Timeline for Finding the Square-Root of a Non-diagonalizable Positive Matrix
Current License: CC BY-SA 2.5
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 27, 2014 at 21:29 | answer | added | cdh | timeline score: 1 | |
| May 27, 2014 at 20:45 | answer | added | Michael | timeline score: 0 | |
| May 27, 2014 at 8:11 | answer | added | Federico Poloni | timeline score: 0 | |
| May 27, 2014 at 7:41 | answer | added | J Tyson | timeline score: 1 | |
| Feb 4, 2010 at 6:20 | answer | added | algori | timeline score: 9 | |
| Feb 4, 2010 at 4:50 | comment | added | Yemon Choi | @Mariano: well, in my book "positive complex matrix" can mean only one thing. I'm still slightly stunned by the juxtaposition of this question with some of Abtan's others. | |
| Feb 4, 2010 at 4:47 | comment | added | Mariano Suárez-Álvarez | I think we are waiting for Abtan to clarify what he means by positive... | |
| Feb 4, 2010 at 4:40 | answer | added | S. Carnahan♦ | timeline score: 2 | |
| Feb 4, 2010 at 4:14 | comment | added | Pete L. Clark | Would someone please step up and give their comments as an answer? :) | |
| Feb 4, 2010 at 3:59 | comment | added | Mariano Suárez-Álvarez | It should be noted that, in general, algori's procedure gives many different square roots.. | |
| Feb 4, 2010 at 3:55 | comment | added | Mariano Suárez-Álvarez | Wikipedia has a little section on non symmetric/hermitian positive matrices at en.wikipedia.org/wiki/… | |
| Feb 4, 2010 at 3:54 | comment | added | algori | For a Jordan cell $A$ with eigenvalue $t\neq 0$ write $A=tB$. $B$ has 1's on the main diagonal and $1/t$'s immediately above it. $N=B-I$ is nilpotent, so a square root $C$ of $B=I+N$ can be found using the binomial formula (which gives a finite sum). Then $\sqrt{t}C$ will be a square root of $A$. If $A$ is a Jordan $n$ by $n$ cell with eigenvalue 0, then $A$ has no square roots for $n>1$ (for rank reasons). The case when $A$ is arbitrary follows from the above. | |
| Feb 4, 2010 at 3:48 | comment | added | Jonas Meyer | @Leonid: Yes, thank you for the correction. I was only thinking of symmetric matrices and not the condition you were referring to. | |
| Feb 4, 2010 at 3:46 | comment | added | Yemon Choi | @Leonid: should that be $\pi/6$ ? But good point, I'll think about that a bit more. | |
| Feb 4, 2010 at 3:31 | comment | added | Yemon Choi | @Jonas Oh yes, you're right, I was for some reason thinking of the results for diagonalizability of unitary/orthogonal matrices. As you say, the argument that does the complex case does the real case, just because it's a theorem about inner products, IIRC | |
| Feb 4, 2010 at 3:30 | comment | added | Yemon Choi | @Abtan: each self-adjoint (complex) matrix has an orthonormal spanning set of eigenvectors. See en.wikipedia.org/wiki/Spectral_theorem or a lin alg textbook | |
| Feb 4, 2010 at 3:29 | comment | added | Jonas Meyer | (My last comment was for Yemon.) Abtan, see here: en.wikipedia.org/wiki/Spectral_theorem | |
| Feb 4, 2010 at 3:29 | comment | added | Jonas Meyer | You don't need to go to the complex case; real positive matrices are orthogonally diagonalizable. | |
| Feb 4, 2010 at 3:26 | comment | added | Abtan Massini | Really, why does positivity imply a spanning set of eigenvectors? | |
| Feb 4, 2010 at 3:24 | history | edited | Yemon Choi | CC BY-SA 2.5 |
corrected spelling and removed unnecessary OA tag
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| Feb 4, 2010 at 3:21 | comment | added | Yemon Choi | Leonid: while I'm not 100% sure of the details off the top of my head, you don't get much new. If you just want to find a square root, then you can just diagonalize over the complex field, take the unique square root of your diagonal matrix, and conjugate back again. This will give us a real matrix because we stay inside the real algebra generated by the original matrix. | |
| Feb 4, 2010 at 3:07 | comment | added | Yemon Choi | If, by positive, you mean that $\langle Ax,x\rangle$ is non-negative for all vectors $x$, then the matrix is diagonalizable. So your question might need rethinking | |
| Feb 4, 2010 at 3:02 | history | asked | Abtan Massini | CC BY-SA 2.5 |