Timeline for Smallest singular value of $X\mapsto AX^{T}+XA^{T}$
Current License: CC BY-SA 3.0
15 events
when toggle format | what | by | license | comment | |
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Aug 24, 2017 at 21:22 | vote | accept | Richard Zhang | ||
Aug 24, 2017 at 21:17 | answer | added | Will Sawin | timeline score: 5 | |
Aug 24, 2017 at 20:52 | history | edited | Richard Zhang | CC BY-SA 3.0 |
added 21 characters in body
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Aug 24, 2017 at 20:48 | comment | added | Richard Zhang | @AliTaghavi, you're absolutely right. The square case is not interesting. Instead, consider the long-and-skinny case $n\ll m$. Where $n=1$, the minimum is certainly not zero. | |
Aug 24, 2017 at 20:43 | history | edited | Richard Zhang | CC BY-SA 3.0 |
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Aug 24, 2017 at 20:41 | comment | added | Ali Taghavi | @RichardZhang For the square case the operator is not surjective so is not injective then it has non trivial kernel. then that minimum is zero. Am I mistaken?Or your question is some thing else? | |
Aug 24, 2017 at 20:38 | comment | added | Richard Zhang | @Min-Oo: X has the be the same size as A in order for $AX^T + XA^T$ to make sense. | |
Aug 24, 2017 at 20:37 | comment | added | Richard Zhang | @AliTaghavi I mean the Frobenius norm. I've edited the question accordingly. | |
Aug 24, 2017 at 20:36 | history | edited | Richard Zhang | CC BY-SA 3.0 |
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S Aug 24, 2017 at 19:51 | history | suggested | Ali Taghavi |
I add a tag
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Aug 24, 2017 at 19:42 | comment | added | Min-Oo | Is X the same size as A? | |
Aug 24, 2017 at 19:32 | comment | added | Federico Poloni | It would be strange if the rectangular case were easier than the square one... | |
Aug 24, 2017 at 19:17 | review | Suggested edits | |||
S Aug 24, 2017 at 19:51 | |||||
Aug 24, 2017 at 19:16 | comment | added | Ali Taghavi | What do you mean by $\parallel \;. \;\parallel_F$? | |
Aug 24, 2017 at 19:10 | history | asked | Richard Zhang | CC BY-SA 3.0 |