We assume that $\Sigma_Y$ has $n$ distinct eigenvalues which we order in strictly decreasing order $\lambda_1 > \cdots > \lambda_n$. We equip ourselves with a vector $U_0$ such that $\mathbb{V}\left(U_0^\top Y\right) = \max_{U \in C} \mathbb{V}\left(U^\top Y\right)$, where $C = \left\{ U \in \mathcal{M}_{n,1}(\mathbb{R}) \mid U^\top U = 1 \right\}$. We denote $$C' = \left\{ U \in \mathcal{M}_{n,1}(\mathbb{R}) \mid U^\top U = 1 \text{ and } U_0^\top U = 0 \right\}.$$ Justify that $q_Y$ admits a maximum on $C'$.
We assume that $\Sigma_Y$ has $n$ distinct eigenvalues which we order in strictly decreasing order $\lambda_1 > \cdots > \lambda_n$. We equip ourselves with a vector $U_0$ such that $\mathbb{V}\left(U_0^\top Y\right) = \max_{U \in C} \mathbb{V}\left(U^\top Y\right)$, where $C = \left\{ U \in \mathcal{M}_{n,1}(\mathbb{R}) \mid U^\top U = 1 \right\}$. We denote
$$C' = \left\{ U \in \mathcal{M}_{n,1}(\mathbb{R}) \mid U^\top U = 1 \text{ and } U_0^\top U = 0 \right\}.$$
Justify that $q_Y$ admits a maximum on $C'$.