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Suppose we have two matrices, A and B, and

$\left\Vert A\right\Vert _{F}\geq\left\Vert B\right\Vert _{F}$

where $\left\Vert .\right\Vert _{F}$ denotes Frobenius norm. Does it imply

$\left\Vert A\right\Vert \geq\left\Vert B\right\Vert $

for any other norm $\left\Vert .\right\Vert $ ??


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Yes but with a multiplicative constant in the inequality, since all finite dimensional norms are equivalent. – soup Dec 23 '12 at 10:58
up vote 0 down vote accepted

No. It is not the case. For instance, let $A,B$ be $2 \times 2$ matrices. Let the singular values of $A$ be $\sigma_{1A}, \sigma_{2A}$ and that of $B$ be $\sigma_{1B}, \sigma_{2B}$. $$\Vert A \Vert_F \geq \Vert B \Vert_F \implies \sigma_{1A}^2 + \sigma_{2A}^2 \geq \sigma_{1B}^2 + \sigma_{2B}^2$$ This doesn't mean that $$\Vert A \Vert_2 = \sigma_{1A} \geq \sigma_{1B} = \Vert B \Vert_2$$ For instance, let $\sigma_{1A} = \sigma_{2A} = 1$ and $\sigma_{1B} = 1.5, \sigma_{2B} = 0$.

We then have $$\Vert A \Vert_F = \sqrt{2} > \sqrt{1.5} = \Vert B \Vert_F$$ but $$\Vert A \Vert_2 = 1 < \sqrt{1.5} = \Vert B \Vert_2$$

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thanks Marvis :) – aning Dec 23 '12 at 12:26

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