We fix a polynomial $f \in \mathbb{C}[X]$ of degree $n \geq 1$. We consider a complex number $z \in \overline{\mathbb{D}}$ and we define the matrices $M \in \mathcal{M}_{n+1}(\mathbb{C})$ and $P \in \mathcal{M}_{n+1,1}(\mathbb{C})$ by $$M = \left(\begin{array}{cccccc} z & 0 & 0 & \ldots & 0 & \sqrt{1-|z|^2} \\ \sqrt{1-|z|^2} & 0 & 0 & \ldots & 0 & -\bar{z} \\ 0 & & & & & 0 \\ 0 & & & & & 0 \\ \vdots & & I_{n-1} & & & \vdots \\ 0 & & & & & 0 \end{array}\right)$$ and $$P = \left(\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array}\right)$$ Prove Theorem 1: Let $f \in \mathbb{C}[X]$ be a polynomial. Then $$\sup_{z \in \overline{\mathbb{D}}} |f(z)| = \sup_{z \in \mathbb{S}} |f(z)|$$