Singular matrices are Nowhere Dense

Question

I need to show that the subset S of all singular matrices is Nowhere dense in Set of all matrices of order n.

What I tried:-

As I need to show closure of S has no interior point . But S is a closed set so it is enough to show S has no interior point.

Initially what I thought was if I take any arbitray matrix A in S then in every open ball of radius r>0 centred at A will contain a matrix A° with first entry a11+r/2 and rest the same as A.

But it turns out that this matrix may not be non-singular.

How can I fix my argument can anyone help!! Thanks & regards

Answer 1

Putting $A$ in Jordan Normal Form, there exists a diagonal matrix $D$, a matrix $N$ which has only $1$'s on the super-diagonal and $0$'s else, and some invertible matrix $V$ such that $A=V^{-1}(D+N)V$. Then perturb $D$ on its diagonal $0$-entries (by adding $\varepsilon$), name the new matrix $\tilde{D}_{\varepsilon}$. Then, $\textrm{det}(\tilde{D}_{\varepsilon}+N)\neq 0$ for any $\varepsilon>0$ and hence, $$ \tilde{A}=V^{-1}(\tilde{D}_{\varepsilon}+N)V $$ is the matrix you want for $\varepsilon$ sufficiently small (note that, by construction, $\lim_{\varepsilon\to 0} \tilde{D}_{\varepsilon}=D$ in $L^{\infty}$ and all norms on the matrices are equivalent).

Answer 2

The characteristic polynomial $\det(\lambda I -A)$ of $A$ has a zero at $\lambda=0$. It can't vanish in a vicinity of that point. So it's also possible to perturb A along the diagonal to reach a non-singular Matrix.