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