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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.

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    $\begingroup$ Since $A \mapsto A^\ast A - A A^\ast$ is continuous, the set of normal matrices is closed. So it cannot be dense for $n > 1$. $\endgroup$
    – Gerd
    Feb 26, 2023 at 7:49
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    $\begingroup$ If it was dense then every matrix would be normal, as the conjugation is continuous. $\endgroup$ Feb 26, 2023 at 8:43
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    $\begingroup$ @geometricK You might find this post interesting. $\endgroup$ Feb 26, 2023 at 13:30
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    $\begingroup$ 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. $\endgroup$ Feb 26, 2023 at 16:50
  • $\begingroup$ @ronno Ok, I will set it out as an answer. $\endgroup$
    – Gerd
    Feb 28, 2023 at 21:46

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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.

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