$A+A^2B+B=0$ implies $A^2+I$ invertible? Let $A$ and $B$ be two square matrices over a field such that $A+A^2B+B=0$.
Is it true that $A^2+I$ is always invertible ?

By taking the transpose, this is equivalent to asking whether $A + BA^2 + B = 0$ implies $A^2+I$ invertible. $(A^2+I)v = 0$ for some vector. Then $0 = (A + BA^2 + B) v = Av + B(A^2+I)v = Av$, therefore $v = (A^2+I)v = 0$. Therefore $A^2+I$ has trivial kernel and is invertible.
