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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.

SSH
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    $A=\left[\begin{array}{cc}0&1\0&0\end{array}\right]$, $B=\left[\begin{array}{cc}0&0\0&1\end{array}\right]$ is a counterexample – Julian Rosen Nov 10 '14 at 17:45
  • we assume that I-AB is invertible – SSH Nov 10 '14 at 18:06
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    In this example $I-AB=\left[\begin{array}{cc}1&-1\0&1\end{array}\right]$ is invertible – Julian Rosen Nov 10 '14 at 18:07
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    Do you mean that either $A$ or $B$ is invertible? Because then the statement would be correct – Ben Grossmann Nov 10 '14 at 18:12
  • @Omnomnomnom (I-AB) is invertible – SSH Nov 10 '14 at 18:16
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    @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. – Ben Grossmann Nov 10 '14 at 18:17
  • @JulianRosen what about (I-BA) I think it is invertible as well – SSH Nov 10 '14 at 18:18
  • @Salem both of those are invertible in the example given. We have $I -BA = I$ – Ben Grossmann Nov 10 '14 at 18:19
  • @Omnomnomnom I want to prove that (I-BA) is invertible. So my idea is to prove that (I-AB) and (I-BA) are similar – SSH Nov 10 '14 at 18:22
  • @Salem that's a very different question. You should edit your post to indicate this. – Ben Grossmann Nov 10 '14 at 18:24
  • The Hoffman-Kunze problem actually gives a hint: show that $\left(I-BA\right)^{-1} = I + B \left(I-AB\right)^{-1} A$. Notice that this does not require $A$ and $B$ to be square as long as $AB$ is well-defined. – darij grinberg Nov 10 '14 at 18:48
  • @darijgrinberg Yes you are right I'll use the hint, but I want to see if there is any other way to prove it using the determinants properties – SSH Nov 10 '14 at 18:56

3 Answers3

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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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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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Ben Grossmann
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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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