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
In fact, $AB$ and $BA$ have essentially the same characteristic polynomials. That is, assuming $m\le n,$ if the characteristic polynomial of $AB$ is $p(x),$ then the characteristic polynomial of $BA$ is $x^{n-m}p(x).$ The question of whether $AB+\mathbb{I}_m$ or $BA+\mathbb{I}_n$ is invertible is the question of whether $-1$ is a zero of these characteristic polynomials.
However, $-1$ cannot be an eigenvalue of $AB$ since then $AB+\mathbb{I}$ is not invertible, which is an obscurity. Now, for the sake of contradiction assume that $BA+\mathbb{I}$ is not invertible. Then $BA$ must have an eigenvalue of $-1$ and let $\mathbf{v}$ be its corresponding eigenvector. Thus:
$$\left ( BA \right )\mathbf{v}= -\mathbf{v} \Rightarrow AB \left ( A \mathbf{v} \right )= -A\mathbf{v}$$
meaning that $AB$ has an eigenvalue of $-1$, contradiction. This completes our proof.
Solution
In fact, $AB$ and $BA$ have essentially the same characteristic polynomials. That is, assuming $m\le n,$ if the characteristic polynomial of $AB$ is $p(x),$ then the characteristic polynomial of $BA$ is $x^{n-m}p(x).$ The question of whether $AB+\mathbb{I}_m$ or $BA+\mathbb{I}_n$ is invertible is the question of whether $-1$ is a zero of these characteristic polynomials.
However, $-1$ cannot be an eigenvalue of $AB$ since then $AB+\mathbb{I}$ is not invertible, which is an obscurity. Now, for the sake of contradiction assume that $BA+\mathbb{I}$ is not invertible. Then $BA$ must have an eigenvalue of $-1$ and let $\mathbf{v}$ be its corresponding eigenvector. Thus:
$$\left ( BA \right )\mathbf{v}= -\mathbf{v} \Rightarrow AB \left ( A \mathbf{v} \right )= -A\mathbf{v}$$
meaning that $AB$ has an eigenvalue of $-1$, contradiction. This completes our proof.
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