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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
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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
added 106 characters in body
S Aug 24, 2017 at 19:51 history suggested Ali Taghavi
I add a tag
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