By identifying $\text{Mat}_n(\mathbb{C})$ with a direct sum of two copies of $\text{Mat}_n(\mathbb{R})$, or otherwise, show that the map $\text{Mat}_n(\mathbb{C}) \to \text{Mat}_n(\mathbb{C}),\; x \mapsto x^T\bar{x}$ is continuous.
I'm confused as how to approach this question; should I be working straight from the definition or is there a simpler way to see the continuity?
Please help!
EDIT: Let $U \subset \mathbb{R}^m$, $V \subset \mathbb{R}^n$ be subsets and let $\psi : U \to V$ be a map. We say $\psi$ is continuous at $u_0$ if for all $\epsilon > 0$, there exists $\delta > 0$ such that $||u-u_0|| < \delta \implies ||\psi(u) - \psi(u_0)||<\epsilon$.
Both functions$$\begin{array}[t]{ccc}\operatorname{Mat}_n(\mathbb{C})&\longrightarrow&\operatorname{Mat}_n(\mathbb{C})\\x&\mapsto&x^T\end{array}\text{ and }\begin{array}[t]{ccc}\operatorname{Mat}_n(\mathbb{C})&\longrightarrow&\operatorname{Mat}_n(\mathbb{C})\\x&\mapsto&\overline x\end{array}$$are continuous. Inedeed, they are isometries ($\|x^T-y^T\|=\|x-y\|$ and $\|\overline x-\overline y\|=\|x-y\|$). Therefore, their product is a continuous function.