Estimate $\frac{(\frac{1}{2}\lambda_{\text{min}}(A+A^T))^2}{\lambda_{\text{max}}A^TA}<1$

Question

If $A$ is positive definite, (maybe not symmetric), how to prove that $$\frac{(\frac{1}{2}\lambda_{\text{min}}(A+A^T))^2}{\lambda_{\text{max}}A^TA}<1,$$ I know that $\lambda_{\text{max}}A^TA>(\lambda_{\text{max}}A)^2$

Answer 1

Presumably, $A$ is real and by a "positive definite" matrix that is not necessarily symmetric, you mean a matrix such that $x^TAx>0$ for all nonzero vector $x$. Such a matrix is actually the sum of a symmetric positive matrix $P$ and a skew-symmetric matrix $K$.

Therefore, $\frac12(A+A^T)=P$ and $A^TA=(P^T+K^T)(P+K)=P^2+K^TK+P^TK+K^TP$. So, the inequality in question is equivalent to $$ \lambda_\min(P^2)<\lambda_\max(P^2+K^TK+P^TK+K^TP).\tag{$\ast$} $$

Since $P$ is positive definite, the LHS is equal to $\lambda_\min(P)^2$. Let $x$ be a unit eigenvector of $P$ corresponding to $\lambda_\min(P)$. For any symmetric positive semidefinite matrix $M$, we have $\lambda_\max(M)=\max_{\|u\|_2=1}u^TMu$. Therefore, to prove $(\ast)$, it suffices to show that $\lambda_\min(P)^2<x^T(P^2+K^TK+P^TK+K^TP)x$. You may continue from here.