Exercise 7.2.8

Answers

1.

Let

β

be the set

{v_{i}}_{i = 1}^{m}

. Then we have the similar fact

T (c v_{1}) = λc v_{1}

and

T (c v_{i}) = λc v_{i} + c v_{i - 1} .

So the matrix representation does not change and the new ordered basis is again a Jordan canonical basis for $T$ .

2.

Since

(T - λI) (y) = 0

, the vector

T (x + y) = T (x)

does not change. Hence

γ^{'}

is a cycle. And the new basis obtained from

β

by replacing

γ

γ^{'}

is again a union of disjoint cycles. So it i sa Jordan canonical basis for

T

3.

Let

x = (- 1, - 1, - 1, - 1)

and

y = (0, 1, 2, 0)

. Apply the previous argument and get a new Jordan canonical basis

{(\begin{matrix} - 1 \\ 0 \\ 1 \\ - 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 2 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix})} .

jephianlin

2011-06-27 00:00