Why the word "projective" for $PGL_n(\mathbb{F})$? I wrote the title for this question exactly as I had it exactly in my mind. Let me denote by $G=GL_n(\mathbb{F})$ for simplicity; I was working throughout the previous years many times with the quotient $G/Z(G)$ (i.e. the group of inner automoprhisms of $G$) which is known as the "projective linear group" and it is denoted explicitly as $PGL_n(\mathbb{F})$. Now my question as it mentioned in the title is why the name of that group contains the word "projective". Apparently there is a correlation with projective geometry, whilst whether $n=2$ we have the so-called Möbius Transformation group. At the Wiki article the situation (at least in my eyes) isn't that clear. So can you please help me out with this?
 A: The general linear group $GL_n(k)$ is the group of automorphisms (in a suitable sense) of the $n$-space $k^n$. Similarly, the projective linear group $PGL_n(k)$ is the group of automorphisms (in a suitable sense*) of the projective $n$-space $\mathbb{P}^n(k)$.
In projective geometry, two nonzero points are identified if they are on a same line going through the origin. (The relation with projective geometry is well explained at Wikipedia.) So it's natural to think that a transformation that takes any line to itself (i.e. all vectors are eigenvectors) acts "trivially" on the projective space. This explains why we must at least quotient out by the scalar multiple of the identity.
It's then an exercise to show that if a linear map is such that all vectors are eigenvectors, then the linear map is a scalar multiple of the identity. So we don't need to quotient out by anything else.

* The group $PGL_n(k)$ is the set of invertible elements of the monoid of functions $\mathbb{P}^n(k) \to \mathbb{P}^n(k)$ which come from linear maps $k^{n+1} \to k^{n+1}$ under the identification $\mathbb{P}^n(k) = (k^{n+1} \setminus \{0\}) / \sim$.
A: The projective plane over $\mathbb F$ arises from $\mathbb F^3$ by excluding the origin and then identifying vectors that are multiples of each other.
Similarly, $PGL_n(\mathbb F)$ arises from the set of $n\times n$ matrices by excluding the singular matrices and then identifying matrices that are multiples of each other (because the center of the general linear group consists of exactly the scalar multiples of $I$).
