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I know that $\text{SL}(N,\mathbb{C}),\text{SU}(N)$ and $\text{U}(N)$ are path-connected. You can use the $\text{QR}$-decomposition for example to see the $\text{SL}(N,\mathbb{C})$-case.

With that knowledge, how can I see that $\text{GL}(N,\mathbb{C})$ is path-connected?

Consider a $\text{GL}(N,\mathbb{C})\ni A=QR$ with $Q\in\text{U}(N)$ and $R$ upper triangular with $R_{ii}>0$.

Is there a path from $A$ to a matrix lying in the already proven cases?

  • It isn't too hard to construct a path from $A$ to the identity. You just have to make sure you don't pass any eigenvalues from $A$, which isn't too hard since the amount of eigenvalues of $A$ is finite. – Demophilus Nov 08 '17 at 12:23

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$GL(n,\Bbb C)$ is the image of the surjective continuous map $SL(n,\Bbb C)\times \Bbb C^*\to GL(n,\Bbb C)$, $(A,z)\mapsto zA$.

More explicitly, for any two matrices $A,B$, there are $\alpha,\beta\ne 0$ such that $\alpha^n=\det A$ and $\beta^n=\det B$, and paths $\delta:[0,1]\to\Bbb C\setminus\{0\}$ and $\Gamma:[0,1]\to SL(n,\Bbb C)$ such that $\delta(0)=\alpha$, $\delta(1)=\beta$, $\Gamma(0)=\frac1\alpha A$ and $\Gamma(1)=\frac1\beta B$. Your path is then $\phi(t)=\delta(t)\Gamma(t)$.