Prove that the set of diagonalizable matrices is dense in $\mathbb C^{n,n}$ with respect to spectral norm. Is it true for any matrix norm?
My attempt:
Let $\Lambda$ be the set of diagonalizable matrices in $\mathbb C^{n,n}.$ Our aim is to prove $\mathbb C^{n,n}\subset\overline{\Lambda}$. Let $A\in \mathbb C^{n,n}$, by Schur decomposition, we can say that there exists Unitary matrix $U\in \mathbb C^{n,n}$ such that $U^H AU=R.$ where $R$ is an upper triangular matrix. Therefore we can write $A=URU^H$.I know the definition of spectral norm. $||A||=\max\{\sqrt{\lambda}:\lambda$ is an eigenvalue of $A^HA\}. $
Let $R=R^*+D$, where $R^*$ has different elements on the diagonal. $R^*$ is an upper triangular matrix. $D$ is a digonal matrix with eigenvalues of $D^HD<\epsilon.$
Then we can write
$$ ||URU^H-UR^*U^H||=||U DU^H||\leq ||U|| ||D|| || U^H||<\epsilon.$$
We know that $U^HU=I$. So, spectral radius of $U$ and $U^H$ are 1 respectively. This prove that $UR^*U^H$ is the matrix exists in the epsilon neighbourhood of $A$ such a way that $UR^*U^H\in \Lambda$. This proves that every epsilon neighbourhood of $A$ with respect to the spectral norm intersects $\Lambda.$ Hence, $A\in \mathbb C^{n,n}.$
