Context:
Let $\mathbb{K}$ an algebrically closed field of zero caracteristic.
Let $m, n \in \mathbb{N}$ and $k \in \mathbb{N}$ such that $k \leq m$.
Let's denote $\text{Gr}_{m-k}(\mathbb{K}^m)$ the Grassmannian of $\mathbb{K}^m$ (i.e. the set of subspaces of $\mathbb{K}^m$ of dimension $m-k$). In addition, we denote $M_{nm}(\mathbb{K})$ the space of $n \times m$ matrices.
We define the incidence variety by $$\Gamma_k = \{(A,F) \in M_{nm}(\mathbb{K}) \times Gr_{m-k}(\mathbb{K}^m) \, | \, F \subset \ker A\}$$ We want to find the dimension of $\Gamma_k$ as a variety.
How I started: Let's denote $\pi_2 : M_{nm}(\mathbb{K}) \times Gr_{m-k}(\mathbb{K}^m) \to Gr_{m-k}(\mathbb{K}^m)$ the projection to the second factor. In a first time, we will fix $F \in Gr_{m-k}(\mathbb{K}^m)$. The purpose is to calculate $\dim \pi_2^{-1}(\{F\}) \cap \Gamma_k$.
We have \begin{eqnarray*} \pi_2^{-1}(\{F\}) \cap \Gamma_k & = & \{(A,F) \, | \, F \subset \ker A\} \\ & \simeq & \{A \in M_{nm}(\mathbb{K}) \, | \, \forall x \in F, \, Ax = 0\} \end{eqnarray*} So $\pi_2^{-1}(\{F\}) \cap \Gamma_k$ is a vector space. Let's calculate its dimension. We consider $$\begin{array}{ccccc} \varphi_F& : & M_{nm}(\mathbb{K}) & \longrightarrow & \mathbb{K}^{(m-k)n} \\ & & A & \longmapsto & \begin{pmatrix} Au_1 \\ Au_2 \\ \vdots \\ Au_{m-k} \end{pmatrix} \end{array}$$ where $F = \text{vect}(u_1, \dots, u_{m-k})$.
At this point, I wasn't able to show that $\varphi_F$ is well defined and that it is surjective. But assuming that, and by using the rank theorem we have $$nm = \dim \ker \varphi_F + (m-k)n.$$ So $$\dim \ker \varphi_F = nm - nm + nk = nk.$$ We conclude then that $$\dim(\pi_2^{-1}(\{F\}) \cap \Gamma_k) = nk.$$ My questions are: Are the above treatment of the question correct? and how can we deduce the dimension of $\Gamma_k$ ?
Thanks in advance for any hint or help.