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 Post subject: Invertible matrixPosted: Sun Jan 24, 2016 10:40 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 828
Location: Larisa
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.

(Romania, 2012)

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 Post subject: Re: Invertible matrixPosted: Fri Aug 26, 2016 5:07 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 137
Location: Melbourne, Australia
What the question basically asks is if $-1$ is a zero of the essentially same characteristic polynomials. $AB$ and $BA$ have quite similar characteristic polynomials. In fact if denote $p(x)$ 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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