1 problem found
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. {[}You may quote standard results about matrices without proof.{]}
Solution: 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*}