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I thought of creating a path from a matrix $A$ to a matrix $B$ using their traces, but got nowhere.

user21820
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the_dummy
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    No. The sets ${A:\det(A) > 0}$ and ${A: \det(A) < 0}$ are open and disjoint, but their union is the full set. – Ben Grossmann Nov 13 '17 at 01:26
  • Also helpful to think about $GL_1$. – Randall Nov 13 '17 at 01:30
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    Or ${\rm GL}(1,\Bbb R) = \Bbb R \setminus {0}$, if you want to be more extreme. – Ivo Terek Nov 13 '17 at 01:30
  • For the complex case, see https://math.stackexchange.com/questions/139549/how-to-show-path-connectedness-of-gln-mathbbc – Nate Eldredge Nov 13 '17 at 02:16
  • However, the sign of the determinant is the only obstruction; that is, $GL_n(\mathbb{R})$ has two connected components, the subset of matrices with positive determinant and the subset of matrices with negative determinant. This is slightly tricky to prove but various proofs are possible. – Qiaochu Yuan Nov 13 '17 at 03:55

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Hint:

Consider $U = \det^{-1}(-\infty, 0)$ and $V = \det^{-1}(0,\infty)$.

Tom
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If $\det A > 0$ and $\det B < 0$, you can't join them by any continuous arc $\gamma$. If $\gamma(0) = A$ and $\gamma(1) = B$, continuity of the determinant and the intermediate value theorem gives us $0<t_0<1$ such that $\det \gamma(t_0) = 0$. Can you see why this is a problem?

Ivo Terek
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    I like this argument. – Randall Nov 13 '17 at 01:31
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    Strictly speaking this logic shows that the set is not path connected, which in general is weaker than showing it is not connected. (This set happens to be an open subset of Euclidean space, so it is connected iff it is path connected, but that might require proof.) – Nate Eldredge Nov 13 '17 at 02:14
  • Yes, good point. We can't conclude non-connectedness for $n=1$ with this way. Fortunately the $n=1$ situation is trivial here. – Ivo Terek Nov 13 '17 at 02:17
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$\text{Fact}$ : The continuous image of connected set is connected.

Define a function $f:GL_n(\mathbb{R}) \to \mathbb{R}$ by $f(A)=det(A)$ ; where $A$ is $n \times n$ real matrix and $det(A)$ denotes determinant of the matix $A$.

Notice that the map $f$ is continuous , and its image is $\mathbb{R}\setminus \lbrace 0 \rbrace$. If $GL_n(\mathbb{R})$ is connected, then its image, i.e., $\mathbb{R}\setminus \lbrace 0 \rbrace$ must be connected , which is absurd. So, $GL_n(\mathbb{R})$ can't be connected.

risefrominfinite
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