Matrix identities (for invertible matrices) I've got a little question concerning matrix operations. I am supposed to prove the following equation:
$(I + CBC^T)^{-1} = I - C(C^TC + B ^{-1})^{-1}C^T$
B and C are assumed to be invertible real square (n x n) matrices and I is the identity matrix (n x n). Also, $(C^TC + B ^{-1})^{-1}$ is assumed to exist.
Can anybody give me a hint how that's supposed to be done? I know some basic identities for transposed matrices like $(A^T)^T = A$ and $(AB)^T = B^TA^T$ but I don't know how to apply them here.
Thanks for your help!
 A: Let's go by the definition of the inverse:
$$I = (I + CBC^T)(I + CBC^T)^{-1}.$$
Let us first do some simplification:
$$I - C(C^TC + B^{-1})^{-1} C^T = I - (C^{-T} (C^TC) C^{-1} + C^{-T} B^{-1} C^{-1})^{-1} = I - (I + (CBC^T)^{-1})^{-1}.$$
So, we check:
\begin{align*}
(I + CBC^T) &(I - C(C^TC + B^{-1})^{-1} C^T) = (I + CBC^T) (I - (I + (CBC^T)^{-1})^{-1}) \\
&= (I + CBC^T) - (I + CBC^T)(I + (CBC^T)^{-1})^{-1}.\tag{1}
\end{align*}
It would be nice if we had:
$$(I + CBC^T)(I + (CBC^T)^{-1})^{-1} \stackrel{?}{=} CBC^T.$$
So, multiply it by $(CBC^T)^{-1}(CBC^T) = I$:
\begin{align*}
(I + CBC^T)&(I + (CBC^T)^{-1})^{-1} = (I + CBC^T)(I + (CBC^T)^{-1})^{-1}(CBC^T)^{-1}(CBC^T) \\
&= (I + CBC^T) \left( (CBC^T)(I + (CBC^T)^{-1}) \right)^{-1}(CBC^T) \\
&= (I + CBC^T) \left( CBC^T + I \right)^{-1}(CBC^T) = CBC^T.\tag{2}
\end{align*}
Using $(1)$ and $(2)$, we have:
\begin{align*}
(I + CBC^T) &(I - C(C^TC + B^{-1})^{-1} C^T) = (I + CBC^T) - (I + CBC^T)(I + (CBC^T)^{-1})^{-1} \\
&= I + CBC^T - CBC^T = I.
\end{align*}
A: Just as when manipulating real number identities, you can perform the same operation to LHS and RHS preserving the identity/non-identity property if the operation you use is invertible. So, starting with
$$
(I+CBC^T)^{-1}=I-C(C^TC+B^{-1})^{-1}C^T\\
$$
multiply both LHS and RHS on the left by $(I+CBC^T)$
$$
I=(I+CBC^T)(I-C(C^TC+B^{-1})^{-1}C^T)\\
$$
use now the distributive law of the product w.r.t. to sum
$$
I=I+CBC^T-(I+CBC^T)C(C^TC+B^{-1})^{-1}C^T\\
$$
subtract $I+CBC^T $ from both LHS and RHS
$$
-CBC^T=-(I+CBC^T)C(C^TC+B^{-1})^{-1}C^T\\
$$
multiply both LHS and RHS on the right by $(C^T)^{-1}$
$$
-CB=-(I+CBC^T)C(C^TC+B^{-1})^{-1}\\
$$
multiply both LHS and RHS on the right by $(C^TC+B^{-1})$
$$
-CB(C^TC+B^{-1})=-(I+CBC^T)C\\
$$
use the distributive law again
$$
-CBC^TC-C=-C-CBC^TC\\
$$
and you finally get an identity since sum is commutative.
