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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 dont 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 dont see how this could help. I dont 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. – Hans Engler Dec 7 '12 at 4:36
@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 $||A||_F$? – user972276 Dec 7 '12 at 4:42
@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 $||A||_2 \le ||A||_F$. – user972276 Dec 7 '12 at 4:50
$\Vert A\Vert_F^2 = trace(AA^T)$. Now compute the same thing for $AU$ and use the fact that $U$ is orthogonal. – Hans Engler Dec 7 '12 at 12:14

1 Answer

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Write $x=\sum_{j=1}^nc_je_j$, for coefficients $c_1,\ldots,c_n$. Suppose that $\|x\|_2=1$, i.e. $\sum_j |c_j|^2=1$. Then $$ \|Ax\|_2^2=\|\sum_j c_j\,Ae_j\|_2^2\leq\left(\sum_j|c_j|\,\|Ae_j\|_2\right)^{2}\\ \leq\left(\sum_j|c_j|^2\right)\sum_j\|Ae_j\|_2^2=\sum_j\|Ae_j\|_2^2=\|A\|_F^2, $$ where the triangle inequality is used in the first $\leq$ and Cauchy-Schwarz in the second.

As $x$ was arbitrary, we get $\|A\|_2\leq\|A\|_F$.

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is there a way to do it without using Cauchy-Schwartz? – user972276 Dec 7 '12 at 4:54
Not that I can think of. In any case I simplified the argument by using the triangle inequality (basically, the previous version had a built-in proof of the triangle inequality, which by the way I don't think you can prove without CS). – Martin Argerami Dec 7 '12 at 10:58

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