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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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Presumably $A$ and $B$ are real. Your requirement is possible if and only if $1$ and $-1$ are not eigenvalues of $A$.

Suppose $1$ (or $-1$) is an eigenvalue of $A$ and $x$ is a corresponding eigenvector. Then your requirement implies that $B\succ A^TBA$, which is impossible because $x^TBx=x^TA^TBAx$.

Suppose $\pm1$ are not eigenvalues of $A$. By a change of basis ($B$ is a quadratic form and $A$ is a linear transformation; so they undergo the same change of basis), we may assume that $A$ is already in its real Jordan form. Let $B$ be a block diagonal matrix with block sizes conforming to $A$'s. We can solve the problem blockwise.

For each Jordan block $A_0$ of $A$ with eigenvalue modulus greater than $1$, set the corresponding diagonal subblock of $B$ to the negative definite matrix $B_0$, where $-B_0$ is the positive definite solution to the Lyapunov equation $A_0^{-T}(-B_0)A_0^{-1} - (-B_0) - I = 0$. For each Jordan block $A_0$ with eigenvalue modulus $\le1$ (the modulus can be $1$, it's just that the eigenvalues are a conjugate pair of nonreal eigenvalues on the unit circle), set the corresponding diagonal subblock of $B_0$ to a positive semidefinite solution to $A_0^TB_0A_0 - B_0 = -I$. Since $\rho(A)>1$, $B$ always possesses a negative definite diagonal subblock. Therefore $B$ is not positive semidefinite.

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