# Matrices Inequality Proof

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?

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.
• 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. – Hamio Jiang Mar 14 at 12:29
• @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. – user1551 Mar 14 at 14:19
• I mean the $A$ as you defined. So my question is why $A$ is positive semidefinite. Could you explain it further? Thanks. – Hamio Jiang Mar 14 at 14:29
• @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. – user1551 Mar 14 at 15:25