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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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    I see :) then I can't help more than upvoting you2012-02-06

1 Answers 1

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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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    @Ronny: You're welcome.2012-02-08