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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?)

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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 $|\lambda|<1$.

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