Density of matrices with positive determinant Let $V$ be a vector space (not necessarily finite dimensional). 
Is $GL^+(V)$ dense in $GL(V)$? By $GL^+(V)$ I mean the group of endomorphisms with positive determinant (so they are invertible) and by $GL(V)$ I mean the group of invertible endomorphims. 
I am trying to use a good perturbation of the matrix for which I want to build a convergent sequence, but I do not manage to do it. Can somebody helps me?
Thanks!
 A: Posting the comments as an answer. 
In order that the question makes sense, we assume $V$ is a finite-dimensional vector space over $\mathbb R$. Since $\det : M_n (\mathbb R) \to \mathbb R$ is a continuous mapping, if $A$ is any matrix with $\det A < 0$, then there exists a sufficiently small neighborhood of $A$ consisting of only matrices of negative determinant. Therefore, the closure of $GL^+(V)$ contains only matrices with nonnegative determinant. Already this establishes that $GL^+(V)$ is not a dense set. 
But for completeness, we can say a bit more. In fact, the closure of $GL^+(V)$ (EDIT: seen as a subset of the set of all $n \times n$ matrices) is precisely the set of matrices with nonnegative determinant. To prove this, it suffices to establish that if $\det A = 0$, then $A$ is a limit point of $GL^+(V)$. This is not too hard, so I will leave it as an exercise. :) But I think the question talks about the closure in $GL(V)$ and not the set of all matrices. 
EDIT: On the other hand, as a subset of $GL(V)$, the set of matrices with positive determinant and those with negative determinant both form connected components of $GL(V)$ (again, we're working over $\mathbb R$). Thus both these sets are both closed and open. Hence the closure of the set of positive matrices in $GL(V)$ is itself, which is certainly not the whole of $GL(V)$. 
