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Suppose $A \in M_{d\times d}\left(\mathbb{C}\right)$. Let $||A||=\max_{ij} |a_{ij}|$ be the max norm of $A=\left[a_{ij}\right]_{d \times d}$. For a positive integer $n$ is there any relation between $||A^n||$ and $||A||^n$ ?

What I attempted:- We have, \begin{equation} \begin{aligned} ||A^2||&=\max_{ij} |\sum_{k=1}^d a_{ik} a_{kj}| \\ & \le \max_{ij} \sum_{k=1}^d |a_{ik} a_{kj}| \quad (\mbox{Since, by triangle inequality for some $c_i's \in R$, $\hspace{2mm}$ $|\sum_{i=1}^n c_i|\le \sum_{i=1}^n |c_i|$})\\ & \le \max_{ij} \sum_{k=1}^d |a_{ik}||a_{kj}|\quad (\mbox{For $a,b \in R, \hspace{2mm} |ab|\le |a| |b|$}) \quad \dots \dots (1)\\ \\ & \mbox{For any particular $k,\hspace{2mm} 1\le k \le n$}\\ \\ & \quad \max_{ij} |a_{ik}||a_{kj}| \le \max_{ij}|a_{ij}|. \max_{ij}|a_{ij}|=\left(\max_{ij}|a_{ij}|\right)^2=||A||^2 \end{aligned} \end{equation} Thus,\begin{equation} \begin{aligned} ||A^2||&\le \max_{ij} \sum_{k=1}^d |a_{ik}||a_{kj}|\\ & \le \sum_{k=1}^d \max_{ij} |a_{ik}| |a_{kj}|\\ & \le \sum_{k=1}^d ||A||^2 \quad \mbox{From $(1)$}\\ & = d ||A||^2 \end{aligned} \end{equation} It allows us to suspect that $||A^n||\le d^{n-1} ||A||^n$. Suppose the result is true for $n=r$. Thus, if $A^r=\left[b_{ij}\right]$ , then \begin{equation} \begin{aligned} & ||A^r||\le d^{r-1} ||A||^r\\ & \Rightarrow \max_{ij}|b_{ij}|\le d^{r-1} \left(\max_{ij}|a_{ij}|\right)^r \quad \dots \dots (2)\\ \end{aligned} \end{equation} Now, \begin{equation} \begin{aligned} ||A^{r+1}||&=||A^r.A||\\ &= \max_{ij}|\sum_{k=1}^d b_{ik}a_{kj}|\\ & \le \max_{ij} \sum_{k=1}^d |b_{ik}a_{kj}|\\ & \le \max_{ij} \sum_{k=1}^d |b_{ik}||a_{kj}|\\ & \le \sum_{k=1}^d \max_{ij} |b_{ik}||a_{kj}| \\ & \le \sum_{k=1}^d \max_{ij} |b_{ij}| \max_{ij} |a_{ij}|\\ & \le \sum_{k=1}^d d^{r-1} \left(\max_{ij}|a_{ij}|\right)^r \max_{ij} |a_{ij}|\quad (\mbox{From $(2)$})\\ & = d.d^{r-1}\left(\max_{ij}|a_{ij}|\right)^{r+1}\\ &=d^r ||A||^{r+1} \end{aligned} \end{equation}

So, by the principle of mathematical induction, $||A^n||\le d^{n-1} ||A||^n$. Here $d$ is always positive. So, if this claim is true, then of course $||A^n||\le d^{n-1} ||A||^n < d^n ||A||^n$.

So, the exact relationship should look like $||A^n|| < d^n ||A||^n$.

Am I correct?

1 Answers1

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If $A$ and $B$ are $d\times d$ matrices and $\|A\|_\infty=\max_{i,j} |A_{ij}|$ denotes the max-norm of $A$, then $$\|AB\|_{\infty}=\max_{i,j} |(AB)_{ij}|\leq \max_{i,j}\sum_{k}|A_{ik}||B_{kj}|$$ $$\leq d \max_{i,k}|A_{ik}|\max_{k,j}|B_{k,j}| \leq d \|A\|_\infty \|B\|_\infty.$$ By repeatedly using this inequality, one has $\|A^n\|_\infty \leq d^{n-1}\|A\|_\infty^{n}$.

Marco
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