
m
)
,
Im
(
B
) =
Span
(
w
1
, w
2
, . . ., w
m
)
,
Im
(
A
+
B
) =
Span
(
v
1
+
w
1
, v
2
+
w
2
, . . ., v
m
+
w
m
)
.
Now notice that
Span
(
v
1
+
w
1
, v
2
+
w
2
, . . ., v
m
+
w
m
)
⊂
Span
(
v
1
, v
2
, . . ., v
m
, w
1
, w
2
, . . ., w
m
)
.
The dimension of the last span is no bigger than
dim
(
Im
(
A
)) +
dim
(
Im
(
B
))
, and
the rank of
A
+
B
is the dimension of the first span. Therefore, we get
rank
(
A
+
B
)
≤
rank
(
A
) +
rank
(
B
)
.
(We could use problem 47 to get a better estimate.)
*63.
For any
v
,
ABv
=
A
(
Bv
)
, so
Im
(
AB
)
⊂
Im
(
A
)
, so
rank
(
AB
)
≤
rank
(
A
)
.
On the other hand, the image of
AB
is the image of
Im
(
B
)
under the transforma
tion given by
A
. Since the image of a transformation has dimension less than or
equal to the dimension of the domain, we get, once again,
rank
(
AB
)
≤
rank
B
.
*64.
What we need to do to show two sets are different is to find an element of
one that is not an element of the other. Looking at matrix
A
we see that
4
times
the first column plus
5
times the second plus
6
times the fourth gives the fifth. This
tells us that the vector
x
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
4
5
0
6

1
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
is in the kernel of
A
. But
Bx
=
⎡
⎢
⎢
⎣
0
0

1
0
⎤
⎥
⎥
⎦
=
0,
so
x
is not in the kernel of
B
. That shows the kernels are different.