Timeline for Every square complex matrix is similar to a TRIDIAGONAL complex-symmetric matrix?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 20 at 21:32 | comment | added | Igor Khavkine | Since tridiagonal form is reached in a finite number of steps, the tridiagonal matrix and the transformation matrices are finite algebraic expressions of the original matrix elements, for a real symmetric matrix. Analytic continuation gives you the result for free for complex symmetric matrices (same for upper Hessenberg form of non-symmetric matrices). The only issue with the formula is division by zero (of some vector norms), which are avoided by a small change of basis. But I think you've already noticed both those things in your answer. | |
| Mar 20 at 19:46 | answer | added | wlad | timeline score: 2 | |
| Mar 20 at 19:34 | comment | added | wlad | OK, I have an answer | |
| Mar 20 at 19:21 | comment | added | wlad | @IgorKhavkine No it's not straightforward from the wiki page. You misinterpreted the question. | |
| Mar 20 at 18:02 | comment | added | wlad | @IgorKhavkine Hold on. The Lanczos method (unlike Householder reflections) might do it. | |
| Mar 20 at 17:58 | comment | added | wlad | @IgorKhavkine You might be confusing Hermitian with complex-symmetric. It's less clear in the complex-symmetric case. | |
| Mar 20 at 17:26 | comment | added | Igor Khavkine | Every symmetric matrix (real or complex) is similar to a tridiagonal symmetric matrix. One can apply a sequence of Householder transformations or the Lanczos algorithm (en.wikipedia.org/wiki/…, en.wikipedia.org/wiki/…). | |
| Mar 20 at 15:35 | history | asked | wlad | CC BY-SA 4.0 |