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Denote $ \rho(A)$ to be the spectral radius of a matrix $A,$ that is the maximal eigenvalue of $A.$ We say that a matrix $M$ is positive definite, respectively positive semidefinite, if $x^TMx>0$ and $ x^TMx \geq 0$ respectively for all vectors $x$ with nonzero entries.

I want to show that if $ \rho(A)>1,$ then there exists a real symmetric matrix $B$ that is not positive semidefinite such that $A^TBA - B = -C$ holds for some positive definite matrix $C.$

Any hints or proof would be appreciated.

I've shown the part where, if $ \rho(A)<1$ then for every positive definite matrix $C,$ $ A^TBA - B = -C$ has a unique solution $B$ that is also symmetric and positive definite. ;)

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