# Density of normal matrices in $M_n(\mathbb{C})$

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.

• Since $A \mapsto A^\ast A - A A^\ast$ is continuous, the set of normal matrices is closed. So it cannot be dense for $n > 1$.
– Gerd
Feb 26, 2023 at 7:49
• If it was dense then every matrix would be normal, as the conjugation is continuous. Feb 26, 2023 at 8:43
• @geometricK You might find this post interesting. Feb 26, 2023 at 13:30
• Another way of specifying normality is via Schur's Inequality $\big\Vert A\big\Vert_F^2 \geq \sum_{k=1}^n \vert \lambda_k\vert^2$ with equality iff $A$ is normal. So let e.g. $A$ be a non-zero nilpotent matrix. It is not normal so $f(A)=\big\Vert A\big\Vert_F^2 - \sum_{k=1}^n \vert \lambda_k\vert^2 = d \gt 0$. Eigenvalues vary continuously with coordinates of a matrix so e.g. select $\epsilon :=\frac{d}{2}$ and there is a $\delta\gt 0$ ball around $A$ such that every matrix in it is not normal. Feb 26, 2023 at 16:50
• @ronno Ok, I will set it out as an answer.
– Gerd
Feb 28, 2023 at 21:46

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.