We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 2$. Let $F = H$ be a hyperplane of $E_{n}$ and let $N \in E_{n}$ be a unit vector normal to $H$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$. Show that there exists a matrix $B = \begin{pmatrix} B_{1} & B_{2} \\ B_{3} & B_{4} \end{pmatrix}$ with $B_{1} \in \mathcal{M}_{n}(\mathbb{R}), B_{2} \in \mathcal{M}_{n,1}(\mathbb{R}), B_{3} \in \mathcal{M}_{1,n}(\mathbb{R})$, $B_{4} \in \mathcal{M}_{1}(\mathbb{R})$ such that: $$A_{N}B = \begin{pmatrix} I_{n} & 0 \\ N^{\top}A^{-1} & -N^{\top}A^{-1}N \end{pmatrix}$$
We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 2$. Let $F = H$ be a hyperplane of $E_{n}$ and let $N \in E_{n}$ be a unit vector normal to $H$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$.
Show that there exists a matrix $B = \begin{pmatrix} B_{1} & B_{2} \\ B_{3} & B_{4} \end{pmatrix}$ with $B_{1} \in \mathcal{M}_{n}(\mathbb{R}), B_{2} \in \mathcal{M}_{n,1}(\mathbb{R}), B_{3} \in \mathcal{M}_{1,n}(\mathbb{R})$, $B_{4} \in \mathcal{M}_{1}(\mathbb{R})$ such that:
$$A_{N}B = \begin{pmatrix} I_{n} & 0 \\ N^{\top}A^{-1} & -N^{\top}A^{-1}N \end{pmatrix}$$