Bound $\frac{||W^Tx||}{||Wx||}$ for non-symmetric $W$.

Question

Let $W\in\mathbb{R}^{d\times d}$ and consider $n:=\frac{||W^Tx||}{||Wx||}$ for $x\in\mathbb{R}^d$ with $||x||=1$. If $W=W^T$ then $n=1$. Can we say anything about $n$ in terms of the eigenvalues or singular values of $W$ when $W$ is not symmetric?

Answer 1

Using SVD we can get $\frac{||V\Sigma U^Tx||}{||U\Sigma V^Tx||}=\frac{||\Sigma U^Tx||}{||\Sigma V^Tx||}\le \frac{ \sigma_{max} ||U^Tx||}{\sigma_{min}||V^Tx||}=\frac{\sigma_{max}}{\sigma_{\min}}$.

One can construct $W$ and $x$ for which the bound is tight as follows. Pick a diagonal matrix $\Sigma$ with sorted $\Sigma_1\ge ... \ge \Sigma_d$. Then choose $x$ and $U,V$ such that $U^Tx=e_1$ and $V^Tx=e_d$. For $W=U\Sigma V^T$ the bound is tight for $x$, that is, $\frac{||W^Tx||}{||Wx||}=\frac{||\Sigma U^Tx||}{||\Sigma V^Tx||}=\frac{||\Sigma e_1||}{||\Sigma e_d||} =\frac{\sigma_{max}}{\sigma_{min}}$.