$C*$-algebra Identity for Matrices

Question

What's the easiest way to see that the $n \times n$-matrices over $\mathbb{C}$ satisfy the $C^*$-algebra identity $\|aa^*\| = \|a\|^2$?

Answer 1

Let $\langle \cdot, \cdot \rangle$ be the inner product of $\mathbb{C}^n$ (or any Hilbert space), with corresponding norm $\| \cdot \|$ and let $T$ be an $n \times n$ matrix (or bounded linear operator).

Then by properties of inner products, we have $| \langle x, y \rangle| \le \|x\| \|y\|$.

In particular, $\|x \| = \sup \{ | \langle x, y \rangle| : \|y \| \le 1 \}$.

Thus we can write the operator norm of $T$ as $$ \|T\| = \sup \{ \|Tx\| : \|x\| \le 1\} = \sup_{\|x\| \le 1} \sup_{\|y\|\le 1} | \langle Tx, y \rangle|.$$

So \begin{align} \|T^*T\| &=& \sup_{\|x\| \le 1} \sup_{\|y\|\le 1} |\langle T^*Tx, y \rangle | \\ &=& \sup_{\|x\| \le 1} \sup_{\|y\|\le 1} |\langle Tx, Ty \rangle| \\ &\geq& \sup_{\|x\| \le 1} |\langle Tx, Tx \rangle| \\ &=& \|T\|^2 \\ \end{align}

But also \begin{align} \|T^*T\| &=& \sup_{\|x\| \le 1} \sup_{\|y\|\le 1} |\langle T^*Tx, y \rangle| \\ &=& \sup_{\|x\| \le 1} \sup_{\|y\|\le 1} |\langle Tx, Ty \rangle| \\ &\le& \sup_{\|x\| \le 1} \sup_{\|y\|\le 1} \|Tx\| \|Ty\| \\ &=& \|T\|^2. \end{align}