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Exercise 7.3.16

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1.

Let

f (t)

be the characteristic polynomial of

T

. Then we have

f (T) (x) = T_{0} (x) = 0 \in W_{1}

. So there must be some monic polynomial

p (t)

of least positive degree for which

p (T) (x) \in W_{1}

. If

h (t)

is a polynomial for which

h (T) (x) \in W_{1}

, we have

h (t) = p (t) q (t) + r (t)

for some polynomial $q (t)$ and $r (t)$ such that the degree of $r (t)$ is less than the degree of $p (t)$ by Division Algorithm. This means that

r (T) (x) = h (T) (x) - p (T) p (T) (x) \in W_{1}

since $W_{1}$ is $T$ -invariant. Hence the degree of $r (t)$ must be $0$ . So $p (t)$ divides $h (t)$ . Thus $g_{1} (t) = p (t)$ is the unique monic polynomial of least positive degree such that $g_{1} (T) (x) \in W_{1}$ .

2.

This has been proven in the previous argument.

3.

Let

p (t)

and

f (t)

be the minimal and characteristic polynomials of

T

. Then we have

p (T) (x) = f (T) (x) = 0 \in W_{1}

. By the previous question, we get the desired conclusion.

4.

Observe that

g_{2} (T) (x) \in W_{2} \subset W_{1}

. So

g_{1} (t)

divides

g_{2} (t)

by the previous arguments.

jephianlin

2011-06-27 00:00

Exercise 7.3.16

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