Let $A$ and $B$ be nonsingular $n\times n$ real matrices, while $B$ is symmetric positive definite and eigenvalues of $A$ are larger than $1$.
If $\mathrm{vec}(B)=(A\otimes A-I)^{-1}\mathrm{vec}(I)$, can we conclude that $\mathrm{vec}(B^{-1})=(A\otimes A-I)\mathrm{vec}(I)$?
For diagonal case it is obvious. Then if we assume that $A$ is in Jordan canonical form, maybe it also holds? Examples I have tried so far for $3\times 3$ matrices hold.
Here $\otimes$ represents Kronecker product.
Some thoughts:
The equation $\mathrm{vec}(B)=(A\otimes A-I)^{-1}\mathrm{vec}(I)$ means that $B$ solves the equation $$ ABA^T - B = I, $$ which indeed has a unique solution if the eigenvalues of $A$ are all larger than $1$ in magnitude. On the other hand, the equation $\operatorname{vec}(B^{-1}) = (A \otimes A - I)\operatorname{vec}(I)$ implies that $$ B^{-1} = AA^T - I \implies B = (AA^T - I)^{-1}. $$ Plugging this second equation into the first leads us to the conclusion that if this second condition is to hold, it must be the case that $$ A(AA^T - I)^{-1}A^T - (AA^T - I)^{-1} = I \implies\\ A(AA^T - I)^{-1}A^T(AA^T - I) - I = (AA^T - I) \implies\\ A(AA^T - I)^{-1}A^T(AA^T - I) = AA^T. $$ I doubt that this holds in general. However, it is easy to see that the equation will hold whenever $A$ is symmetric (i.e. $A = A^T$). In fact, it is sufficient for $A$ to satisfy $AAA^T = AA^TA$ (i.e. $A$ commutes with $AA^T$). Because $A$ is invertible, this is equivalent to the condition $AA^T = A^TA$ (i.e. $A$ is normal).
