is $GL(n,\mathbb R)$ dense in $M(n,\mathbb R)$ Is $GL(n,\mathbb R)$ dense in $M(n,\mathbb R)$? I have proved it to be open,not closed,not connected but not sure about this property .How to do this?
 A: Since this is posted under the tag metric spaces, I will show that the statement is not true for every metric:
Let $d$ be the discrete metric:
$$
d(A,B) = \begin{cases} 0 & \text{ if } A=B\\ 1 & \text{ if } A\ne B.\end{cases}
$$
Then
$$
d(0,B) =1
$$
for all invertible $B$, hence $GL(n,\mathbb R)$ is not dense in $M(n,\mathbb R)$.
If the metric is induced by a norm, $d(A,B)=\|A-B\|$,  then the statement is true. Let $A$ be not invertible. Then there are invertible matrices $S,T$ such that
$$
SAT = \pmatrix{ I_r & 0 \\ 0 & 0 },
$$
with $r=rank(A)$. Now let $\epsilon\ne0$ be given.
Set
$$
A_\epsilon :=  S^{-1} \pmatrix{ (1+\epsilon)I_r & 0 \\ 0 & \epsilon I_{n-r} } T^{-1} = A + \epsilon S^{-1}T^{-1}.
$$
This matrix is clearly invertible as a product of invertible matrices $S^{-1}$, $\pmatrix{ (1+\epsilon)I_r & 0 \\ 0 & \epsilon I_{n-r} }$, $T^{-1}$.
Then
$$
A-A_\epsilon = S^{-1}\pmatrix{ -\epsilon I_r & 0 \\ 0 & -\epsilon I_{n-r}} T^{-1} = - \epsilon S^{-1}T^{-1},
$$
which shows
$$
\|A-A_\epsilon\| = |\epsilon|\cdot \|S^{-1}T^{-1}\|.
$$
And $GL(n,\mathbb R)$ is dense in $M(n,\mathbb R)$.
A: The way I like writing this is as follows: Consider $A\in M_{n}(\mathbb R)$ and let $\varepsilon>0$. Let $\lambda _1, \ldots, \lambda_n$ be the eigenvalues of $A$. Take $\delta$ such that $0<\delta<\frac{\varepsilon}{n^{1/2}}$ and $\delta\neq \lambda_j$ for every  $j\in \{1, \ldots, n\}$. Define
$$A_\delta:=A-\delta I.$$
Then $A_\delta$ is invertible. In fact, on the contrary,
$$\det(A_\delta)=\det(A-\delta I)=0$$
and therefore $\delta$ would be an eigenvalue of $A$, so that $\delta =\lambda_j$ for some $j\in \{1, \ldots, n\}$. This contradicts the choice of $\delta$. Finally,
$$d_2(A, A_\delta)=\|A-A_\delta\|_2=\|A-(A-\delta I)\|_2=|\delta|\|I\|_2=\delta {n^{1/2}}<\frac{\varepsilon}{n^{1/2}} n^{1/2}=\varepsilon.$$
I assumed here that your metric is induced by norm:
$$\|A\|_2=\left(\sum_{i=1}^n \sum_{j=1}^n a_{ij}^2\right)^{1/2}.$$
A: Yes, $GL(n,\mathbb R)$ is dense in $M(n,\mathbb R)$. 
Using the determinant function $\det:M(n,\mathbb R)\rightarrow \mathbb R$, check that for small values of $\varepsilon$, the matrix $A+\varepsilon I_n$ is invertible.
A: Hint: A matrix has finitely many eigenvalues.  If none of them is $0$, ...
A: Here GLn(R) is subspace as a metric space in set of all metrices of order n. If I take any member in Mn(R): if it is in GLn(R) then clearly we get sequence of GLn(R) that converge to our chosen point. Now if it is not in GLn(R) then by changing exactly one element of the matrix I can make it a member of the set of all invertible matrices, so by this way I can get a sequence of GLn(R) which converges to set of our chosen point.
