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?
Best Answer
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.