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Exercise 7.2.2

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A coloumn in a dot diagram means a cycle. Each cycle has the corresponding eigenvalue as the diagonal entries and $1$ as the upper subdiagonal entries. So we have

J_{1} = (\begin{matrix} 2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \end{matrix}),

J_{2} = (\begin{matrix} 4 & 1 & 0 & 0 \\ 0 & 4 & 1 & 0 \\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 4 \end{matrix}),

and

J_{3} = (\begin{matrix} - 3 & 0 \\ 0 & - 3 \end{matrix}) .

And the Jordan canonical form $J$ of $T$ is

J = (\begin{matrix} J_{1} & O & O \\ O & J_{2} & O \\ O & O & J_{3} \end{matrix}),

where $O$ is the zero matrix with appropriate size.

jephianlin

2011-06-27 00:00

Exercise 7.2.2

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