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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year, 4 months |
| seen | Apr 21 at 4:28 | |
| stats | profile views | 9 |
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Feb 8 |
comment |
On matrix norm equivalence Oh, nevermind. I think i thought of normal matrices, though i wanted to proof it for regular matrices using eigenvalues. My mistake. Thanks for the counter example and the explanations. |
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Feb 8 |
comment |
On matrix norm equivalence 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. |
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Feb 7 |
comment |
On matrix norm equivalence 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? |
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Feb 6 |
awarded | Supporter |
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Feb 6 |
awarded | Scholar |
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Feb 6 |
accepted | On matrix norm equivalence |
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Feb 6 |
comment |
On matrix norm equivalence Thank you, that sounds really neat, haven't hat SVD in mind on that one. |
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Feb 6 |
awarded | Editor |
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Feb 6 |
revised |
On matrix norm equivalence Mentioned, that $\|\cdot\|_F$ is the Frobenius norm. |
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Feb 6 |
comment |
On matrix norm equivalence 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. |
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Feb 6 |
awarded | Student |
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Feb 6 |
asked | On matrix norm equivalence |
