How to prove that $ {\mathbf{GL}_{n}}(\mathbb{R}) $ is dense in $ {\mathbf{M}_{n}}(\mathbb{R}) $ This question is inspired by an old question on here.

Prove:
$ {\mathbf{GL}_{n}}(\mathbb{R}) $ is dense in $ {\mathbf{M}_{n}}(\mathbb{R}) $

Proof:
$ {\mathbf{GL}_{n}}(\mathbb{R}) = \{A \in {\mathbf{M}_{n}}(\mathbb{R}) : \det A \neq 0\}$ where $ {\mathbf{M}_{n}}(\mathbb{R}) $ consists of all $n \times n$ matrices with real entries.
We want to prove that $ {\mathbf{GL}_{n}}(\mathbb{R}) $ is dense in $ {\mathbf{M}_{n}}(\mathbb{R}) $.
$ {\mathbf{GL}_{n}}(\mathbb{R}) $ is dense in $ {\mathbf{M}_{n}}(\mathbb{R}) $ if every matrix $A \in {\mathbf{M}_{n}}(\mathbb{R}) $ either belongs to $ {\mathbf{GL}_{n}}(\mathbb{R}) $ or is a limit point of $ {\mathbf{GL}_{n}}(\mathbb{R}) $.
A matrix $P$ is a limit point of $ {\mathbf{GL}_{n}}(\mathbb{R}) $ in $ {\mathbf{M}_{n}}(\mathbb{R}) $ if and only if every open set $U$ containing $P$ also contains  a point of $ {\mathbf{GL}_{n}}(\mathbb{R}) $ different from
$P$.

And now when we should start to think I am not sure what to do as I have never worked with $ {\mathbf{GL}_{n}}(\mathbb{R}) $ or $ {\mathbf{M}_{n}}(\mathbb{R}) $ before.
I would appreciate a hint or a few on how to continue.
 A: The set $X$ of matrices with zero determinant is a submanifold of codimension $1$ (with some singularities). Then by Sard's theorem it has measure zero. It follows that its complement is dense.
A: (1) If $A=[a_1\cdots a_n]$ with ${\rm rank}\ A=k,\ k<n $, then we can
assume that $\{ a_1,\cdots,a_k\}$ is linearly independent. If $
\{a_1,\cdots,a_k, b_{k+1},\cdots, b_n\}$ is linearly independent and if $B:= [ 0\cdots 0\ b_{k+1}\ \cdots\
b_n]$, then ${\rm det}\ (A+ \epsilon B)  = {\rm det} \ [a_1\cdots a_k \ \epsilon b_{k+1}\ \cdots\
\epsilon b_n]\neq 0 $ 
(2) $M_n({\bf R})$ has a inner product, i.e., $|X|^2={\rm trace}\ XX^T$ and we have a metric $d(X,Y)=|X-Y|$. So since $|A-(A+ \epsilon B)|=\epsilon |B|$ Hence $A$ has an invertible matrix within arbitrary small distance. That is $Gl_n({\bf R})$ is dense.
A: Let $A \in M_n(K)$, with $K=\mathbb{R}$ or $\mathbb{C}$, then $\chi_A(\lambda)=\det\left(A-\lambda I_n \right)$ is a nonzero polynomial, so it has a finite number of roots. So it exits a $p \in \mathbb{N}$ such that $\forall k>p \text{, }$ $\chi_A(1/k)\neq 0$ and then the sequence $$\left(A-\frac{1}{k} \cdot I_n\right)_{k>p}\longrightarrow A$$
is made of invertible matrices. 
QED.
A: Let $A\in M_n$. Choose $\epsilon >0$ such that $\epsilon$ is less than the smallest positive eigenvalue of $A$ (if any exists, otherwise any positive $\epsilon$ will do). Observe that $A-\epsilon I$ is invertible (otherwise, $\epsilon$ is an eigenvalue for $A$), therefore $A_\epsilon \in GL_n$.
A: I particularly like this argument. Take any matrix $A$ and any invertible matrix $B$. Consider $$p(t) = \det((1-t)A+tB)$$ which is clearly a polynomial in one variable, and it is not zero polynomial because $p(1)\neq 0$. Thus, it has finitely many zeroes. That means that you can find arbitrarily small $t$ such that $C_t = (1-t)A+tB$ is invertible and we have $$\|A-C_t\| = |t|\|A-B\|<\varepsilon,\quad |t|<\frac{\varepsilon}{\|A-B\|}$$ Geometrically, we consider a line segment through matrices $A$ and $B$ and can find invertible matrix on it arbitrarily close to $A$.
A: The function $\det g$ is a polynomial in the $g_{ij}$. In particular, it's equal to its Taylor series everywhere. It follows that the closed set $X = \{g\in M_n(\mathbb{R}):\, \det g = 0\}$ has empty interior, and thus its complement in $GL_n(\mathbb{R})$ is dense in $\mathbb{R}^n$.
