Timeline for Singular values of $X$ in $AX+XA=C$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 18, 2019 at 18:31 | comment | added | Richard Zhang | The unique choice of $X$ satisying $Q^TXQ = \hat{X}$ as above and $Q^TXP = (P^TXQ)^T = (Q^TAQ)^{-1}(Q^TCP)$, however, is indeed the least-squares solution, which you can implicitly compute in time linear to dimension $d$ but cubic to rank $k$ | |
| Sep 18, 2019 at 18:28 | comment | added | Richard Zhang | This is not the least-squares solution. Let $P$ be the orthogonal complement of $Q$. Observe that $Q^TCP \ne 0$ in general, but we can always choose $Q^TXP$ such that $Q^TAXP = Q^TCP$ because $Q^TAXP=Q^TAQQ^TXP$ and $Q^TAQ$ is nonsingular by hypothesis. | |
| Sep 16, 2019 at 21:18 | comment | added | Yaroslav Bulatov | BTW, I found something relevant in Golub/Loan 5.4.7: Complete Orthogonal Decomposition -- you can get $\hat{A}$ by applying QR with pivoting followed by regular QR | |
| Sep 16, 2019 at 6:11 | comment | added | Federico Poloni | @YaroslavBulatov I would call it a "low-rank decomposition", generically. It doesn't have a name, as far as I know, also because it's not uniquely defined: you can choose to take $\hat{A}$ diagonal with nonnegative elements, and then it becomes a thin/economy-sized SVD, for instance. I think it's a sort of "folklore idea" in linear algebra. | |
| Sep 15, 2019 at 23:29 | comment | added | Yaroslav Bulatov | Thanks, that approach makes sense. Is there a name for this decomposition of A? For my application, I actually just need the trace(X)/norm(X) for some solution X, which I suspect is the same for all X satisfying the constraints | |
| Sep 15, 2019 at 6:47 | history | answered | Federico Poloni | CC BY-SA 4.0 |