I read this in my textbook. I have tried working through it. I keep getting max $\| Ax \|_2$, which is just the magnitude of $Ax$. How should I prove it?
Please do note that this is not for homework. I'm just trying to understand why the textbook's claim is true, as no proof is provided.
From your question, I assume you mean the operator norm with respect to the 2-norm. Let $A = {\rm diag}(\lambda_1, \ldots, \lambda_n)$ a diagonal matrix. We have \begin{align*} \|A\| &= \max_{\|x\|_2 = 1} \|Ax\|_2\\ &= \max_{\|x\|_2 = 1} \left(\sum_{i=1}^n \lambda_i^2x_i^2\right)^{1/2}\\ &\le \max_{\|x\|_2 = 1} \max_i|\lambda_i| \left(\sum_{i=1}^n x_i^2\right)^{1/2}\tag 1\\ &= \max_i|\lambda_i| \cdot \max_{\|x\|_2 = 1} \|x\|_2\\ &= \max_i|\lambda_i| \end{align*} For (1), we can argue as follows: For each $i$: we have $x_i^2 \ge 0$, hence multiplying the inequality $|\lambda_i|^2 \le \max_i |\lambda_i|^2$ by $x_i^2$, we get $x_i^2|\lambda_i|^2 \le (\max_i|\lambda_i|)^2 x_i^2$. Now sum and take the square root of both sides.
On the other hand, let $x$ be an eigenvector, corresponding to the largest eigenvalue. Then $$ \|Ax\|_2 = \max_i|\lambda_i| \cdot \|x\|_2 $$ This gives $\|A\| = \max_i |\lambda_i|$.
This also follows from the fact that for any diagonal matrix D, the elements on the diagonal are just the matrix's singular values and the 2-norm of any matrix can be shown to equal its largest singular value.
Let $d_1, d_2, \dots, d_n \in {\Bbb R}$. Build the diagonal matrix ${\bf D} := \operatorname{diag} \left( d_1, d_2, \dots, d_n \right)$. Note that
$$ {\bf D} := \sum_{i=1}^n d_i {\bf e}_i {\bf e}_i^\top = \sum_{i=1}^n | d_i | \left( \operatorname{sgn} \left( d_i \right) \, {\bf e}_i \right) {\bf e}_i^\top = \sum_{i=1}^n | d_i | \, {\bf e}_i \left( \operatorname{sgn} \left( d_i \right) \, {\bf e}_i \right)^\top $$
where $\operatorname{sgn}$ denotes the sign function. Thus, the singular values of a real diagonal matrix are the absolute values of the entries on the main diagonal, whereas the left and right singular vectors are $\pm 1$ ("signed") combinations of the vectors of the (same) canonical basis. Hence, the spectral norm of a diagonal matrix is the $\infty$-norm of its main diagonal
$$ \| {\bf D} \|_2 := \sigma_{\max} ({\bf D}) = \color{blue}{\max_{1 \leq i \leq n} | d_i | } $$
What if $d_1, d_2, \dots, d_n \in {\Bbb C}$?