Show that $(A+BC^{T})^{-1}=A^{-1}-A^{-1}B\left(I+C^{T}A^{-1}B\right)^{-1}C^{T}A^{-1}$

Question

Suppose $A\in \mathbb{R}^{n\times n}$ is invertible, $B,C\in\mathbb{R}^{n\times m}$ and $\left(I+C^{T}A^{-1}B\right)^{-1}$ exists. Show that $$ \begin{equation} (A+BC^{T})^{-1}=A^{-1}-A^{-1}B\left(I+C^{T}A^{-1}B\right)^{-1}C^{T}A^{-1} \end{equation} $$

I am having a lot of trouble with what appears to be simple algebraic manipulation. I have tried to approach this problem in many different ways (too many to write out), but here is the approach I used that seems to have got me the "furthest":

$$ \begin{align} A^{-1} &= \left(A+BC^{T}\right)^{-1}\left(A+BC^{T}\right)A^{-1} \\ &= \left(A+BC^{T}\right)^{-1}\left(I + BC^{T}A^{-1}\right) \\ &= \left(A+BC^{T}\right)^{-1} + \left(A+BC^{T}\right)^{-1}BC^{T}A^{-1} \\ \\ A^{-1} - \left(A+BC^{T}\right)^{-1}&= \left(A+BC^{T}\right)^{-1}BC^{T}A \\ \left(A+BC^{T}\right)^{-1}&=A^{-1} - \left(A+BC^{T}\right)^{-1}BC^{T}A^{-1} \\ &= A^{-1}-A^{-1}A\left(A+BC^{T}\right)^{-1}BC^{T}A^{-1} \\ &= A^{-1} - A^{-1}\left[\left(A+BC^{T}\right)A^{-1}\right]^{-1}BC^{T}A^{-1} \\ &= A^{-1}-A^{-1}\left(I+BC^{T}A^{-1}\right)^{-1}BC^{T}A^{-1} \end{align} $$

What do you get when you make the product $$\big( A+BC^t \big) \big( A^{-1}-A^{-1}B(I+C^tA^{-1}B)^{-1}C^tA^{-1} \big)$$ ? If its the identity matrix, necessarily $$\big( A+BC^t \big)^{-1} = \big( A^{-1}-A^{-1}B(I+C^tA^{-1}B)^{-1}C^tA^{-1} \big)$$ — azif00, Aug 07 '19 at 01:50
When you wrote $A^{-1} = \left(A+BC^{T}\right)^{-1}\left(A+BC^{T}\right)A^{-1}$, you needed first to verify the invertibility of $\left(A+BC^{T}\right)$. — Feng, Aug 07 '19 at 02:07
@Azif00 Thank you, yes I have been able to do this and show that the product is indeed equal to the identity, but how can I derive the result directly (ie. starting from $(A+BC^{T})^{-1}$ (and assuming that it is invertible)? — Andrew, Aug 07 '19 at 02:08
@Andrew If you'd like to derive the result directly, I suggest following the blockwise elimination proof given on wikipedia — Ben Grossmann, Aug 07 '19 at 15:29
@Andrew the proof can be framed similarly to how I prove the Sherman Morrison formula here — Ben Grossmann, Aug 07 '19 at 15:34

Show that $(A+BC^{T})^{-1}=A^{-1}-A^{-1}B\left(I+C^{T}A^{-1}B\right)^{-1}C^{T}A^{-1}$

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