Timeline for Is this inequality involving the Frobenius norm right?
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Nov 18, 2016 at 13:35 | comment | added | Federico Poloni | @RodrigodeAzevedo It might be the case that we have different definitions of singular values. With my definition, $\begin{bmatrix}1\\0\end{bmatrix}$ has one singular value $1$, and for $A=\begin{bmatrix}0&1\end{bmatrix}$ one has $0 = \|AG\|_F$ but $\sigma_{\min}(G)\|A\|_F=1.$ | |
| Nov 18, 2016 at 12:02 | comment | added | Rodrigo de Azevedo | Doesn't the inequality $$\|AG\|_F \geq \sigma_{\min}(G) \|A\|_F \geq n \, \sigma_{\min}(G) \|A\|$$ always hold? However, $\sigma_{\min}(G) > 0$ only if $G$ has full row rank. $G$ being fat does not imply it has full row rank. | |
| Nov 17, 2016 at 13:32 | history | edited | Federico Poloni | CC BY-SA 3.0 |
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| Nov 17, 2016 at 13:14 | history | edited | Federico Poloni | CC BY-SA 3.0 |
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| Nov 17, 2016 at 13:10 | history | undeleted | Federico Poloni | ||
| Nov 17, 2016 at 13:06 | history | deleted | Federico Poloni | via Vote | |
| Nov 17, 2016 at 13:06 | history | answered | Federico Poloni | CC BY-SA 3.0 |