Why is the operator $2$-norm of a diagonal matrix its largest value? I read this in my textbook have tried working through it - I keep getting max 2-norm(Ax), which is just the magnitude of Ax. 
How should I do this proof? (note, this is not for homework, I'm just trying to understand why as no proof is provided). 
 A: 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 $\|A\| = \max_i |\lambda_i|$.
A: 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.
