Why is $GL_2(\mathbb C)$ connected? I have proved that the set of all $2\times 2$ non-singular matrices over $\mathbb R$ is disconnected as $\{A:\det A>0\}$ and $\{A:\det A<0\}$ forms a disconnection of $GL_2(\mathbb R)$.
What I can't understand is that why is $GL_2(\mathbb C)$ connected?
How should I approach?
 A: Well, you can break the problem into pieces:


*

*If $D$ is a diagonal invertible matrix, then one can connect it to the identity by providing a path $\gamma_i(t)$ from each diagonal entry $d_{ii}$ to $1$, and considering the path $D(t) \in GL_2(\mathbb{C})$ whose diagonal entries are the $\gamma_i$.

*If $B$ is an upper triangular invertible matrix, you can do the same thing while multiplying each non-diagonal entry by $(1-t)$.

*If $A$ is any invertible matrix, you can conjugate it to an upper triangular matrix and conjugate the entire path obtained in (2).
Hope this makes sense!
A: This might be a bit of an overkill here, but I thought I would also mention an elegant proof coming the theory of Lie groups, using the existence of a logarithm for any invertible matrix. Let $a, b \in \operatorname{GL}_n(\mathbb{C})$, and pick $A, B \in \mathcal{M}_n(\mathbb{C})$ two logarithms of $a$ and $b$ respectively (which means here : matrices such that $e^A = a$ and $e^B = b$).
Then we can define a continuous path $\Gamma : t \in [0,1] \mapsto \exp(tB + (1-t)A)$ which is always in $\operatorname{GL}_n(\mathbb{C})$ (since the exponential of a matrix is always invertible) and goes from $a$ to $b$.
A: Let $(A,B)\in GL(n,\mathbb{C})^2$ and let define:
$$f:z\in\mathbb{C}\mapsto\det((1-z)A+zB).$$
One has $f\in\mathbb{C}[z]$ and $f\neq 0$, since $f(0)=\det(A)$. From there, let $\{z_1,\ldots,z_r\}$ be the distinct roots of $f$. Notice that $\mathbb{C}\setminus\{z_1,\ldots,z_r\}$ is path connected and since $0,1\in\mathbb{C}\setminus\{z_1,\ldots,z_r\}$, there exists $\gamma:[0,1]\rightarrow\mathbb{C}\setminus\{z_1,\ldots,z_r\}$ a continuous function such that $\gamma(0)=0$ and $\gamma(1)=1$. Now, let define: $$\Gamma:t\in[0,1]\mapsto (1-\gamma(t))A+\gamma(t)B\in GL(n,\mathbb{C}).$$
$\Gamma$ is continuous and $\Gamma(0)=A$ and $\Gamma(1)=B$. Hence, $GL(n,\mathbb{C})$ is path connected and therefore connected.
