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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).

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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 $\|A\| = \max_i |\lambda_i|$.

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  • $\begingroup$ so that line where you introduce the inequality - what inequality are you using to do that? $\endgroup$ – praks5432 Jun 6 '13 at 22:32
  • $\begingroup$ oh or is it just because you take the max lambda? $\endgroup$ – praks5432 Jun 6 '13 at 22:34
  • $\begingroup$ @praks5432 Added something, hope this helps. $\endgroup$ – martini Jun 6 '13 at 22:36
  • $\begingroup$ @praks5432 Yes, exactly $\endgroup$ – martini Jun 6 '13 at 22:36
  • $\begingroup$ How can we be sure that $\lambda_i$ is an eigenvalue $\endgroup$ – Tyler Hilton Dec 3 '14 at 22:42

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