Timeline for Some but not all eigenvectors mutually orthogonal
Current License: CC BY-SA 4.0
19 events
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| S Jun 21 at 20:01 | history | bounty ended | Michael Hardy | ||
| S Jun 21 at 20:01 | history | notice removed | Michael Hardy | ||
| Jun 16 at 5:46 | comment | added | Pietro Majer | (so I guess there may be more complicated relations of orthogonality between eigenvectors to consider, than just “$k$ of them mutually orthogonal “ ) | |
| Jun 16 at 5:40 | comment | added | Pietro Majer | Also consider a matrix $M$ [or more generally $U^*MU$] direct sum of arbitrary blocks $M_k$: the eigenvectors of each block can be whatever, but eigenvectors of different blocks are orthogonal. | |
| Jun 16 at 2:38 | answer | added | Buzz | timeline score: 1 | |
| Jun 15 at 21:36 | comment | added | Federico Poloni | Note that OP's original example can be generalized to $\begin{bmatrix}0 & 0\\ B & symmetric\end{bmatrix}$. | |
| Jun 15 at 19:53 | history | edited | YCor |
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| Jun 15 at 19:41 | comment | added | Michael Hardy | @GerryMyerson : You proposal of $\displaystyle \left[ \begin{array}{c|c} {\text{diag-} \atop \text{onal}} & A \\ \hline 0 & B \end{array} \right]$ is obviously sufficient to imply there will be some mutually orthogonal eigenvectors, just as it is obvious that the eigenvectors of a diagonal matrix are mutually orthogonal. So next we want a proposition that bears the same relation to your proposal that the orthogonality of eigenvectors of a real symmetric matrix bears to the orthogonality of eigenvectors of an orthogonal matrix. | |
| Jun 15 at 19:27 | comment | added | Michael Hardy | @GerryMyerson : . . . . and specifically what patterns are hinted at by your example was not clear from the example alone. | |
| S Jun 15 at 19:17 | history | bounty started | Michael Hardy | ||
| S Jun 15 at 19:17 | history | notice added | Michael Hardy | Draw attention | |
| Jun 15 at 19:16 | comment | added | Michael Hardy | @GerryMyerson : My example was not intended to hint at any pattern that could be part of an answer, but only to show that cases can exist were the matrix is almost symmetric and almost all of the eigenvectors are mutually orthogonal. | |
| Jun 13 at 22:23 | comment | added | Gerry Myerson | I'm aware that I have only given one example, although I think it points the way to many more examples; a diagonal matrix $D$ in the upper left corner, zeroes below $D$, and some more-or-less arbitrary columns to the right (keeping the matrix upper triangular, for simplicity). It doesn't answer your question, but it adds more information than any other user has added so far (and may I point out that you, too, have just mentioned one particular example?). | |
| Jun 13 at 18:55 | comment | added | Michael Hardy |
@TheTiler : I haven't yet understood your comment. But here we see a MathJax hazard: The code (n—u_r) gets rendered as $(n—u_r),$ whereas the code (n-u_r) gets rendered as $(n-u_r).$ Which shows you why the latter is considered correct in MathJax (and in LaTeX).
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| Jun 13 at 18:52 | comment | added | Michael Hardy | @GerryMyerson : You example just mentions one particular example, of which the conclusion is true. But the question is about what sorts of assumptions entail that conclusion. | |
| Jun 13 at 12:02 | comment | added | The Tiler | See : archive.org/details/… "...of the coefficients of system (105) to be equal to $(n — u_{r})$, where py is the multiplicity of the root A, of the secular equation. When this condition is satisfied, system (105) defines ux linearly independent vectors ..." | |
| Jun 13 at 11:44 | comment | added | Gerry Myerson | A smaller example would be $$\pmatrix{1&0&1\cr0&2&2\cr0&0&3\cr}$$ with orthogonal eigenvectors $(1,0,0)$ and $(0,1,0)$, and nonorthogonal eigenvector $(1,4,2)$. I don't know whether that's interesting. Perhaps interest is in the eye of the beholder. | |
| Jun 13 at 2:03 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| Jun 10 at 15:53 | history | asked | Michael Hardy | CC BY-SA 4.0 |