Operator norm of $\frac{x^TW^Tx\cdot I-Wxx^T}{||Wx||^2}$.

Question

Let $x\in S^{d}=\{x\in \mathbb{R}^d \mid \| x\|=1\}$ and $W\in\mathbb{R}^{d\times d}$ how tight can we bound $\sigma_{\max} \left(\frac{x^TW^Tx\cdot I-Wxx^T}{||Wx||^2}\right)$ with respect to $\lambda_\min(W)$ and $\lambda_{max}(W)$?

Answer 1

Operator norm is infinite. Take $W$ such that $\|W\|=1$ and $x$ is the right eigenvector with eigenvalue $\lambda<1$

$$Wx=\lambda x$$

Applying operator gives you the following matrix with norm $1/\lambda$ $$\frac{I-xx'}{\lambda}$$