We consider the space $E = \mathcal{M}_{k,d}(\mathbb{R})$ equipped with the inner product defined by $$\forall (A, B) \in E^{2}, \quad \langle A \mid B \rangle = \operatorname{tr}\left(A^{\top} \cdot B\right)$$ We denote by $\|\cdot\|_{F}$ the associated Euclidean norm. We fix a vector $u = (u_{1}, \ldots, u_{d})$ in $\mathbb{R}^{d}$ with $\|u\| = 1$, and define $$g : \left\lvert \, \begin{aligned} & \mathcal{M}_{k,d}(\mathbb{R}) \rightarrow \mathbb{R} \\ & M \mapsto \|M \cdot u\| \end{aligned} \right.$$ Show that $C = \left\{M \in \mathcal{M}_{k,d}(\mathbb{R}) \mid g(M) \leqslant r\right\}$ is a convex and closed subset of $\mathcal{M}_{k,d}(\mathbb{R})$.
We consider the space $E = \mathcal{M}_{k,d}(\mathbb{R})$ equipped with the inner product defined by
$$\forall (A, B) \in E^{2}, \quad \langle A \mid B \rangle = \operatorname{tr}\left(A^{\top} \cdot B\right)$$
We denote by $\|\cdot\|_{F}$ the associated Euclidean norm. We fix a vector $u = (u_{1}, \ldots, u_{d})$ in $\mathbb{R}^{d}$ with $\|u\| = 1$, and define
$$g : \left\lvert \, \begin{aligned} & \mathcal{M}_{k,d}(\mathbb{R}) \rightarrow \mathbb{R} \\ & M \mapsto \|M \cdot u\| \end{aligned} \right.$$
Show that $C = \left\{M \in \mathcal{M}_{k,d}(\mathbb{R}) \mid g(M) \leqslant r\right\}$ is a convex and closed subset of $\mathcal{M}_{k,d}(\mathbb{R})$.