What if $\|A^2\|_\infty=\|A\|^2_\infty$?

Question

I have this exercise:

For the matrix norm $$\|A\|_\infty := \max_{1\leq i\leq n}{\sum_{j=1}^n{|a_{ij}|}}$$ show or refute $$\|AB\|_\infty=\|A\|_\infty\|B\|_\infty$$ What happens in the special case $\|A^2\|_\infty=\|A\|^2_\infty$?

I already refuted that $\|AB\|_\infty=\|A\|_\infty\|B\|_\infty$ with $A=\begin{bmatrix}0&1\\1&1\end{bmatrix}$ and $B=\begin{bmatrix}1&2\\3&4\end{bmatrix}$, where $AB=\begin{bmatrix}3&4\\4&6\end{bmatrix}$, $\|AB\|_\infty=10$, $\|A\|_\infty\cdot\|B\|_\infty=2\cdot7=14$.

But, what happens when $\|A^2\|_\infty=\|A\|^2_\infty$? I appreciate any hint.

Answer 1

Consider the 2-Nilpotent matrices. I.e. matrices $A\neq0$ such that $A^2=0$. Can $|A|^2_\infty=0$? What about $|A^2|_\infty$?

Edit: Clearly, for a scalar $\lambda\in\mathbb R$, if a matrix $A\neq 0$ satisfies $A^2=\lambda A$ (there are such matrices, e.g. $\lambda E_n$, or the Nilpotents if $\lambda=0$) then $$|A^2|_\infty-|A|_\infty^2 =(\lambda-|A|_\infty)|A|_\infty$$ is nonzero whenever $|A|_\infty\neq \lambda$.

This is just an example to show how complicated it could get to classify the solutions.

However, if I am not mistaking this definition of $|\cdot|_\infty$ then one can show that for all matrices $A,B$ it holds that $|AB|_\infty\leq |A|_\infty\cdot|B|_\infty$.

Answer 2

Try with $$ A=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$