A matrix $A\in M_n(\mathbb{C})$ is said to be normal if $A^*A=AA^*$, where $A^*$ is the Hermitian conjugate. Consider $M_n(\mathbb{C})$ with its norm topology.
Question: Is the space of normal matrices dense in $M_n(\mathbb{C})$?
Thoughts: I know that the space of all diagonalisable matrices is dense in $M_n(\mathbb{C})$, and that a matrix is normal if and only if it is unitarily diagonalizable. So the question amounts to asking whether the unitarily diagonalizable matrices still form a dense subset.
Best Answer
The mappings $(A,B) \mapsto AB$ and $A \mapsto A^\ast$ are continuous. Hence $\Phi:M_n(\mathbb{C}) \to M_n(\mathbb{C})$, $\Phi(A)= A^\ast A-AA^\ast$ is continuous. The set of normal matrices in $M_n(\mathbb{C})$ is $\Phi^{-1}(\{0_{n\times n}\})$, and is therefore closed. If $n > 1$, for example each nontrivial nilpotent matrix is not normal and is therefore an inner point of $M_n(\mathbb{C}) \setminus \Phi^{-1}(\{0_{n\times n}\})$. Thus $\Phi^{-1}(\{0_{n\times n}\})$ is not dense.