Invertible matrix

Consider the matrices $A \in \mathcal{M}_{m \times n}$ and $B \in \mathcal{M}_{n \times m}$ . If $AB +\mathbb{I}_m$ is invertible prove that $BA+\mathbb{I}_n$ is also invertible.

Solution

So we have to answer the question if $-1$ is a zero of the essentially same characteristic polynomials. $AB$ and $BA$ have quite similar characteristic polynomials. In fact if $p(x)$ denotes the polynomial of $AB$ , then the polynomial of $BA$ will be $q(x)= x^{n-m} p(x)$ . It is easy to see that $-1$ cannot be an eigenvalue of the $AB$ matrix, otherwise it wouldn’t be invertible. Now, let us assume that $BA$ is not invertible. Then it must have an eigenvalue of $-1$ and let $\mathbf{x}$ be the corresponding eigenvector. Hence: