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Exercise 5.3.5

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First we see three special matrices,

P = (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}), Q = (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}), R = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) .

The limit of power of these three matrices is themselves respectly, because they has the property that their square are themselves. However, we know that

\frac{1}{\sqrt{2}} P \cdot \frac{1}{\sqrt{2}} P = R .

So we pick $A = \frac{1}{\sqrt{2}} P$ , $B = \frac{1}{\sqrt{2}} Q$ , and $C = R$ . Thus we have the limit of power of $A$ and $B$ are the zero matrix. And the limit of power of $C$ is $C$ itself.

jephianlin

2011-06-27 00:00

Exercise 5.3.5

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