Timeline for An inequality concerning the solution of a Lyapunov equation
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2018 at 17:19 | comment | added | Mahdi - Free Palestine | Oh, I supposed that $Q$ is invertible. | |
| Sep 11, 2018 at 15:31 | comment | added | Ludwig | @Mahdi: It seems to me that, for two matrices $X>0$, $Y\ge 0$, we have $\mathrm{tr}(X^{-1}Y)=\mathrm{tr}(Y)/\lambda_{\max}(X)$, for any choice of $X$ when $Y=v_{\min}v_{\min}^\top$, with $v_{\min}$ being the eigenvector corresponding to the smallest eigenvalue of $X^{-1}$. | |
| Sep 11, 2018 at 14:44 | comment | added | Mahdi - Free Palestine | For every Hermitian matrices $A$ and $B$, we have $\langle \lambda^{\downarrow}(A),\lambda^{\uparrow}(B) \rangle \leq tr(AB)$. So, if there exists $Q, P$ such that $tr(P^{-1}Q) = tr(Q) / \lambda_{max}(P)$, then all eigenvalues of $P$ are equal. am I missing something? | |
| Sep 11, 2018 at 14:28 | comment | added | Ludwig | @Mahdi: could you please elaborate a little more your comment? | |
| Sep 11, 2018 at 13:49 | history | edited | Ludwig | CC BY-SA 4.0 |
edit question for clarification
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| Sep 11, 2018 at 13:44 | comment | added | Ludwig | @AbhishekHalder: Yes, $f(Q)$ is indeed constant; however the RHS of ($\star$) depends on $Q$. I will edit the OP in order to clarify this fact. | |
| Sep 11, 2018 at 7:49 | comment | added | Abhishek Halder | I am surely missing something: why is $f(Q)$ not constant? Pre-multiplying both sides of Lyapunov equation with $P^{-1}$ and taking trace gives $f(Q) = -2\text{tr}(A)$? | |
| Sep 11, 2018 at 2:27 | history | edited | Ludwig | CC BY-SA 4.0 |
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| Sep 11, 2018 at 2:22 | history | edited | Ludwig | CC BY-SA 4.0 |
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| Sep 11, 2018 at 1:59 | history | asked | Ludwig | CC BY-SA 4.0 |