Best way to show that this is a subvariety of the Grassmannian Let $k$ be a field and let $\text{Gr}(r,k^n)$ be the Grassmannian of $r$-dimensional subspaces of $k^n$.  Fix a linear map $\phi\colon k^n \rightarrow k^n$.  Define $X_{\phi}$ to be the subspace of $\text{Gr}(r,k^n)$ consisting of $r$-dimensional subspaces $V \subset k^n$ such that $\phi(V) \subset V$.  Question: what is the best way to see that $X_{\phi}$ is a closed subvariety of $\text{Gr}(r,k^n)$?
One way of doing this is by working in charts: Letting $W \subset k^n$ be an $(n-r)$-dimensional subspace, the subspace of $\text{Gr}(r,k^n)$ consisting of $V$ such that $V \cap W = 0$ is an open affine subspace that is isomorphic to $\mathbb{A}_k^{r(n-r)}$, and it is not hard to see that the intersection of $X_{\phi}$ with this open affine is closed in the Zariski topology.  However, this is not particularly elegant, so surely there is a better way!
The answer, of course, will depend first on which approach one takes to endowing $\text{Gr}(r,k^n)$ with the structure of an algebraic variety.  I am completely agnostic here: use whatever approach makes this problem transparent!  Perhaps my underlying issue here is that I don't know a great way to specify subvarieties of $\text{Gr}(r,k^n)$.  For $r=1$ (i.e. projective space), I can use homogeneous polynomials, but I don't know anything quite as nice for $r>1$.

Let $0 \to U \subset O^{\oplus n} \to Q \to 0$ be the tautological exact sequence of bundles on the Grassmannian. Consider the composition
$$
U \to O^{\oplus n} \stackrel\phi\to O^{\oplus n} \to Q.
$$
Then $X_\phi$ is the zero locus of this morphism, hence is a closed subscheme.
