Matrix norm inequalities for 2-norm and $\infty$-norm

Question

I am doing the following exercise: https://i.stack.imgur.com/Ye9C3.png

I proved parts $a$ and $b$, but I got a little confused with parts $c$ and $d$. Here is my proof for part $(c)$:

Let $x \in \mathbb{R}^{n}$ and view $A$ as a $mn$-dimensional vector. Then,

$||Ax||_{\infty} \leq ||Ax||_2$ by part $(a)$.

We know

$||Ax||_2 \leq ||A||_2 ||x||_2$ by norm property.

By part $(b)$ we now have

$\cdots \leq ||A||_2 \sqrt{n}||x||_\infty$

Then since the above was written with arbitrary $x$, let it be the unit vector. This yields what was desired: $$||A||_\infty \leq \sqrt{n} \cdot ||A||_2$$

Regarding part $(d)$, doesn't this follow immediately from part $(b)$ since we can view $A$ as an $mn$-dimensional vector?

Answer 1

The operator (equivalently matrix) norm is defined as $$\left\lVert A\right\rVert_{\Box}=\sup\limits_{x\in B\setminus\{0\}}\frac{\left\lVert Ax\right\rVert_{\Box}}{\left\lVert x\right\rVert_{\Box}}=\sup\limits_{\substack{x\in B\\ \left\lVert x\right\rVert_{\Box}=1}}\left\lVert Ax\right\rVert_{\Box}$$ where $B$ is your favorite (vector) space to work on and $\Box$ can be any symbol depending on your choice of vector spaces, simply fit accordingly.

Your answer to $c)$ is correct, but you can be a little bit more precise with your redaction (as this seems to be an introduction exercise, we shall make any steps very clear in our minds).

I suggest the following:

Let $x\in\mathbb{K}^m\setminus\{0\}$ ($\mathbb{K}$ being your favorite field, for example $\mathbb{R}$ or $\mathbb{C}$). Then we have \begin{align} \left\lVert Ax\right\rVert_{\infty} &\leqslant\left\lVert Ax\right\rVert_{2}\\ &\leqslant \left\lVert A\right\rVert_{2}\left\lVert x\right\rVert_2\\ &\leqslant \left\lVert A\right\rVert_2\sqrt{n}\left\lVert x\right\rVert_{\infty} \end{align} where we used the fact that $Ax\in\mathbb{K}^n$ so that we can use the previous inequalities: $a)$ for the first one, an operator norm property for the second and $b$ for the third one. Note here that I have used the $\infty$ symbol for both $Ax$ and $x$ where in fact they lie in different spaces (unless $n=m$), we could be more rigorous and distinguish both norms, but usually we are lazy and we do not.

This implies that $$\frac{\left\lVert Ax\right\rVert_\infty}{\left\lVert x\right\rVert_\infty}\leqslant \sqrt{n}\left\lVert A\right\rVert_2$$

As the right hand side does not depend on $x\in\mathbb{K}^m\setminus\{0\}$, we have $$\left\lVert A\right\rVert_{\infty}=\sup\limits_{x\in B\setminus\{0\}}\frac{\left\lVert Ax\right\rVert_{\infty}}{\left\lVert x\right\rVert_{\infty}}\leqslant \sqrt{n}\left\lVert A\right\rVert_2$$ which is the desired result.

Simply use the same logic for $d)$ and you'll be done.