Timeline for vector to diagonal matrix [closed]
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2016 at 12:28 | history | edited | Stefan Kohl♦ |
Added top-level tag (bump the question -- maybe someone wishes to cast the remaining delete vote).
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| S Apr 24, 2015 at 16:33 | history | suggested | isarandi | CC BY-SA 3.0 |
This question indeed make sense.
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| Apr 24, 2015 at 16:11 | review | Suggested edits | |||
| S Apr 24, 2015 at 16:33 | |||||
| Feb 18, 2011 at 8:07 | history | closed |
Theo Johnson-Freyd Yemon Choi Denis Serre David Roberts♦ Andrew Stacey |
not a real question | |
| Feb 18, 2011 at 7:25 | answer | added | Tom De Medts | timeline score: 16 | |
| Feb 18, 2011 at 6:43 | comment | added | Yemon Choi | Jerry, if I can't call it a method I am certainly not going to call it a notation. A notation cannot transform anything to anything. I find it hard to intuit what your actual question is | |
| Feb 18, 2011 at 5:14 | comment | added | Theo Johnson-Freyd | I think this question should be improved before it is appropriate for MO. As is, I have voted to close, but I hope that instead OP rewrites it to clarify (you can modify the question by clicking the little "edit" button). Please see mathoverflow.net/howtoask . | |
| Feb 18, 2011 at 4:34 | comment | added | Anadim | notation-wise that's $diag({\bf x})$. proof-wise, however, I agree with the first comment, you jump from a $1$-dimensional space to an $n$-dimensional space, so no linear operator can get you there. From the diagonal you can definitely go to the vector. Just multiply it with the all ones vector. | |
| Feb 18, 2011 at 4:22 | comment | added | Jerry | Well you can hardly call it a method, it doesn't really do any multiplications, its basically just writing it in a way that implies its shape. | |
| Feb 18, 2011 at 4:17 | comment | added | Yemon Choi | When you say "notation", do you mean "method"? | |
| Feb 18, 2011 at 4:11 | comment | added | Allen Knutson | I have worried about this, and think the answer is basically no, largely because a column vector is rank 1 and the diagonal matrix is larger rank. But I wouldn't say that's a proof. | |
| Feb 18, 2011 at 4:00 | history | asked | Jerry | CC BY-SA 2.5 |