Serge Ballif
MATH 535 Homework 8
October 26, 2007
The goal of this homework set is to establish the, perhaps surprising, analogue for finite dimensional
vector spaces for the problems of last weeks homework. For this set, we fix a linear transformation
L
on
n
given by a square matrix
A
.
This produces what we shall call an
action
of
[
X
] on
V
: (
P
(
x
)
, v
)
7→
P
(
A
)
v
:=
∑
d
i
=0
a
i
A
i
v
, when
P
(
X
) =
∑
d
i
=0
a
i
X
i
.
Notice that it is easy to verify that
•
1
v
=
v
, for al
v
∈
V
.
•
P
(
X
)(
v
1
+
v
2
) =
P
(
X
)
v
1
+
P
(
X
)
v
2
, for all
P
(
X
)
∈
[
X
],
v
1
, v

2
∈
V
.
•
(
P
(
X
) +
Q
(
X
))
v
=
P
(
X
)
v
+
Q
(
X
)
v
, for all
P
(
X
)
, Q
(
X
)
∈
[
X
],
v
∈
V
.
•
(
P
(
X
)
Q
(
X
))
v
=
P
(
X
)(
Q
(
X
)
v
), for all
P
(
X
)
, Q
(
X
)
∈
F
[
X
],
v
∈
V
.
This is summarized by saying that
V
is a module over
F
[
X
]. We say that a nonzero
v
∈
V
has order
P
(
X
)
if
P
(
X
) is the monic polynomial in
[
X
] of least degree such that
P
(
X
)
v
= 0.
1. A subspace
U
of
V
is said to be
cyclic
if, for some
u
∈
U
,
U
=
u
+
Au
+
· · ·
+
A
k
u
. Show that the subspace
U
of
V
is cyclic if and only if it contains an element
u
whose order has degree equal to the dimension of
U
. Such an
element is called a cyclic vector for
U
.
(
⇒
) Suppose that
U
is cyclic with minimal polynomial
M
(
x
).
Then for some
u
∈
U
,
U
=
u
+
Au
+
· · ·
+
A
d
u
. In particular,
0 =
a
0
u
+
a
1
Au
+
· · ·
+
a
k

1
A
d

1
u
+
A
d
u
for suitable choice of
a
i
.
Define
Q
(
x
) =
∑
d
i
=0
a
i
X
i
to be the polynomial that
gives the above action on
u
. Then
Q
(
x
) is a monic polynomial of degree dim
U
such that
Q
(
A
)
u
= 0. We claim that
Q
(
x
) is the order of
u
. This is clear because
the linear independence of the elements
u, Au, . . . , A
d
u
guarantees that there is
no polynomial
P
(
x
) of degree less than
d
for which
P
(
x
)
u
= 0.
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 Fall '08
 BROWNAWELL,WOODRO
 Math, Linear Algebra, Algebra, Vector Space, minimal polynomial

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