I know that the $PGL(n, \mathbb{R})$ or $PGL(n, \mathbb{C})$ is Lie group, because $PGL(n, F) = GL(n, F) / Z(n, F)$, where $Z(n, F)$ - scalar transformation and $F$ is $\mathbb{C}$ or $\mathbb{R}$.
I think, that $PGL(n, \mathbb{C})$ is connected because $GL(n, \mathbb{C})$ is connected and $Z(n, F)$ is set of constant. I want to say that the path from $GL(n, \mathbb{C})$ descends to the quotient - $PGL(n, \mathbb{C})$. (or we can use that projection $P: GL(n, \mathbb{C}) \rightarrow GL(n, \mathbb{C}) / Z(n, \mathbb{C})$ is continuous map, and the image of connected space is also connected)
But what to do in the case of $\mathbb{R}$ and how to find all the connected components?
Thank you so much!
Hint By the QR decomposition there is a deformation retract $GL(n, \Bbb R) \to O(n, \Bbb R)$ (cf. deformation retract of $GL_n^{+}(\mathbb{R})$), and $O(n, \Bbb R)$ has two connected components, thus, so does $GL(n, \Bbb R)$, namely the preimages $GL_{\pm}$ of $\{\pm x > 0\}$ under the map $GL(n, \Bbb R) \to \Bbb R^*$.
So, the number of connected components of $PGL(n, \Bbb R)$ is either $1$ or $2$, depending on whether the quotient map $GL(n, \Bbb R) \to PGL(n, \Bbb R)$ identifies elements from $GL_+$ with elements from $GL_-$.
As Jason Devito mentioned in his comment, the answer depends on the parity of $n$.