How to prove **using SVD** that $\mathbb{C}^{n \times n}$ is dense for nonsingular matrices?

Question

How to show using SVD that the set of nonsingular matrices is dense in $\mathbb{C}^{n \times n}$? That is, for any $A \in \mathbb{C}^{n \times n}$, and any given $\varepsilon > 0$, there exists a nonsingular matrix $A_\varepsilon \in \mathbb{C}^{n \times n}$ such that:
$\left \| A-A_\varepsilon \right \| \le \varepsilon$.

Answer 1

In finite dimensional space all norms are equivalent. We choose an algebra norm.

Let $A=U\Sigma V^*$ the singular value decomposition where $U$ and $V$ are unitary matrices that's $||U||=||V||=1$ and $\Sigma=diag(\sigma_n,\ldots,\sigma_1)$ with $$\sigma_n\geq\cdots\geq\sigma_1\geq0.$$

Let $\Sigma_p=diag(\sigma_n+1/p,\ldots,\sigma_1+1/p)$ and $A_p=U\Sigma_pV^*\in GL_n(\mathbb{C})$ then we have: $$||A-A_p||=||U(\Sigma-\Sigma_p)V^*||\leq||U||||\Sigma-\Sigma_p||||V^*||=||\Sigma-\Sigma_p||=\frac{1}{p}||I_n||,$$ so $$\lim_{p\to\infty}||A-A_p||=0,$$ and we conclude.

Answer 2

Let $A=U\Sigma V^\ast$ be a full SVD of $A$. Just modify the zero diagonal entries of $\Sigma$ to small nonzero ones.