Exercise 5.6.3

For $A$ and $M$, we can surely find matrices $B,C$ such that $\left|\right|\mathit{BA}{B}^{-1}\left|\right|<1$ and $\left|\right|\mathit{CM}{C}^{-1}\left|\right|<1$, because the eigenvalues of $A$ and $M$ are below 1 in absolute value. And as presented on page 317, we can find $B,C$ to make the norms less than 1.

If $M$ is a Markov matrix, it must have an eigenvalue of 1, which makes it impossible to have norm less than 1.

We can choose $B=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 8\hfill \end{array}\right]$, so $\mathit{BA}{B}^{-1}=\left[\begin{array}{cc}\hfill \frac{1}{2}\hfill & \hfill \frac{1}{8}\hfill \\ \hfill 0\hfill & \hfill \frac{1}{2}\hfill \end{array}\right]$, which has a norm of $\sqrt{\frac{33}{64}}$

• From $M$, notice $M=\mathit{LU}=\mathit{LS}{L}^{-1}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0.9\hfill & \hfill 0.1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 1\hfill \end{array}\right]$, so we have $L^{-1}\mathit{ML}=\left[\begin{array}{cc}\hfill 0.9\hfill & \hfill 0.1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=S$, let $D=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right]$, we have $D^{-1}\mathit{SD}=\left[\begin{array}{cc}\hfill 0.9\hfill & \hfill 0.1d\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$, to make the Frobenius norm less than 1, we pick $d=4$, then $C=D^{-1}{L}^{-1}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -\frac{1}{4}\hfill & \hfill \frac{1}{4}\hfill \end{array}\right]$

$\mathit{CM}{C}^{-1}=D^{-1}{L}^{-1}\mathit{MLD}=\left[\begin{array}{cc}\hfill 0.9\hfill & \hfill 0.4\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ will have a norm less than 1.