<s>[INST] $A+B$ is non singular and $C=(A+B)^{-1}(A-B)$,then prove that $C^TAC=A$ If $A$ is a symmetric and $B$ is a skew symmetric matrix and $A+B$ is non singular and $C=(A+B)^{-1}(A-B)$,then prove that $C^TAC=A$.

My Attempt:
$C^T=((A+B)^{-1}(A-B))^T=(A-B)^T((A+B)^{-1})^T=(A^T-B^T)((A+B)^T)^{-1}$
$C^T=(A+B)(A-B)^{-1}$
$C^TAC=(A+B)(A-B)^{-1}A(A+B)^{-1}(A-B)$
But i am stuck here and could not solve further.Please help me.Thanks.
 [/INST] First of all, we show that $C^T(A+B)C=A+B.$ To do that:
$$\begin{align}(A+B)(A-B)^{-1}(A+B)(A+B)^{-1}(A-B) & \\ & \underbrace{=}_{(A+B)(A+B)^{-1}=I}(A+B)(A-B)^{-1}(A-B)\\ & \underbrace{=}_{(A-B)(A-B)^{-1}=I}A+B.\end{align}$$ 
In a similar way we can show that $C^T(A-B)C=A-B.$ Indeed:
$$\begin{align}(A+B)(A-B)^{-1}(A-B)(A+B)^{-1}(A-B) & \\ & \underbrace{=}_{(A-B)(A-B)^{-1}=I}(A+B)(A+B)^{-1}(A-B)\\ & \underbrace{=}_{(A+B)(A+B)^{-1}=I}A-B.\end{align}$$ 
Finally adding both equalities we get the desired result.
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