Now a bit of a disclaimer, its been two years since I last took a math class, so I have little to no memory of how to construct or go about formulating proofs. My tutor unfortunately is very and excruciatingly slow at figuring them out and this one the tutor cannot seem to solve.
What I was taught is to start with "What do we want?" and "What do we know?"
What we want:
$||A||_2 \leq \sqrt{n}||A||_\infty$
What we know:
$||A||_2^2$ = $\lambda _\max(A^tA)$
$\sqrt{n}||A||_\infty = \sqrt{n} * max_i | \Sigma a_{ij} y_j |$ Where j= $1 \dots n$. Otherwise the maximum absolute row sum.
Thus:
$\sqrt{n}^2||A||_\infty^2 = n * max_i | \Sigma a_{ij} y_j |^2$ Where j= $1 \dots n$.
Or max row sum squared times n.
I feel like that the square of the max times n is likely going to be bigger than the largest positive eigenvalue of A, but I have no means of showing/knowing this and nothing in the notes indicates how to find the bounds other than maybe Banach-Lemma but I think that's only if A is invertible no? And I don't think that's given.
Was a homework question, but I'm way past due anyways and just working on it to understand it/prepare for the first mid term.
Best Answer
The matrix $p$-norm is the operator norm: $$ \|A\|_p=\max_{x\neq 0}\frac{\|Ax\|_p}{\|x\|_p}. \tag{1} $$ The fact that $$\tag{2} \|A\|_2=\rho(A^*A)\quad\text{and}\quad\|A\|_{\infty}=\max_i\sum_j|a_{ij}| $$ is just the consequence of (1). The $2$-norm and $\infty$-norm satisfy $$\tag{3} \|x\|_{\infty}\leq\|x\|_2\leq\sqrt{n}\|x\|_{\infty} $$ for any $n$-vector $x$ and the fact that $\|A\|_2\leq\sqrt{n}\|A\|_{\infty}$ follows by plugging (3) to (1): $$ \|A\|_2=\max_{x\neq 0}\frac{\|Ax\|_2}{\|x\|_2}\leq \max_{x\neq 0}\frac{\sqrt{n}\|Ax\|_\infty}{\|x\|_2} \leq \max_{x\neq 0}\frac{\sqrt{n}\|Ax\|_\infty}{\|x\|_{\infty}}=\sqrt{n}\|A\|_{\infty}. $$
Assuming that you know nothing about (1) and take (2) as the definitions (such an approach in a class is a bit odd because it makes showing a lot of things quite complicated), you can proceed as follows: