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For finite dimensional spaces, all norms are equivalent, i.e. there exist constants say $A,B$ such that for all matrices from the $\mathbf M \in R^{d\times d}$ (let $d$ be a fixed positive integer) such that

$$A \|\mathbf M\|_2 \leq \|\mathbf M \|_F \leq B\|\mathbf M\|_2\text{,} $$

where $\|\cdot\|_2$ denotes the spectral norm and $\|\cdot\|_F$ (edit: i forgot the second part...) denotes the Frobenius norm.

My question is now, whether you know anything specific about $A$ and $B$, for example whether there exists a analytical expression for let's say $B$.

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Oh thanks, i forgot to mention, that $\|\cdot\|_F$ is (as the letter might indicate) the Frobenius norm. And yes you're right, the supremum should do. I think $B$ should be finite for the (now two) mentioned norms, but i haven't found anything. – Ronny Feb 6 '12 at 13:34
I see :) then I can't help more than upvoting you – Ilya Feb 6 '12 at 13:46
up vote 7 down vote accepted

Both of these norms are invariant if $\mathbf M$ is left- or right-multiplied by a unitary matrix. Thus we can form the singular value decomposition of $\mathbf M$, drop the two unitary matrices and keep only the diagonal part. For a given spectral norm, the lowest Frobenius norm will be achieved if only one singular value has this value and the others are zero, and the highest Frobenius norm will be achieved if all singular values have this value; thus

$$\|\mathbf M\|_2 \leq \|\mathbf M \|_{\mathrm F} \leq \sqrt d\,\|\mathbf M\|_2\;,$$

that is, $A=1$ and $B=\sqrt d$.

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Thank you, that sounds really neat, haven't hat SVD in mind on that one. – Ronny Feb 6 '12 at 19:16
Shouldn't that also be possible – if $\mathbf{M}$ is regular – to take your proof and use then the unitary basis of the eigenspaces, hence choose $\mathbf{Q}$ in the diagonalizable representation $\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}$ unitary and get – for that the case of regular matrices – also the proof? – Ronny Feb 7 '12 at 16:08
I'm not sure I follow you. How are you using the term "regular"? Do you mean "normal"? If so, the answer is yes, simply because unitary diagonalization is a special case of singular value decomposition. – joriki Feb 8 '12 at 5:15
For normal matrices, your argumentation is clear, using eigenvalues. But with regular i meant, that the determinant of §\mathbf M$ is nonzero or in other words the matrix has full rank. – Ronny Feb 8 '12 at 5:50
@Ronny: Then I don't understand what you mean by "the unitary basis of the eigenspaces" and "the diagonalizable representation". An invertible matrix doesn't necessarily have full eigenspaces and isn't necessarily diagonalizable; take for example $$\pmatrix{1&1\\0&1}\;.$$ – joriki Feb 8 '12 at 6:31

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