Part (b) is similar, except that the containments are reversed: if
~
y
∈
Im
(
A
)
,then
~
y
=
A
~
x
for some
~
x
;if
~
y
∈
Im
(
A
2
)
~
y
=
A
2
~
w
=
A
(
A
~
w
)=
A
~
x
for
~
x
=
A
~
w
.
So
Im
(
A
2
)
⊂
Im
(
A
)
. In general,
Im
(
A
)
⊃
Im
(
A
2
)
⊃
Im
(
A
3
)
⊃
...
and once again equality is possible but not required.
*48.
This is basically the same as one of the problems in the midterm. Part (a)
wasexact
lythesameasoneo
ftheprob
lemsinthem
idterm
:i
f
~
w
∈
Im
(
A
)
A
~
w
=
~
w
. See the midterm solutions for details.
For part (b), note ±rst that the only matrix with rank zero is the zero matrix.
If, on the other hand,
A
has rank two, it is invertible (since it’s a
2
×
2
matrix).
Multiplying
A
2
=
A
by
A

1
on both sides gives
A
=
I
.Sotheon
lyranktwomatr
ix
that satis±es
A
2
=
A
is the identity. (Another way to see this: if
A
has rank two,
then the image of
A
is all of
R
2
. By part (a), it follows that
A
~
w
=
~
w
for every
~
w
,
so
A
is the identity.)
Finally, suppose
A
has rank one. Then both the kernel and the image of
A
are
spanned by one vector, i.e., they are lines in
R
2
.G
i
v
e
n
~
x
∈
R
2
,le
t
~
w
=
A
~
x
,
~
v
=
~
x

A
~
x
.Then
~
w
∈
Im
(
A
)
,
~
v
∈
ker
(
A
)
,
~
x
=
~
w
+
~
v
,and
A
~
x
=
~
w
. This is what
problem 2.2.33 de±nes as “the projection on the image along the kernel.”
b. Section
3
.
2
, problems
1
,*
2
,
3
4
6
,
8
,
10
,
12
14
16
20
32
34
,
*
36
37
.
1.
Not a subspace, since
⎡
⎣
0
0
0
⎤
⎦
is not in
W
.
*2.
Not a subspace. The easiest way to see this is to note that if
x<y<z
then
⎡
⎣
x
y
z
⎤
⎦
∈
W
but
(
1
)
⎡
⎣
x
y
z
⎤
⎦
/
∈
W
. (Multiplying by

1
reverses inequalities!)
3.
In this case,
W
=
Span
⎛
⎝
⎡
⎣
1
4
7
⎤
⎦
,
⎡
⎣
2
5
8
⎤
⎦
,
⎡
⎣
3
6
9
⎤
⎦
⎞
⎠
,
so it is a subspace.
*4.
Yes, as we checked in class. If you don’t see how to do this, come ask.
*6.
The intersection is always a subspace, but the union hardly ever is.
Few men speak humbly of humility, chastely of chastity, skeptically of skepticism. – Blaise Pascal