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Given $A \in R^{m\times n}$, I have these three norm equivalence equations:

  1. $\|A\|_2 \le \|A\|_F \le \sqrt {n}\|A\|_2$
  2. $\frac {1} {\sqrt n}\|A\|_{\infty} \le \|A\|_2 \le \sqrt {m} \|A\|_{\infty}$
  3. $\frac {1} {\sqrt m}\|A\|_1 \le \|A\|_2 \le \sqrt {n} \|A\|_1$

I cannot use the following:

$$||A||_F = \sqrt {Tr(A^TA)}$$

I need to prove each of them. I do not even know where to begin on this so ANY help would be great!

The norms are the matrix norms. You can find more at Essentially the 1 norm is just adding up each element of a column and finding the largest column sum. The infinity norm is the same as 1 norm except you add up the elements in a row and find the largest row sum. The F norm (frobenius norm) is adding up the squares of each element in A and then taking the square root.

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What exactly are each of these norms? – Tom Oldfield Dec 4 '12 at 2:56
These are matrix norms: – echoone Dec 4 '12 at 3:03
up vote 1 down vote accepted

(not doing all, but here's a few)

  1. The key here is to notice that $$ \|A\|_F^2=\text{Tr}_n(A^TA). $$ Then $$ \|A\|_2^2=\lambda_\max(A^TA)\leq\text{Tr}_n(A^TA)\leq n\lambda_\max(A^TA), $$ so $\||A\|_2\leq\|A\|_F\leq\sqrt n\|A\|_2$.

  2. Here, writing $e=(1,\ldots,1)^T\in\mathbb R^n$, $$ \|A\|_\infty=\max_i\sum_j|A_{ij}|=\|Ae\|_\infty\leq\|Ae\|_2\leq\|A\|_2\|e\|_2=\sqrt n\|A\|_2 $$

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I am not suppose to use that the frobenius norm is equal to the sqrt of the trace. I should of stated that, sorry! – user972276 Dec 4 '12 at 3:42

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