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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.

Vinod Kumar Punia
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1 Answers1

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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.

mfl
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