$x^T(Q-I)x\geq 0 \Rightarrow x^T(Q^{-1}-I)x) \leq 0$ verification

Question

We have the condition that $Q$ is an $n\times n$ invertible matrix with all entries positive and it is NOT symmetric. It is given that $x^T(Q-I)x \geq 0\,\forall x\in \mathbb{R}^n$. My doubt is, can we show that $x^T(Q^{-1}-I)x\leq 0$.

I tried proving this in the following manner:

As $Q$ is invertible, its square root exists, then $\forall\, x,\,\exists \,y$ such that $x=Q^{-1/2}y$.
Thus, $y^TQ^{{-1/2}^T}(Q-I)Q^{-1/2}y\geq 0\,\forall y \in \mathbb{R}^n$ (this is where I am worried, can we say $y \in \mathbb{R}^n$??)
$\Rightarrow y^T(I-Q^{-1})y\geq 0\,\forall y \in \mathbb{R}^n$.

In short, my doubt is if $Q$ is an invertible matrix with positive entries, does square root exist in the set of real matrices?

Answer 1

Yes. Let $S=Q+Q^T-I$ and $K=Q-Q^T$. Then \begin{aligned} \forall x,\ x^T(Q-I)x\ge0 &\Leftrightarrow (Q-I)+(Q^T-I)\succeq0\\ &\Leftrightarrow S\succeq I\\ &\Leftrightarrow \lambda_\min(S)\ge1 \end{aligned} and \begin{aligned} \forall x,\ x^T(Q^{-1}-I)x\le0 &\Leftrightarrow (Q^{-1}-I)+(Q^{-1}-I)^T\preceq0\\ &\Leftrightarrow Q(Q^{-1}-I)Q^T+Q(Q^{-1}-I)^TQ^T\preceq0\\ &\Leftrightarrow Q+Q^T-2QQ^T\preceq0\\ &\Leftrightarrow (2Q-I)(2Q-I)^T\succeq I\\ &\Leftrightarrow (S+K)(S+K)^T\succeq I\\ &\Leftrightarrow \sigma_\min(S+K)\ge 1. \end{aligned} Therefore it suffices to show that $\lambda_\min(S)\ge1\Rightarrow\sigma_\min(S+K)\ge 1$ whenever $S$ is symmetric and $K$ is skew-symmetric.

Suppose $\lambda_\min(S)\ge1$. Let $u$ and $v$ be respectively a left and a right unit singular vector corresponding to the minimum singular value of $S+K$, so that $(S+K)v=\sigma_\min(S+K)u$. Then $$ 1\le\lambda_\min(S)\le v^TSv=v^T(S+K)v=\sigma_\min(S+K)v^Tu\le\sigma_\min(S+K). $$ Now we are done.