3
$\begingroup$

Recently, I read a paper and there is a step which turns out not obvious to me. The statement is as follows:

All matrices here are real matrices. $F$ is an arbitrary square matrix. $\Psi$ is a symmetric positive definite matrix. Let $$\lambda_{\max}(A)\equiv\text{The maximum eigenvalue of symmetric matrix A}$$ (The ambiguity comes when $A$ is not symmetric. Here I guess if $A$ is not symmetric, then $\lambda_{\max}(A)=\sqrt{\text{Maximum eigenvalue of }A^TA}$ ). Then the following inequality holds:

For all $x\in \mathbb R^n$ $$x^T(I-F)^T\Psi(I-F)x\le\lambda_\max(\Psi^{-1}(I-F)^T\Psi(I-F))x^T\Psi x $$ rewrite it, $$x^T\Big[(I-F)^T\Psi(I-F)-\lambda_\max(\Psi^{-1}(I-F)^T\Psi(I-F))\Psi\Big]x\le0 $$ or $$x^T\Psi\Big[\Psi^{-1}(I-F)^T\Psi(I-F)-\lambda_\max(\Psi^{-1}(I-F)^T\Psi(I-F))I\Big]x\le0\tag{*} $$ and if $\Psi$ commutes with $(I-F)^T\Psi(I-F)$, then, $\Psi^{-1},\,(I-F)^T\Psi(I-F)$ can be simultaneously diagnolized. Then $$\Psi^{-1}(I-F)^T\Psi(I-F)-\lambda_\max(\Psi^{-1}(I-F)^T\Psi(I-F))I $$ is negatively semi-definite and diagnolized under certain basis, same as $\Psi$. Then under the basis, since $\Psi$ is positive definite, $\Psi\Big[\Psi^{-1}(I-F)^T\Psi(I-F)-\lambda_\max(\Psi^{-1}(I-F)^T\Psi(I-F))I\Big]$ is negative semi-definite$\Rightarrow$ the inequality holds.

However, in general, $\Psi$ may not commute with $(I-F)^T\Psi(I-F)$. Are there any answers to that?

$\endgroup$
1
$\begingroup$

Let $A=\Psi^{-1/2}(I-F)^T\Psi(I-F)\Psi^{-1/2}$ and $y=\Psi^{1/2}x$. Then $\Psi^{-1}(I-F)^T\Psi(I-F)=\Psi^{-1/2}A\Psi^{1/2}$ is similar to $A$ and hence the inequality in question can be rewritten as $$ y^TAy\le\lambda_\max(A)y^Ty. $$ Now the inequality holds because $A$ is positive semidefinite.

$\endgroup$
  • $\begingroup$ Hello, thanks for the hint, but I have a question. Here $A$ needs not to be positive semidefinite since $(I-F)$ is arbitrary. However, as long as $A$ is a symmetric real matrix, it can be diagonalized under a certain base. Then $A-\lambda_\max(A)I$ is a negative semidefinite matrix. $\endgroup$ – Hamio Jiang Mar 14 at 12:29
  • $\begingroup$ @HamioJiang Does the paper you read have an $A$? If so, the $A$ in my answer is not your $A$, but the matrix defined in the first sentence of my answer. The value of $F$ is irrelevant. As long as $\Psi$ is positive definite, $A$ is semidefinite. $\endgroup$ – user1551 Mar 14 at 14:19
  • $\begingroup$ I mean the $A$ as you defined. So my question is why $A$ is positive semidefinite. Could you explain it further? Thanks. $\endgroup$ – Hamio Jiang Mar 14 at 14:29
  • 1
    $\begingroup$ @HamioJiang $A$ is clearly symmetric. For any nonzero vector $x$, let $v=(I-F)\Psi^{-1/2}x$. Then $x^TAx=v^T\Psi v\ge0$ because $\Psi$ is positive definite (note that $v$ may be zero because $I-F$ can be singular; therefore, we can only say that $v^T\Psi v\ge0$ but not that $v^T\Psi v>0$), meaning that $A$ is positive semidefinite. $\endgroup$ – user1551 Mar 14 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.