Proof if $I+AB$ invertible then $I+BA$ invertible and $(I+BA)^{-1}=I-B(I+AB)^{-1}A$ 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!
 A: 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:
$$
\begin{array}{rcl}
(I + BA)(1 - B (1 + AB)^{-1}A) &=& I + BA - B(I + AB)^{-1}A - BAB(I + AB)^{-1}A \\
&=& I + BA - B((I+AB)(I+AB)^{-1})A \\
&=& I + BA - BIA \\
&=& I + BA - BA \\
&=& I
\end{array}
$$
A: $\large{\text{Neumann}}:$ (supposing $\rho (AB) <1$)
$$(I+AB)^{-1}=I-AB+(AB)(AB)-(AB)(AB)(AB)+\cdots$$
$$B(I+AB)^{-1}A=BA-(BA)(BA)+(BA)(BA)(BA)+(BA)(BA)(BA)(BA)+\dots$$
$$B(I+AB)^{-1}A=-(I+BA)^{-1}+I$$
$$(I+BA)^{-1}=I-B(I+AB)^{-1}A$$
A: 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*}
A: Instead of multiplying the two matrices to get the identity matrix, we can also 'derive' the inverse by simple matrix multiplication. This way it would be clear why the inverse looks like the way it does.
It has been already shown by OP that $I+BA$ is invertible $\Leftrightarrow I+AB$ is invertible.
Let $ D={I+BA}$ $\qquad($$I,B$ and $A$ have orders such that $D$ is defined$)$
$\Rightarrow {DB}= B+{BAB}= B (I+{AB})$
$\Rightarrow (DB)(I+{AB})^{-1}= B$
$\Rightarrow ({DB})(I+{AB})^{-1} A={BA}$
$\Rightarrow  I+({DB})(I+{AB})^{-1} A=I+{BA}= D$
$\Rightarrow  I= D (I-{B}(I+{AB})^{-1} A)$
$\Rightarrow {D^{-1}}=I-{B}(I+{AB})^{-1} A$
$\text{i.e.,}\quad  \boxed{( {I+BA})^{-1}=I-{B}(I+{AB})^{-1} A}$
