Proof if $I+AB$ invertible then $I+BA$ invertible and $(I+BA)^{-1}=I-B(I+AB)^{-1}A$

Question

I have the following question :

Proof if $I+AB$ invertible then $I+BA$ invertible and $(I+BA)^{-1}=I-B(I+AB)^{-1}A$

I managed to proof that $I+BA$ invertible

My proof :

We know that $AB$ and $BA$ has the same eigenvalues, and Since $I+AB$ invertible $-1$ is not an eigenvalue for $I+AB$ since if $-1$ is an eigenvalue then $I+AB$ is singular which is a contradiction. and since $AB$ and $BA$ has the same eigenvalues then $-1$ is also not an eigenvalue for $I+BA$ therefore $I+BA$ is also invertible.

But how do I show that $(I+BA)^{-1}=I-B(I+AB)^{-1}A$

I tried to "play" with the equations to reach one end to other end meaning that $(I+BA)^{-1}=...=...=...=I-B(I+AB)^{-1}A$

Or to show that $ I=(I+BA)^{-1}(I-B(I+AB)^{-1}A)$

But wasn't successful.

Any ideas?

Thank you!

Answer 1

I take it that $A$ and $B$ are $n \times n$ matrices here. In that context, $X$ is invertible with inverse $Y$ iff $XY = I$. Here we have: $$ \begin{array}{rcl} (I + BA)(1 - B (1 + AB)^{-1}A) &=& I + BA - B(I + AB)^{-1}A - BAB(I + AB)^{-1}A \\ &=& I + BA - B((I+AB)(I+AB)^{-1})A \\ &=& I + BA - BIA \\ &=& I + BA - BA \\ &=& I \end{array} $$

Answer 2

$\large{\text{Neumann}}:$ (supposing $\rho (AB) <1$) $$(I+AB)^{-1}=I-AB+(AB)(AB)-(AB)(AB)(AB)+\cdots$$ $$B(I+AB)^{-1}A=BA-(BA)(BA)+(BA)(BA)(BA)+(BA)(BA)(BA)(BA)+\dots$$ $$B(I+AB)^{-1}A=-(I+BA)^{-1}+I$$ $$(I+BA)^{-1}=I-B(I+AB)^{-1}A$$

Answer 3

Just make the product: \begin{align*} (I+BA)(I-B(I+AB)^{-1}A) ={}& I-B(I+AB)^{-1}A + BA - BAB(I+AB)^{-1}A={} \\ {}={}& I-B(I+AB)^{-1}A + BA - B(I + AB - I)(I+AB)^{-1}A ={} \\ {}={}& I-B(I+AB)^{-1}A + BA - B((I + AB)(I+AB)^{-1} - (I+AB)^{-1})A ={} \\ {}={}& I-B(I+AB)^{-1}A + BA - B(I - (I+AB)^{-1})A {} \\ {}={}& I-B(I+AB)^{-1}A + BA - BA + B(I+AB)^{-1}A = I \end{align*}