Throughout this part, $\mathcal{A}$ is a subalgebra of $\mathcal{M}_{n}(\mathbb{R})$ strictly contained in $\mathcal{M}_{n}(\mathbb{R})$ and we denote by $d$ its dimension. We denote by $\mathcal{A}^{\perp}$ the orthogonal complement of $\mathcal{A}$ in $\mathcal{M}_{n}(\mathbb{R})$ and we denote by $r$ its dimension. We fix a basis $(A_{1}, \ldots, A_{r})$ of $\mathcal{A}^{\perp}$. Let $M \in \mathcal{M}_{n}(\mathbb{R})$. Show that $M$ belongs to $\mathcal{A}$ if and only if, for all $i \in \llbracket 1, r \rrbracket$, $\langle A_{i} \mid M \rangle = 0$.
Throughout this part, $\mathcal{A}$ is a subalgebra of $\mathcal{M}_{n}(\mathbb{R})$ strictly contained in $\mathcal{M}_{n}(\mathbb{R})$ and we denote by $d$ its dimension. We denote by $\mathcal{A}^{\perp}$ the orthogonal complement of $\mathcal{A}$ in $\mathcal{M}_{n}(\mathbb{R})$ and we denote by $r$ its dimension. We fix a basis $(A_{1}, \ldots, A_{r})$ of $\mathcal{A}^{\perp}$.
Let $M \in \mathcal{M}_{n}(\mathbb{R})$. Show that $M$ belongs to $\mathcal{A}$ if and only if, for all $i \in \llbracket 1, r \rrbracket$, $\langle A_{i} \mid M \rangle = 0$.