Timeline for Is there half an iteration of the QR algorithm?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 5, 2022 at 16:48 | answer | added | José C Ferreira | timeline score: 1 | |
| Mar 26, 2022 at 16:28 | comment | added | wlad | Uniqueness proof: $QR = Q'R'$ implies $Q^TQ' = R(R')^{-1}$. The LHS is orthogonal, and the RHS is upper triangular. By the orthogonality property, the inverse of the RHS is its transpose. But the inverse of an upper triangular matrix is upper triangular. Therefore the transpose of the RHS is upper triangular. Therefore the RHS is diagonal. A diagonal matrix which is orthogonal has $\pm 1$ along its diagonal. Therefore $R = R' \operatorname{diag}(\pm 1, \pm 1, \dotsc, \pm 1)$. Recall the claim about what $R$ is unique up to. From this, uniqueness of $Q$ is easy to show under the same conditions | |
| Mar 26, 2022 at 16:16 | comment | added | wlad | @GottfriedHelms Make sure that the matrix R you get from your different algorithms has positive entries along its diagonal. | |
| Mar 26, 2022 at 16:00 | comment | added | wlad | @GottfriedHelms The QR decomposition is only unique subject to the requirement that the diagonal entries of R be positive. This should be easy to fix in Matlab or Numpy, or whatever software you're using | |
| Mar 21, 2022 at 4:51 | history | became hot network question | |||
| Mar 21, 2022 at 0:17 | comment | added | Todd Trimble | Does anyone know how to describe in formal terms how the "Francis function" is well-behaved over PSD matrices? | |
| Mar 20, 2022 at 22:31 | answer | added | Federico Poloni | timeline score: 3 | |
| Mar 20, 2022 at 21:15 | comment | added | Paul Tupper | Is your Francis function uniquely defined if $M$ is singular? | |
| Mar 20, 2022 at 21:06 | history | edited | wlad |
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| Mar 20, 2022 at 20:57 | history | edited | wlad | CC BY-SA 4.0 |
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| Mar 20, 2022 at 20:51 | history | asked | wlad | CC BY-SA 4.0 |