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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!

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    What does "determinant" mean if $V$ isn't finite-dimensional? And this is clearly false when $V$ is finite-dimensional.2011-06-11
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    As Qiaochu said this is false. In fact determinant is a continuous mapping (BTW, I'm assuming $V$ is a real or complex finite dimensional vector space, endowed with its unique normed space topology). So a sequence of positive endomorphism may only converge to a positive endomorphism or to a degenerate one.2011-06-11
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    You are absolutely right, thank you for your answers.2011-06-11

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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 $GL(V)$.

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    More can be said. The set of all matrices with positive determinants and the set of all matrices with negative determinants are the connected components of $GL(V)$ (working over $\mathbb{R}$). Thus both sets are both open and closed. So the closure of the set of positive matrices is itself (definitely not all of $GL(V)$).2011-11-02
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    @Bill Indeed. I will add that to the answer.2011-11-02
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    @Bill Actually, the closure of $GL^+$ is the set of matrices with nonnegative determinant, which anyway is not the full space.2011-11-02
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    I think Bill is talking about the closure within $GL(V)$, Srivatsan about the closure within $M_n({\bf R})$. The latter has the matrices with zero determinant, the former, not.2011-11-02
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    @Gerry I see. Somehow I did not really notice it. Thanks for the comment.2011-11-02
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    @Bill, Pinging so that you will see Gerry's comment. I will update the answer.2011-11-02