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How do I prove this inequality / equivalence about matrix p-norms? It appears on the wikipedia and mathworld.wolfram pages on matrix norms without proof.

$\|A\|^2_2 \leq \|A\|_1 \|A\|_\infty$

Maybe a proof of this would look a lot like a proof of Holder's inequality?

Neal Lawton
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    Just write the expanded sum of $tr (A^T A)=\sum_{i,j} |A_{ij}|^2$ and use the simplest upper bound you have – Bertrand R Sep 08 '13 at 10:49
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    I think you might be mistaking the matrix 2-norm for something else. In fact what you're describing is called the Frobenius Norm, which is different. The matrix 2-norm is NOT the sum of the square entries. It's defined as $|A|2 = \max_x \frac{|Ax|_2}{|x|_2}$, where on the right hand side we're using the usual vector 2-norm (sum of square entries). It can be proven that $|A|_2 = \sigma{max}(A)$, the largest singular value of A. – Neal Lawton Sep 08 '13 at 17:38
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    I don't know if this is still actual but $|A|2^2=\rho(A^A)\leq|A^A|_1\leq|A^*|_1|A|_1=|A|{\infty}|A|_1$ :-) – Algebraic Pavel Feb 09 '15 at 08:50
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    Yes, I think that works, @AlgebraicPavel; for anyone wondering, the first inequality comes from the fact that $\rho(A) \leq |A|$ for any matrix norm. – Neal Lawton Jul 07 '15 at 00:59

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