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. Let $\mathcal{A}^{\top} = \{ M^{\top} \mid M \in \mathcal{A} \}$.
Show that $\mathcal{A}^{\top}$ is a subalgebra of $\mathcal{M}_{n}(\mathbb{R})$ of the same dimension as $\mathcal{A}$.