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Is there an $M \in \mathbb{R}$ such that $\forall A \in \mathbb{M}^{n \times m},||A||_{frob} \leq M||A||_{op}$?

Research effort We can assume that $A \neq 0$. $$ \frac{||A||_{\mathrm{frob}}}{||A||_{\mathrm{op}}} \leq \frac{\sum_i{|Ae^i|}}{\sup_{||v||=1}||Av|| } = \frac{\sum_i{|Ae^i|}}{\sup_{||v||=1}|| \sum_i v_i Ae^i||} $$ And... that's where I got stuck. If I'd do something with the triangular inequality in the denominator, the expression would only become bigger. Can you please give me a hint?

Davide Giraudo
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Koenraad van Duin
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    Yes, since all norms on finite dimensional spaces are equivalent. – Alex Youcis Sep 08 '13 at 09:17
  • Write the Frobenius norm as the square root of the sum of squared norms of the columns and bound them from above with the operator norm. Similarly can do with rows. Goal: have $M=\min{m,n}$. Actually it can be made $M=\mathrm{rank}(A)$ but this involves "harder" stuff like orthogonal invariance and SVD. – Algebraic Pavel Sep 09 '13 at 00:36

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