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How can you prove that: $$\|A\|_2 \le \|A\|_F$$ I cannot use: $$\|A\|_2^2 = \lambda_{max}(A^TA)$$ It makes sense that the 2-norm would be less than or equal to the Frobenius norm but I don't know how to prove it. I do know:

$$\|A\|_2 = \max_{\|x\|_2 = 1} {\|Ax\|_2}$$

and I know I can define the Frobenius norm to be:

$$\|A\|_F^2 = \sum_{j=1}^n {\|Ae_j\|_2^2}$$ but I don't see how this could help. I don't know how else to compare the two norms though.

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    Let $\Vert A\|_2 = \|Av\|$ for some $v$ with $\|v\| = 1$. Define an orthogonal matrix $U$ for which $v$ is the first column. Compute $\Vert AU\Vert_F $ and show that it is equal to $\Vert A\Vert_F $, then compare it to $\Vert Av\Vert$ by a direct computation.2012-12-07
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    @HansEngler I dont exactly understand your answer. How would I compute $||AU||_2$ and if it is equal to $||A||_2$ then how does that prove anything about $||A||_F$?2012-12-07
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    @HansEngler I dont understand how to show $||AU||_F = ||A||_F$. I understand that the first column of AU is just Av and so I can show that $||A||_2 \le ||A||_F$.2012-12-07
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    $\Vert A\Vert_F^2 = trace(AA^T)$. Now compute the same thing for $AU$ and use the fact that $U$ is orthogonal.2012-12-07

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