What's an easy way to show that $GL(n,\mathbb C)$ is connected? [duplicate]

I think I've to show it's path connected, but can't figure out the path functions explicitly. Can anyone give these path maps?

• the intuition is that almost all matrices are inversibles Oct 15, 2014 at 11:28
• @Denis Well that intuition is also true for $GL(n,\mathbb R)$, which is not connected. Oct 15, 2014 at 11:32

Proof 1 It's not hard to find a path $$B(t)$$ in $$GL(n, \mathbb{C})$$ that connects an arbitrary Jordan-form matrix $$J$$ with nonzero eigenvalues to the identity matrix $$I$$: For a diagonal matrix $$J$$ with entries $$\lambda_a$$, one can simply pick any paths $$\gamma_a$$ in $$\mathbb{C} - \{0\}$$ (which in particular is path-connected) such that $$\gamma_a(0) = 1 \qquad\text{and}\qquad \gamma_a(1) = \lambda,$$ so that $$B(t) = \text{diag}(\gamma_1(t), \ldots, \gamma_n(t))$$ satisfies $$B(0) = I \qquad\text{and}\qquad B(1) = J.$$ Note that $$\det B(t) = \prod \gamma_a(t) \neq 0$$, so the image of $$B(t)$$ is indeed contained in $$GL(n, \mathbb{C})$$.
If $$J$$ is not diagonal, one can form the path $$B(t)$$ that (a) as above connects $$I$$ to the matrix $$J'$$ with the same diagonal entries as $$J$$ but zeros elsewhere, and then (b) connects $$J'$$ to $$J$$ by a line segment in $$M(n, \mathbb{C})$$. The determinant is constant on the latter segment, and so again $$B(t)$$ is contained in $$GL(n, \mathbb{C})$$.
Now given any matrix $$A \in GL(n, \mathbb{C})$$, form its Jordan decomposition $$A = P^{-1} J P$$, and take $$B(t)$$ to be a curve as above. Then, the curve $$\Gamma: t \mapsto P^{-1} B(t) P$$ satisfies $$\Gamma(0) = P^{-1} B(0) P = P^{-1} I P = I\\ \Gamma(1) = P^{-1}B(1) P = P^{-1}JP = A.$$ So, every point in $$GL(n, \mathbb{C})$$ can be connected to $$I$$ with a path, and hence the space is path-connected.
Proof 2 One can show that the exponential map $$\exp: \mathfrak{gl}(n, \mathbb{C}) \to GL(n, \mathbb{C})$$ is surjective. So, for any $$A \in GL(n, \mathbb{C})$$ there is an element $$a \in \mathfrak{gl}(n, \mathbb{C})$$ such that $$\exp a = A$$. Then, the curve $$\Gamma: t \mapsto \exp(at)$$ satisfies $$\Gamma(0) = \exp(a \cdot 0) = \exp 0 = I$$ and $$\gamma(1) = \exp (a \cdot 1) = \exp a = A$$.