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Let $\textbf{M} = I-\mathbf{A}$ (with $\mathbf{A}^T = -\mathbf{A}$) be a matrix and $\textbf{M}^{-1}$ its inverse. From the identities:

\begin{equation} \textbf{M}^{-1}\textbf{M} = I\\ \textbf{M}\textbf{M}^{-1} = I. \end{equation} it is easy to see that: \begin{equation} \textbf{M}\textbf{M}^{-1} \left(\textbf{M}^{-1}\right)^{T}\textbf{M}^T = I. \end{equation}

Now I have to show that $B = (1+\textbf{A})(1-\textbf{A})^{-1}$ is orthogonal. To do so I wanted to show that $\mathbf{B}\mathbf{B}^T = I$, so I have to show that: $\mathbf{M}^T\mathbf{M}^{-1}\left(\mathbf{M}^{-1}\right)^T\mathbf{M}=I$ holds, but I've been unable to do it. What am I missing?

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    $\begingroup$ Are you looking for $M(M^TM)^{-1}M^T=I$? Or in general $A(A^TA)^{-1}A^T=\pi_A$ is the projection onto the image of $A$. This is a building block of the Moore-Penrose pseudo-inverse. $\endgroup$ – LutzL May 9 '18 at 22:26
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This identity is not correct. If $M = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ and thus $M^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$, then $M^TM^{-1}(M^{-1})^TM = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} $.

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  • $\begingroup$ Hi @Hans Engler... I've edited the question. I turns out that with the lack of information what you say is true. Now the thing is that $\mathbf{M} = I - \mathbf{A}$ and $\mathbf{A}$ is asymmetric. $\endgroup$ – Edgar Ortiz May 10 '18 at 12:22
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Just insert everything and simplify $$ B=(I+A)(I-A)^{-1}=2(I-A)^{-1}-I~\text{ where } A=-A $$ then $$ B^T=(I-A^T)^{-1}(I+A^T)=(I+A)^{-1}(I-A)=\left((I-A)^{-1}(I+A)\right)^{-1} $$ The last observation you need is that as $I$ and $A$ commute, so do $(I+A)$, $(I-A)$ and $(I-A)^{-1}$. Thus one concludes that $$B^T=B^{-1}.$$

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