Solutions to Homework 11.
Math 110, Fall 2006.
Prob 5.4.1.
(a) False,
{
0
}
and the whole space are
T
invariant for any
T
. (b) True (Theorem 5.21). (c)
False: just take
w
to be a nonzero multiple of
v
, then the corresponding
T
cyclic subspaces are the same,
but the vectors are not. (d) False, span
{
v,Tv,T
2
v,.
..
}
is in general diﬀerent from span
{
Tv,T
2
v,T
3
v,.
..
}
.
(e) True (by the CayleyHamilton theorem). (f) True – it is the characteristic polynomial of its
companion
matrix
introduced on p. 316. (g) True (Theorem 5.25).
Prob 5.4.4.
If
TW
⊆
W
, then
T
2
W
=
T
(
TW
)
⊆
TW
⊆
W
, and analogously
T
k
W
⊆
W
for all powers
k
= 0
,
1
,
2
,...
. Also, since
W
is a subspace, we get the following chain of inclusions:
(
a
0
I
+
a
1
T
+
···
+
a
n
T
n
)
W
⊆
a
0
W
+
a
1
(
TW
) +
···
+
a
n
(
T
n
W
)
⊆
a
0
W
+
a
1
W
+
···
+
a
n
W
⊆
W
+
W
+
···
+
W
=
W.
Prob 5.4.5.
Suppose each
W
j
is a
T
invariant subspace of
V
. Take
x
∈ ∩
j
W
j
. Since
x
∈
W
j
for each
ﬁxed
j
, the
T
invariance of
W
j
implies
Tx
∈
W
j
. So,
Tx
is in all subspaces
W
j
, hence
Tx
∈ ∩
j
W
j
. Since
x
was arbitrary, this shows that
∩
j
W
j
is
T
invariant.
Prob 5.4.17.
First let us prove that
A
k
∈
span
{
I,A,A
2
,...,A
n

1
}
for all
k
∈
IN
.
Indeed, this holds for
k < n
trivially and for
k
=
n
since
A
satisﬁes its characteristic polynomial
A
n
+
a
n

1
A
n

1
+
···
+
a
0
I
= 0
,
(1)
and hence
A
n
∈
span
{
I,A,.
..,A
n

1
}
. Now argue by induction: assume that all the powers of
A
up to
A
n
+
m
are already shown to be in the span
{
I,A,.
..,A
n

1
}
. Multiply (1) by
A
m
+1
to obtain
A
n
+
m
+1
+
a
n

1
A
n
+
m
+
···
+
a
0
A
m
+1
= 0
.
This shows that
A
n
+
m
+1
is a linear combination of the preceding powers of
A
. Since those are already
proved to be in span
{
I,A,.
..,A
n

1
}
, so must be
A
n
+
m
+1
. This proves that
span
{
A
j
:
j
∈
ZZ
+
}
= span
{
I,A,.
..,A
n

1
}
.
Now, since the space span
{
I,A,.
..,A
n

1
}
is spanned by
n
matrices, its dimension is at most
n
, QED.
Prob 5.4.18.
(a)
Since
f
(
t
) = det(
A

tI
), by evaluating both sides at 0, we get det
A
=
f
(0) =
a
0
. So,
A
is invertible if
and only if det
A
6
= 0 if and only if
a
0
6
= 0.
(b)
If
a
0
6
= 0, then the characteristic equation can be rewritten as

1
a
0
(
(

1)
n
A
n

1
+
a
n

1
A
n

2
+
···
+
a
1
I
)
·
A
=
I.
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 Spring '06
 Holtz
 Math, Linear Algebra, Algebra, Characteristic polynomial, Jordan normal form, J R, PROB

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