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This question already has an answer here:

If $A$, $B$ are square matrices and $I-AB$ is invertible how do I prove that $I-BA$ is invertible?

This is exercise 8 of section 6.2 in Linear Algebra by Hoffman and Kunze.


My thoughts.

If $A$ and $B$ are invertible then $AB$ and $BA$ are similar, so we can use that to show that $I-AB$ and $I-BA$ are similar, and hence if $I-AB$ is invertible then so is $I-BA$.

However, $A$ and $B$ are not given to be invertible, so I am not able to apply this idea to show that $I-AB$ and $I-BA$ will be similar in general. Can anyone give me a hint to prove this in the general case?


EDIT: As pointed out in the comments below, it is not true in general that $I-AB$ and $I-BA$ are similar. @JulianRosen gave the example $$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}. $$ So, my original approach was completely incorrect. However, it is still true that $I-AB$ invertible implies $I-BA$ invertible.

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marked as duplicate by Brahadeesh, Adrian Keister, Leucippus, max_zorn, metamorphy Jan 4 at 2:47

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    $\begingroup$ $A=\left[\begin{array}{cc}0&1\\0&0\end{array}\right]$, $B=\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$ is a counterexample $\endgroup$ – Julian Rosen Nov 10 '14 at 17:45
  • $\begingroup$ we assume that I-AB is invertible $\endgroup$ – SSH Nov 10 '14 at 18:06
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    $\begingroup$ In this example $I-AB=\left[\begin{array}{cc}1&-1\\0&1\end{array}\right]$ is invertible $\endgroup$ – Julian Rosen Nov 10 '14 at 18:07
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    $\begingroup$ Do you mean that either $A$ or $B$ is invertible? Because then the statement would be correct $\endgroup$ – Omnomnomnom Nov 10 '14 at 18:12
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    $\begingroup$ @Salem then Julian's counterexample applies, and there's nothing we can do. If the question is correct, then there must be some part of the question that you're not sharing with us. $\endgroup$ – Omnomnomnom Nov 10 '14 at 18:17
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Hint: Note that the two matrices have the same determinant. There are several proofs of this, such as those given here.

Note that if either of $A$ and $B$ are invertible, we may further state that the two matrices are similar.

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Write $X = (I - AB)^{-1}$ and set $Y = I + BXA$. It is easy to check that $Y$ is the inverse of $I - BA$, but of course this solution will make you wonder how you come up with this in the first place.. see the mathoverflow question here.

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A linear operator on a finite-dimensional vector space is invertible if and only if it is injective, and thus if and only if its kernel is trivial. An $n \times n$ matrix over a field $F$ defines a linear operator on $F^n$.

Suppose $X \in F^n$ such that $(I-BA)X = 0$. We want to show that $X = 0$. Left-multiplying by $A$, we get $AX - AB(AX) = 0$. If we let $Y = AX$, then this just says that $(I - AB)Y = 0$. Since $I-AB$ is given to be invertible, this implies that $Y = 0$, that is, $AX = 0$. Now, from $(I-BA)X = 0$ we get $X = B(AX) = B0 = 0$. Hence proved.

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