I am not an expert and this is probably not an answer, but your question is essentially a classical problem. You may refer to Hatcher's proof in his book or notes by Michael Hopkins in here. I am sorry that I could not address your question on $Gr(2,4)$ as I have not done the computation for years. The essential tool would be celluar decomposition, calculation of the degree, and Poincare duality. This used to be my final problem when I was learning algebraic topology years ago.
There may be slicker ways to carry out the computation of Schubert calculus but as far as I know it is usually a troublesome process and has only been solved by Ravi Vakil by his paper in Annals. This is also related to some extent with Schubert polynomials, which I personally know very little. Generally Schubert calculus offer a way of getting 'universal' objects that might represent specific objects.
Suppose that the $n \times n$ Hermitian matrix $A$ has orthonormal eigenvectors $e_i$ corresponding to $\lambda_i$ by the spectral theorem, where $\lambda_1 \geq \dots \geq \lambda_n$. Then, for any $v = \sum_{i=1}^{n} c_i e_i$ with $c_1,\dots,c_n \in \mathbb{C}$, we have $$ v^{*}Av = \sum_{i=1}^{n} \lambda_i |c_i|^2 . $$
Proof of "$\mathrm{LHS\leq RHS}$".
We can choose $U_j = \mathrm{span}_\mathbb{C}\{ e_i ： 1 \leq i \leq i_j \}$, and thus $U_1 \subset \cdots \subset U_k$ form a partial flag. Pick any $k$-dimensional $W \in X(U_1,\dots,U_k)$. To obtain an orthonormal basis $v_1,\dots,v_k$ of $W$ such that $v_j \in U_j$ for $1\leq j\leq k$, we proceed as following:
- Write $W_k = W$ and $U_0 = \{\mathbf{0}\}$ for convenience;
- Let $\ell_k = \min\{1\leq j\leq k : W_k \subset U_j\} - 1$, which is finite since $W_k \subset U_k$;
- Pick some unit vector $v_k \in \{v\in W_k : v\perp(U_{\ell_k}\cap W_k)\}$, which exists since $U_{\ell_k}\cap W_k \neq W_k$;
- Let $W_{k-1} = \{ w\in W_k : w \perp v_k \}$, which lies in $X(U_1,\dots,U_{k-1})$ since $U_{\ell_k}\cap W_k \subset W_{k-1} \subset U_{\ell_k + 1}$;
- Repeat step 2.~4. with $k \leftarrow k-1$, until $k$ decreases to $0$.
Therefore, noting that $v_j^{*}Av_j \geq \lambda_{i_j}$, we have $$ \mathrm{tr}(A|_W) = \sum_{j=1}^{k} v_j^{*}Av_j \geq \sum_{j=1}^{k} \lambda_{i_j} .$$
Proof of "$\mathrm{LHS\geq RHS}$".
It suffices to show that for any partial flag $V_1 \subset \cdots \subset V_k$, there exists some $W \in X(V_1,\dots,V_k)$ so that $$\mathrm{tr}(A|_W) \leq \sum_{j=1}^{k} \lambda_{i_j}.$$ To that end, we perform induction on the dimension $n$. Observe that the $n=1$ case is trivial, so we may assume $n \geq 2$. Let $m = \max(i_1-1, 1)$ and define $$E = \mathrm{span}_\mathbb{C}\{ e_i ： m+1 \leq i \leq n \},$$ then $\dim(E) = n-m < n$. Also, the restriction $A|_E$ of $A$ to $E$ has eigenvalues $\lambda_i' = \lambda_{m+i}$ for $1\leq i\leq n-m$. To obtain a new partial flag, we proceed as following:
- Pick some $(i_k - m)$-dimensional $V_k' \subset V_k \cap E$, which exists since $$\dim(V_k \cap E) = \dim(V_k) + \dim(E) - \dim(V_k + E) \geq i_k + (n-m) - n = i_k - m ;$$
- For $j = k-1,\cdots,1$, pick some $(i_j - m)$-dimensional $V_j' \subset V_j \cap V_{j+1}'$, which exists since $$\dim(V_j \cap V_{j+1}') = \dim(V_j) + \dim(V_{j+1}') - \dim(V_j + V_{j+1}') \geq i_j + (i_{j+1}-m) - i_{j+1} = i_j - m .$$
If $i_1 > 1$, then $m = i_1 - 1$ and $\dim(V_1') = 1$, so $V_1'\subset\cdots\subset V_k'$ form a partial flag of $E$. Applying the induction hypothesis to $A|_E$, we find some $W \in X(V_1',\dots,V_k') \subset X(V_1,\dots,V_k)$ as desired. Otherwise, $i_1 = 1$ and $m = 1$. By the induction hypothesis, we can find some $W' \in X(V_2',\dots,V_k')$ such that $\mathrm{tr}(A|_{W'}) \leq \sum_{j=2}^{k} \lambda_{i_j}$. Let $\ell = \max\{j: V_j \subset W'\}+1 \leq k$ and pick some $v \in V_{\ell}\setminus W'$, then $W = W' + \mathrm{span}_\mathbb{C}\{v\}$ satisfies our demand, which can be partly known
from that $\lambda_{i_1} = \lambda_1 \geq w^{*}Aw$ for any unit vector $w$ in the orthogonal complement of $W'$ in $W$.
Best Answer
The cohomology ring of $G(k,n)$ is isomorphic to $$\frac{\mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$ where $T$ and $Q$ are respectively the tautological and the quotient bundle. The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas: