Inverse of $ I-BA $ is $(I-BA)^{−1}=I−B(I-AB)^{−1}A$

Question

Assume $(I-AB)$ is invertible, prove then that $(I-BA)$ is invertible and $$(I-BA)^ { −1 }=I−B(I-AB)^{−1}A$$

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I am unsure of the steps I must take to solve this. Do I just multiply on the left than the right by $(I-BA)$

Answer 1

My original attempt: It suffices to simplify the product $(I - BA)[I−B(I-AB)^{−1}A]$ and show that this expression is equal to $I$. In particular, we have $$ \begin{align} (I - BA)&[I−B(I-AB)^{−1}A] = I - BA - B(I - AB)^{-1}A + BAB(I-AB)^{−1}A \\ & = I - BA - B(I - AB)^{-1}A + B[(AB - I) + I](I-AB)^{−1}A \\ & = I - BA - B(I - AB)^{-1}A + \color{red}{B(AB - I)(I-AB)^{−1}A} + B(I-AB)^{−1}A \\ & = I - BA - B(I - AB)^{-1}A + \color{red}{BA} + B(I - AB)^{-1}A = I \end{align} $$

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As J.G. Notes, we should have $(I - BA) = I \color{red}+ B(I-AB)^{−1}A$.

Here's an attempt at an alternative proof.

Claim: $B(I - AB)^{-1}A$

Proof of Claim: $AB$ and $BA$ have the same non-zero eigenvalues. Note that if $AB$ is diagonalizable, then one can use this to find a polynomial $f(x)$ such that $f(AB) = (I - AB)^{-1}$ and $f(BA) = (I - BA)^{-1}$, and the statement follows. The general statement can be proved using continuity.

It follows that $$ I + B(I - AB)^{-1}A \\= I + (I - BA)^{-1}BA \\ = I + (I - BA)^{-1}[(BA - I) + I] \\ = I - I + (I - BA)^{-1} \\ = (I - BA)^{-1}. $$

Answer 2

@Omnomnomnom's strategy works. The issue is that the original problem statement had a sign error. In fact, we want to prove$$(I-BA)[I+B(I-AB)^{-1}A]=I.$$Indeed,$$\begin{align}(I-BA)[I+B(I-AB)^{-1}A]&=I-BA+B(I-AB)^{-1}A-BAB(I-AB)^{-1}A\\&=I-BA+B(I-AB)^{-1}A+B(I-AB-I)(I-AB)^{-1}A\\&=I-BA+B(I-AB)^{-1}A+BA-B(I-AB)^{-1}A\\&=I.\end{align}$$

Answer 3

Since $I+B(I-AB)^{-1}A$ is in the form of a Schur complement, we may also approach the problem with block matrices. Note that \begin{aligned} \underbrace{\pmatrix{I&B\\ A&I}}_Y\ \underbrace{\pmatrix{I&-B\\ 0&I}}_X &=\underbrace{\pmatrix{I&0\\ A&I-AB}}_Z,\\ \underbrace{\pmatrix{I&-B\\ 0&I}}_X\ \underbrace{\pmatrix{I&B\\ A&I}}_Y &=\underbrace{\pmatrix{I-BA&0\\ A&I}}_W.\\ \end{aligned} Thus $I-AB$ is invertible if and only if $I-BA$ is invertible, and \begin{aligned} \pmatrix{(I-BA)^{-1}&0\\ -A(I-BA)^{-1}&I} &=W^{-1} =Y^{-1}X^{-1} =XZ^{-1}X^{-1}\\ &=\pmatrix{I&-B\\ 0&I}\pmatrix{I&0\\ -(I-AB)^{-1}A&(I-AB)^{-1}}\pmatrix{I&B\\ 0&I}\\ &=\pmatrix{I&-B\\ 0&I}\pmatrix{I&\ast\\ -(I-AB)^{-1}A&\ast}\\ &=\pmatrix{I+B(I-AB)^{-1}A&\ast\\ \ast&\ast}. \end{aligned} Hence $(I-BA)^{-1}=I+B(I-AB)^{-1}A$.