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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.


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:

    \[\left ( BA \right )\mathbf{x}= -\mathbf{x} \Rightarrow AB \left ( A \mathbf{x} \right )= -A\mathbf{x}\]

meaning that AB has an eigenvalue of -1 which is a contradiction. The result follows.

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