proof of cyclic vector theorem

proof of cyclic vector theorem

First, let’s assume

f

has a

cyclic vector

v

. Then

B

=

{

v

,

f

⁢

(

v

)

,

…

,

f

n

-

1

⁢

(

v

)

}

is a basis for

V

. Suppose

g

is a

linear transformation

which commutes with

f

. Consider the

coordinates

(

α

0

,

…

,

α

n

-

1

)

of

g

⁢

(

v

)

in B, that is

g

⁢

(

v

)

=

∑

i

=

0

n

-

1

α

i

⁢

f

i

⁢

(

v

)

.

Let

P

=

∑

i

=

0

n

-

1

α

i

⁢

X

i

∈

k

⁢

[

X

]

.

We show that

g

=

P

⁢

(

f

)

.
For

w

∈

V

, write

w

=

∑

j

=

0

n

-

1

β

j

⁢

f

j

⁢

(

v

)

,

then

g

⁢

(

w

)

=

∑

j

=

0

n

-

1

β

j

⁢

g

⁢

(

f

j

⁢

(

v

)

)

=

∑

j

=

0

n

-

1

β

j

⁢

f

j

⁢

(

g

⁢

(

v

)

)

=

∑

j

=

0

n

-

1

β

j

⁢

f

j

⁢

(

∑

i

=

0

n

-

1

α

i

⁢

f

i

⁢

(

v

)

)

=

∑

j

=

0

n

-

1

β

j

⁢

∑

i

=

0

n

-

1

α

i

⁢

f

j

+

i

⁢

(

v

)

=

∑

j

=

0

n

-

1

∑

i

=

0

n

-

1

β

j

⁢

α

i

⁢

f

j

+

i

⁢

(

v

)

=

∑

i

=

0

n

-

1

∑

j

=

0

n

-

1

β

j

⁢

α

i

⁢

f

j

+

i

⁢

(

v

)

=

∑

i

=

0

n

-

1

α

i

⁢

f

i

⁢

(

∑

j

=

0

n

-

1

β

j

⁢

f

j

⁢

(

v

)

)

=

∑

i

=

0

n

-

1

α

i

⁢

f

i

⁢

(

w

)

Now, to finish the proof, suppose

f

doesn’t have a cyclic vector (we want to see that there is a linear transformation

g

which commutes with

f

but is not a polynomial evaluated in

f

). As

f

doesn’t have a cyclic vector, then due to the

cyclic decomposition theorem

V

has a basis of the form

B

=

{

v

1

,

f

⁢

(

v

1

)

,

…

,

f

j

1

⁢

(

v

1

)

,

v

2

,

f

⁢

(

v

2

)

,

…

,

f

j

2

⁢

(

v

2

)

,

…

,

v

r

,

f

⁢

(

v

r

)

,

…

,

f

j

r

⁢

(

v

r

)

}

.

Let

g

be the linear transformation defined in

B

as follows:

g

⁢

(

f

k

⁢

(

v

1

)

)

=

{

0

for every

⁢

k

=

0

,

…

,

j

1

f

k

i

⁢

(

v

i

)

for every

⁢

i

=

2

,

…

,

r

⁢

and

⁢

k

i

=

0

,

…

,

j

i

.

The fact that

f

and

g

commute is a

consequence

of

g

being defined as zero on one

f

-

invariant subspace

and as the

identity

on its

complementary

f

-invariant subspace. Observe that it’s enough to see that

g

and

f

commute in the basis

B

(this fact is

trivial

). We see that, if

k

=

0

,

…

,

j

1

-

1

, then

(

g

f

)

(

f

k

(

v

1

)

)

=

g

(

f

k

+

1

(

v

1

)

)

=

0

and

(

f

g

)

(

f

k

(

v

1

)

)

=

f

(

g

(

f

k

(

v

1

)

)

=

f

(

0

)

=

0

.

If

k

=

j

1

, we know there are

λ

0

,

…

,

λ

j

1

such that

f

j

1

+

1

⁢

(

v

1

)

=

∑

k

=

0

j

1

λ

k

⁢

f

k

⁢

(

v

1

)

,

so

(

g

⁢

f

)

⁢

(

f

j

1

⁢

(

v

1

)

)

=

∑

k

=

0

j

1

λ

k

⁢

g

⁢

(

f

k

⁢

(

v

1

)

)

=

0

and

(

f

⁢

g

)

⁢

(

f

j

1

⁢

(

v

1

)

)

=

f

⁢

(

0

)

=

0

.

Now, let

i

=

2

,

…

,

r

and

k

i

=

0

,

…

,

j

i

-

1

, then

(

g

f

)

(

f

k

i

(

v

i

)

)

=

g

(

f

k

i

+

1

(

v

i

)

)

=

f

k

i

+

1

(

v

i

)

and

(

f

g

)

(

f

k

i

(

v

i

)

)

=

f

(

g

(

f

k

i

(

v

i

)

)

=

f

k

i

+

1

(

v

i

)

.

In the case

k

i

=

j

i

, we know there are

λ

0

,

i

,

…

,

λ

j

i

,

i

such that

f

j

i

+

1

⁢

(

v

i

)

=

∑

k

=

0

j

i

λ

k

,

i

⁢

f

k

⁢

(

v

i

)

then

(

g

⁢

f

)

⁢

(

f

j

i

⁢

(

v

i

)

)

=

g

⁢

(

f

j

i

+

1

⁢

(

v

i

)

)

=

∑

k

=

0

j

i

λ

k

,

i

⁢

g

⁢

(

f

k

⁢

(

v

i

)

)

=

∑

k

=

0

j

i

λ

k

,

i

⁢

f

k

⁢

(

v

i

)

=

f

j

i

+

1

⁢

(

v

i

)

,

and

(

f

g

)

(

f

j

i

(

v

i

)

)

=

f

(

g

(

f

j

i

(

v

i

)

)

=

f

(

f

j

i

(

v

i

)

)

=

f

j

i

+

1

(

v

i

)

.

This proves that

g

and

f

commute in

B

. Suppose now that

g

is a polynomial evaluated in

f

. So there is a

P

=

∑

k

=

0

h

c

k

⁢

X

k

∈

K

⁢

[

X

]

such that

g

=

P

⁢

(

f

)

. Then,

0

=

g

⁢

(

v

1

)

=

P

⁢

(

f

)

⁢

(

v

1

)

, and so the

annihilator polynomial

m

v

1

of

v

1

divides

P

. But then, as the

annihilator

m

v

2

of

v

2

divides

m

v

1

(see the cyclic decomposition theorem), we have that

m

v

2

divides

P

, and then

0

=

P

⁢

(

f

)

⁢

(

v

2

)

=

g

⁢

(

v

2

)

=

v

2

which is absurd because

v

2

is a vector of the basis

B

. This finishes the proof.

Title

proof of cyclic vector theorem

Canonical

name

ProofOfCyclicVectorTheorem

Date of creation

2013-03-22 14:14:42

Last modified on

2013-03-22 14:14:42

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

12

Author

CWoo (3771)

Entry type

Proof

Classification

msc 15A04

Generated on Fri Feb 9 18:22:05 2018 by

LaTeXML