Timeline for a question of ranks of matrices
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 18, 2013 at 8:00 | answer | added | Andreas Maurischat | timeline score: 4 | |
| Aug 10, 2013 at 12:15 | comment | added | Benjamin Steinberg | Removed arithmetic geometry tag | |
| Aug 10, 2013 at 11:38 | history | edited | Benjamin Steinberg |
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| Aug 10, 2013 at 8:28 | comment | added | Wilberd van der Kallen | If $v_i$ spans the image of $A_i$ then your hypothesis implies one cannot find $t+1$ independent $v_i$. To see this, suppose one could find them. Replace every $A_i$ with $SA_iS^t$, with $S$ chosen so that the $t+1$ vectors $Sv_i$ are part of the standard basis. | |
| Aug 10, 2013 at 6:14 | history | edited | kiseki | CC BY-SA 3.0 |
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| Aug 10, 2013 at 6:14 | comment | added | kiseki | Sorry, I have a typo | |
| Aug 10, 2013 at 6:10 | comment | added | Gerry Myerson | You may assume $m$ is as big as you like; my remark still holds. The rank of the sum is $1$; the sum of the ranks is $m$; $m\le1$ is false. | |
| Aug 10, 2013 at 6:09 | comment | added | kiseki | @Gerry: we assume $m>n$. | |
| Aug 10, 2013 at 5:53 | comment | added | Gerry Myerson | If the $A_i$ are all scalar multiples of each other, then you can take $t=1$, while the sum of the ranks will be $m$, contradiction. | |
| Aug 10, 2013 at 5:48 | history | asked | kiseki | CC BY-SA 3.0 |