Timeline for About reducing the rank of a matrix by substract a diagonal matrix
Current License: CC BY-SA 3.0
10 events
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Dec 29, 2017 at 5:54 | history | edited | YCor |
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Dec 29, 2017 at 4:42 | vote | accept | Minji Kim | ||
Dec 29, 2017 at 3:49 | history | edited | Minji Kim | CC BY-SA 3.0 |
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Dec 28, 2017 at 9:31 | answer | added | Aaron Meyerowitz | timeline score: 3 | |
Dec 28, 2017 at 4:16 | comment | added | Gerry Myerson | OK, that's an example where you can reduce the rank by more than 1 (a simpler example would be $$\pmatrix{1&0&0\cr0&2&0\cr0&0&3\cr}$$ but what about in general? | |
Dec 28, 2017 at 2:28 | comment | added | Minji Kim | 1) For example, think of the matrix $\begin{bmatrix} 2 &2& 1\\ 2 & 6 & 2\\1 & 2 & 4\end{bmatrix}$. We cannot reduce its rank more than one by just substracting $\lambda I$, but if we substract $\begin{bmatrix} 1 &0& 0\\ 0 & 2 & 0\\0 & 0 & 3\end{bmatrix}$, we can reduce the rank by 2. 2) Yes, but I am especially interested in the case where $Q$ is positive definite. | |
Dec 27, 2017 at 19:45 | answer | added | Alexandre Eremenko | timeline score: 4 | |
Dec 27, 2017 at 16:05 | comment | added | Gerry Myerson | Good question. But, 1) is there any reason to think that, in general, you can reduce the rank by more than 1? 2) why the restriction to positive definite matrices? isn't the question of interest for all matrices? | |
Dec 27, 2017 at 12:36 | review | First posts | |||
Dec 27, 2017 at 12:38 | |||||
Dec 27, 2017 at 12:34 | history | asked | Minji Kim | CC BY-SA 3.0 |