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Exercise 7.4.4

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1.

We may write an element in

R (ϕ (T))

ϕ (T) (x)

for some

x

. Since

{(ϕ (T))}^{m} (v) = 0

for all

v

, we have

{(ϕ (T))}^{m - 1} (ϕ (T) (x)) = {(ϕ (T))}^{m} (x) = 0

2.

The matrix

(\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})

has minimal polynomial ${(t - 1)}^{2}$ . Compute $R (L_{A} - I) = span {(1, 0, 0)}$ . But $(0, 0, 1)$ is an element in $N (L_{A} - I)$ but not in $R (L_{A} - I)$ .

3.

We know that the minimal polynomial

p (t)

of the restriction of

T

divides

{(ϕ (t))}^{m}

by Exercise 7.3.10. Pick an element

x

such that

{(ϕ (T))}^{m - 1} (x) \neq 0

. Then we know that

y = ϕ (T) (x)

is an element in

R (ϕ (T))

and

{(ϕ (T))}^{m - 2} (y) \neq 0

. Hence

p (t)

must be

{(ϕ (t))}^{m - 1}

jephianlin

2011-06-27 00:00

Exercise 7.4.4

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