# Stabilizer of an element in the Grassmannian

Let $Gr_{n,k} = \{W\leq \mathbb{C}^n:\mbox{dim(W)}=k \}$ be the set of $k$ dimensional subspaces of $\mathbb{C}^n$.

We have a group action: $$GL_n(\mathbb{C})\times Gr_{n,k} \to Gr_{n,k} \ , \ (g,W)\mapsto g.W=\{gv:v\in W\}.$$

I have shown that this action is transitive. In particular, this follows from the fact that we have a group of rotations $O(n, \mathbb{C})$ and group of scalar matrices inside $GL_n(\mathbb{C}).$

Let $e_1,...,e_n$ be the standard basis of $\mathbb{C}^n$. Let $W$ be a subspace spanned by $e_1,...,e_k$. I want to compute the stabilizer of $W$. By definition, this is just $$G_W = \{g\in GL_n(\mathbb{C}):g.W=W\}.$$

I think I need to use the fact about why this action is transitive. But I don't seem to find the right way to do this. Any hint to compute the stabilizer will be appreciated.

If $\begin{pmatrix} A & B\\ C & D\end{pmatrix}$ is a block matrix and $\begin{pmatrix} v \\ w\end{pmatrix}$ is a vector, then the product is $\begin{pmatrix} Av + Bw\\ Cv + Dw\end{pmatrix}$.
In your case, you want to let $\begin{pmatrix} v \\ w\end{pmatrix}$ vary over all elements in the span of $e_1,\ldots,e_k$, where $v$ has $k$ components and $w$ has $n-k$ components. In this case, we see that $C$ must be zero, and there are no restrictions on $A,B,D$ (except $A$ and $D$ need to be invertible for the block matrix to be invertible).