# Spectral radius of $B$ if $W-B^TWB$ is positive definite

Problem:

Suppose that $$W = S^TS$$ for some square matrix $$S$$, and that $$W-B^TWB$$ is positive definite. Show that the Spectral Radius of $$B$$ is less than $$1$$.

Attempt:

$$W = S^TS$$ is symmetric, so that $$W-B^TWB$$ is also symmetric. It follows that

$$W-B^TWB = P^TDP$$

where $$D = \text{diag}(\lambda_1,\dots,\lambda_n)$$ where $$\lambda_i>0$$ are the eigenvalues of $$W-B^TWB$$.

...and then I'm stuck. I'm not seeing the connection between $$W-B^TWB$$ and $$B$$. Any hints?

Does it help that the spectral radius of a matrix is equal to its $$2$$-norm? (or is this even true?)

Hint. Presumably $$B$$ is a real matrix, so that $$B^T=B^\ast$$. Suppose $$\rho(B)>0$$. Let $$v\in\mathbb C^n$$ be an eigenvector of $$B$$ corresponding to the largest-sized eigenvalue $$\lambda$$ of $$B$$. By considering $$v^\ast(W-B^TWB)v$$, prove that $$v^\ast Wv$$ is positive and $$|\lambda|<1$$.