We fix an element $A$ of $\mathcal{K}$ such that $\det A = \sup_{B \in \mathcal{K}} \det B$. Show that $\det A > 0$ and that there exists $x \in \mathcal{C}$ such that $\|Ax\| = 1$.
We fix an element $A$ of $\mathcal{K}$ such that $\det A = \sup_{B \in \mathcal{K}} \det B$.\\
Show that $\det A > 0$ and that there exists $x \in \mathcal{C}$ such that $\|Ax\| = 1$.