Problem source

For any square matrix $\mathbf{A}$ such that $\mathbf{I-A}$ is non-singular
		(where $\mathbf{I}$ is the unit matrix), the matrix $\mathbf{B}$
		is defined by 
		\[
		\mathbf{B}=(\mathbf{I+A})(\mathbf{I-A})^{-1}.
		\]
		Prove that $\mathbf{B}^{\mathrm{T}}\mathbf{B}=\mathbf{I}$ if and
		only if $\mathbf{A+A}^{\mathrm{T}}=\mathbf{O}$ (where $\mathbf{O}$
		is the zero matrix), explaining clearly each step of your proof.

		{[}\textit{You may quote standard results about matrices without proof.}{]}

Solution source

We use the following properties: $(\mathbf{AB})^T = \mathbf{B}^T\mathbf{A}^T$, $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$, and $(\mathbf{A}^T)^{-1} = (\mathbf{A}^{-1})^T$

\begin{align*}
&&\mathbf{I} &= \mathbf{B}^{\mathrm{T}}\mathbf{B} \\
\Leftrightarrow && {\mathbf{B}^{T}}^{-1} &= \mathbf{B} \\
\Leftrightarrow && (\mathbf{I+A})(\mathbf{I-A})^{-1} &= {((\mathbf{I+A})(\mathbf{I-A})^{-1})^{T}}^{-1}   \\
&&&= {((\mathbf{I+A})(\mathbf{I-A})^{-1})^{-1}}^{T} \\
&&&=  ((\mathbf{I-A})(\mathbf{I+A})^{-1})^{T} \\
&&&= {(\mathbf{I+A})^{-1}}^T(\mathbf{I-A})^T \\
&&&= {(\mathbf{I+A}^T)}^{-1}(\mathbf{I-A}^T) \\
\Leftrightarrow && (\mathbf{I+A}^T)(\mathbf{I+A}) &= (\mathbf{I-A}^T)(\mathbf{I-A}) \\
\Leftrightarrow && \mathbf{I}+\mathbf{A}+\mathbf{A}^T+\mathbf{A}^T\mathbf{A} &= \mathbf{I}-\mathbf{A}-\mathbf{A}^T+\mathbf{A}^T\mathbf{A} \\
\Leftrightarrow && 2( \mathbf{A}^T+\mathbf{A}) &= \mathbf{O} \\
\Leftrightarrow &&  \mathbf{A}+\mathbf{A}^T &= \mathbf{O}
\end{align*}