Question 111
Let $A_1, A_2, \dots, A_m$ be $n \times n$ matrices with real elements, where $m, n \in \mathbb{N}^*$, and $b_1, b_2, \dots, b_m$ be real numbers. Consider the set
$$
\mathcal{B} = \{B \in \mathcal{M}_n(\mathbb{R}) \mid \operatorname{tr}(BA_i^T) = b_i, \forall i = \overline{1, m}\}.
$$
Suppose that the matrix $C = (c_{ij})_{i,j=\overline{1,m}} \in \mathcal{M}_m(\mathbb{R})$, $c_{ij} = \operatorname{tr}(A_i A_j^T)$ has $\det(C) \ne 0$. Prove that there exists a unique matrix $X \in \mathcal{B}$ so that
$$
\operatorname{tr}(XX^T) \le \operatorname{tr}(YY^T), \forall Y \in \mathcal{B}.
$$
$$
\mathcal{B} = \{B \in \mathcal{M}_n(\mathbb{R}) \mid \operatorname{tr}(BA_i^T) = b_i, \forall i = \overline{1, m}\}.
$$
Suppose that the matrix $C = (c_{ij})_{i,j=\overline{1,m}} \in \mathcal{M}_m(\mathbb{R})$, $c_{ij} = \operatorname{tr}(A_i A_j^T)$ has $\det(C) \ne 0$. Prove that there exists a unique matrix $X \in \mathcal{B}$ so that
$$
\operatorname{tr}(XX^T) \le \operatorname{tr}(YY^T), \forall Y \in \mathcal{B}.
$$