Timeline for Trace of a nonlinear matrix equation (cont'd)
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33 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Sep 21, 2016 at 16:40 | vote | accept | Ludwig | ||
| Sep 21, 2016 at 11:30 | answer | added | fedja | timeline score: 5 | |
| Sep 20, 2016 at 23:02 | comment | added | fedja | @NawafBou-Rabee Yeah, but that case is trivial because the trace is not less than the $n$-th root of the determinant (AM-GM), so it just blows up geometrically. | |
| Sep 20, 2016 at 22:55 | comment | added | Nawaf Bou-Rabee | The small font size makes it a bit difficult to read these comments. I see this point made in your earlier comment. With a slight abuse of terminology, I regard the case $\det(A)>1$ as also degenerate, in the sense that the dynamics is asymptotically unstable in this regime. | |
| Sep 20, 2016 at 22:41 | comment | added | fedja | @NawafBou-Rabee As I said from the very beginning, $\det A<1$ is the only interesting case. Note that any power of a positive number is still positive, so no degenerate situations occur in that case. | |
| Sep 20, 2016 at 22:23 | comment | added | Nawaf Bou-Rabee | How can the entire sequence of operators converge to a degenerate operator given the dynamic and initial conditions, which imply that $\det(X_k) = \det(X_0) \det(A)^k$? My impression is that $\det(A)=1$ (in order to avoid degenerate situations), and thus, degenerate fixed points are inaccessible by this dynamic even asymptotically. | |
| Sep 20, 2016 at 21:45 | comment | added | fedja | @NawafBou-Rabee Yes, we can show that the entire sequence converges to a degenerate operator (degenerate fixed points are plentiful). What's next? | |
| Sep 20, 2016 at 21:38 | comment | added | Nawaf Bou-Rabee | If that is the case, and noting that the only fixed point occurs when $A=I_n$, then we get a contradiction because the sequence is clearly bounded (alternatively, the set of unit trace positive definite matrices is compact) and so there must be a convergent subsequence by Bolzano-Weierstrass. | |
| Sep 20, 2016 at 21:17 | comment | added | fedja | Certainly. Assume $A=diag(a_1,\dots,a_n)$. For a self-adjoint matrix $Z$, let $v(Z)$ be the vector whose coordinates are the diagonal entries of $Z$. Assume $Y^2=XAX$. Then $v(Y^2)=(X\circ X^T)v(A)$ while $v(X^2)=(X\circ X^T)v(I)$ where $Z\circ T$ is the Schur (element-wise) product of $Z$ and $T$. Since $X\circ X^T$ is positive-definite, we get $\langle v(Y^2)-v(X^2),v(A)-v(I)\rangle>0$ if $v(A)\ne v(I)$, so we drift in one direction every time with no chance to return. | |
| Sep 20, 2016 at 20:03 | comment | added | Ludwig | @fedja: Could you please explain how you proved that cycles of any length are impossible if $A\neq I$? (Even though it's not a solution, it might give more insights on the problem...) | |
| Sep 20, 2016 at 15:07 | comment | added | Ludwig | @NawafBou-Rabee: Yes, of course replacing the principal square root with the Cholesky square root makes the problem trivial. However the principal square root has remarkable properties that the Cholesky factor does not have (the most obvious one is that the principal square root is positive (semi)definite). In the problem I'm investigating, I need such properties. | |
| Sep 20, 2016 at 14:50 | comment | added | Nawaf Bou-Rabee | Why does the given iteration rule involve the square root factorization of the current state as opposed to the Cholesky factorization? More precisely, why not replace ($\star$) with $$X_{k+1} = L_k B (L_k B)^T \quad \text{where $X_k = L_k L_k^T$ and $A=B B^T$}$$ which admits a global solution given by $X_{k} = (L_0 B^{k}) (L_0 B^{k})^T$ that may be used to conclude (e.g., by contradiction) that $A=I_n$. | |
| Sep 20, 2016 at 8:26 | history | edited | Ludwig | CC BY-SA 3.0 |
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| Sep 20, 2016 at 7:32 | history | edited | Ludwig | CC BY-SA 3.0 |
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| Sep 20, 2016 at 7:16 | history | edited | Ludwig | CC BY-SA 3.0 |
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| Sep 20, 2016 at 6:50 | history | edited | Ludwig | CC BY-SA 3.0 |
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| Sep 20, 2016 at 4:47 | comment | added | fedja | @FedorPetrov I can show that cycles of any length are impossible if $A\ne I$. Alas, this doesn't help much with the original problem unless $X_j$ are uniformly non-degenerate (which is never true in the interesting case $\det A<1$), so we are still nowhere... | |
| Sep 19, 2016 at 19:03 | history | edited | Ludwig | CC BY-SA 3.0 |
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| Sep 19, 2016 at 6:13 | comment | added | Ludwig | @FedorPetrov: Yes, still it's not straightforward to me to find a pair of unit trace $X_0$, $X_1$ satisfying the above constraint for $A\neq I$. (In case of diagonal $X_0$, $X_1$ it's not possible, I would say). | |
| Sep 18, 2016 at 22:22 | comment | added | Fedor Petrov | Of course $A>0$ automatically: we need $X_0^{-1/2}X_1X_0^{-1/2}=A=X_1^{-1/2}X_0X_1^{-1/2}. | |
| Sep 18, 2016 at 22:17 | comment | added | Ludwig | @FedorPetrov: Since $A>0$ it looks improbable to me, but I could be wrong. | |
| Sep 18, 2016 at 22:04 | comment | added | Fedor Petrov | May there be a cycle of length 2, that is, $X_2=X_0$? I do not see why not. | |
| Sep 17, 2016 at 15:33 | answer | added | Nawaf Bou-Rabee | timeline score: 2 | |
| Sep 17, 2016 at 11:14 | answer | added | Ludwig | timeline score: 2 | |
| Sep 17, 2016 at 8:25 | history | edited | Ludwig | CC BY-SA 3.0 |
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| Sep 17, 2016 at 6:28 | history | edited | Ludwig |
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| Sep 16, 2016 at 19:35 | history | edited | Ludwig | CC BY-SA 3.0 |
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| Sep 16, 2016 at 19:26 | history | edited | Ludwig | CC BY-SA 3.0 |
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| Sep 16, 2016 at 19:21 | history | edited | Ludwig | CC BY-SA 3.0 |
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| S Sep 16, 2016 at 19:17 | history | suggested | Nawaf Bou-Rabee |
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| Sep 16, 2016 at 19:15 | review | Suggested edits | |||
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| Sep 13, 2016 at 14:49 | history | asked | Ludwig | CC BY-SA 3.0 |