Proving the a matrix is bounded for every power I want to show that the following matrix satisfies $\|A^n(z)\| \leq C$ for every $z \in \mathbb{C}$ with $Re(z) \leq 0$, for $n= 0,1,2,...$, and some constant $C$. How do I approach this kind of problems?
$$A(z) = \begin{bmatrix} \frac{z^2+6z+8}{2(2-z)^2} & -\frac{z^2+10z+24}{8(2-z)^2}  \\ \frac{2z(z+2)}{(2-z)^2} & -\frac{z(z+6)}{2(2-z)^2}   \end{bmatrix}$$
 A: Observe that $A(z)$ is a rank-one matrix:
$$
A(z)
=\frac{1}{2(z-2)^2}
\begin{bmatrix}\frac{1}{4}(z+4)\\ z\end{bmatrix}
\begin{bmatrix}4(z+2)&-(z+6)\end{bmatrix}.
$$
Therefore
$$
\|A(z)^n\|=|\operatorname{tr}(A(z))|^{n-1}\,\|A(z)\|.
$$
To prove that $\|A(z)^n\|$ is uniformly bounded for all $n\ge0$ and for all $z\in H=\{z:\operatorname{Re}(z)\le 0\}$, it suffices to prove that

*

*$\operatorname{tr}(A(z))=\frac{4}{(2-z)^2}$ is uniformly bounded for all $n\ge0$ and $z\in H$, and

*$\|A(z)\|$ is uniformly bounded for all $z\in H$.

Condition 1 is easy because
$$
\operatorname{Re}(z)\le 0
\quad\Rightarrow\quad|z-2|\ge2
\quad\Rightarrow\quad\left|\frac{4}{(2-z)^2}\right|\le1
\quad\Rightarrow\quad\left|\frac{4}{(2-z)^2}\right|^n\le1 \text{ for all } n.
$$
I will leave the verification of condition 2 to you.
A: Given matrix $M=\{M_{ij}\}$ and $N=\{N_{ij}\}$ where $M_{ij}\in\Bbb C$ and $N_{ij}\in\Bbb R_{>0}$, write $M\prec N$ if $|M_{ij}|<N_{ij}$ for all $i,j$. It is straightforward to verify that if $M\prec N$ and $M'\prec N'$, then $MM'\prec NN'$.
It is enough to prove the following claim.
Claim: Let $z \in \mathbb{C}$ with $\Re(z) \leq 0$. Then $A^n(z) \prec \begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}$ for all $n=1,2,3,\cdots$.
Proof:
Since $\Re(z)<0$, we have $|z+2|<|2-z|$, $\ |z+4|<|2-z|$ and $|z+6|<|2-z|$. $$A(z) = \begin{bmatrix} \frac{(z+2)(z+4)}{2(2-z)^2} & -\frac{(z+4)(z+6)}{8(2-z)^2}  \\ \frac{2z(z+2)}{(2-z)^2} & -\frac{z(z+6)}{2(2-z)^2}   \end{bmatrix}\prec \begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix},$$ which means the claim is true for $n=1$.
Suppose the claim is true for $n$, i.e., $A^n(z)\prec \begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}$. Then
$$A^{n+1}(z)=A^n(z)A(z)\prec \begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}\begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}=\begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}.$$ Hence, the claim is true for $n+1$, too. $\quad\checkmark$.
