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Exercise 7.2.3

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1.

It’s upper triangular. So the characteristic polynomial is

{(2 - t)}^{5} {(3 - t)}^{2}

2.

Do the inverse of what we did in Exercise 7.2.2. For

λ = 2

, the dot diagram is

\begin{matrix} ∙ & ∙ \\ ∙ & ∙ \\ ∙ \end{matrix} .

For $λ = 3$ , the dot diagram is

\begin{matrix} ∙ & ∙ \end{matrix} .

3.

The eigenvalue

λ

with it corresponding blocks are diagonal is has the property

E_{λ} = K_{λ}

4.

The integer

p_{i}

is the length of the longest cycle in

K_{λ_{i}}

. So

p_{2} = 3

and

p_{3} = 1

5.

By Exercise 7.1.9(b), the matrix representations of

U_{2}

and

U_{3}

are

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

and

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

So we have $rank (U_{2}) = 3$ , $rank (U_{2}^{2}) = 1$ and $rank (U_{3}) = rank (U_{3}^{2}) = 0$ . By Dimension Theorem, we have $null (U_{2}) = 2$ , $null (U_{2}^{2}) = 4$ and $null (U_{3}) = null (U_{3}^{2}) = 2$ .

jephianlin

2011-06-27 00:00

Exercise 7.2.3

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