Exercise 5.6.4

Answers

$Z$ has both eigenvalues of 1. If there exists $B$ such that $M = BZ B^{- 1}$ and $∥ M ∥ \leq 1$ , then we have $Z = B^{- 1} MB$ , we compute $Z^{n} = B^{- 1} M^{n} B$ , and $∥ Z^{n} ∥ = ∥ B^{- 1} M^{n} B ∥ \leq ∥ B^{- 1} ∥ ∥ M^{n} ∥ ∥ B ∥ \to 0$ because $∥ M ∥ \leq 1$ . However, by induction, we can see that $Z^{n} = [\begin{matrix} 1 & n \\ 0 & 1 \end{matrix}]$ , so $∥ Z^{n} ∥ \to \infty$ , This contradicts. So it’s impossible to find $B$ .
$Y$ on the other hand has $Y^{n} = I$ when $n$ is even and $Y^{n} = Y$ when $n$ is odd. So its norm is finite. If we can find some $M$ with $∥ M ∥ = 1$ , then $∥ Y^{n} ∥ = ∥ C^{- 1} M^{n} C ∥ \leq ∥ C^{- 1} ∥ ∥ C ∥$ , which is possible.

niuers

2020-03-20 00:00