Stabilizer of action $PGL_n(\mathbb{R})$ on $\mathbb{PR}^{n-1}$

Question

I want to find stabilizer of action $PGL_n(\mathbb{R})$ on $\mathbb{PR}^{n-1}$ at point $[1: 0: \ldots:0]$. Well, since this point corresponds to $x_1$ coordinate axes, then every transformation which sends this line into itself, and acts arbitrary way on $\mathbb{R}^{n-1}$ - space with coordinates $x_2, \ldots, x_n$, so stabilizer is $GL_{n-1}(\mathbb{R}) /\mathbb{R}^* = PGL_{n-1}(\mathbb{R})$. Is this correct?

Answer 1

The stabilizer is a subgroup of $PGL_n(\mathbb R)$, but what you have described is not a subgroup of $PGL_n(\mathbb R)$, so no, what you write is not correct.

Instead, the start of the solution should be to consider an arbitrary invertible $n \times n$ matrix of real numbers $M$ --- which thus determines a point $[M] = \{tM \mid t \ne 0\} \in PGL_n(\mathbb R)$.

Now, you have already given a correct geometric condition, namely that $M$ sends the $x_1$ coordinate axis to itself. But now you need to describe explicit conditions on the $n^2$ coordinates of $M$ that are necessary and sufficient for $M$ to take the $x_1$ axis to itself.

For example, letting $e_1$ be the column vector with $1$ in the first entry and $0$'s elsewhere, the requirement that $M$ takes the $x_1$ axis to itself is equivalent to $M e_1 = t e_1$ for some $t \ne 0$, which is equivalent to saying that the first column of $M$ has a nonzero entry at the top and zeroes below that. So this gives you a necessary condition on the coordinates of $M$.

So to finish this problem you might want ask whether there are any further necessary conditions, and hopefully you will eventually compile a list of necessary conditions that is also sufficient.