Let $X\_r\subset Mat\_{n\times n}$ denote the matrices of rank at most
$r$, and let $S\_{\pi}C^n$ denote the irreducible $GL\_n$-module corresponding
to the partition $\pi$. 
One can check that
degree($X\_r$)=dim($S\_{(n-r)^{n-r}}C^n)$=$\Pi\_{i=0}^{n-r-1}\frac{(n+i)!i!}{(r+i)!(n-r+i)!}$
 Does anyone have a geometric (or any) explanation for this equality? Had it been observed previously?
Here $(n-r)^{n-r}=(n-r,...,n-r)$ has Young diagram the square box.



Why does the degree of the variety of rank at most $r$ $n\times n$ matrices equal dim$S_{(n-r)^{n-r}}C^n$?
There are a number of better statements (i.e. yes, this is "well-known").


To begin with, wouldn't you rather have the *character* of this representation, instead of just its dimension? You can get that as the class $[X\_r] \in H^\*\_{T^n}(Mat\_{n\times n})$. This improves on your statement, because for $X \subseteq V$ a cone in a vector space, $[X] = deg(X) t^{{\rm codim}(X)} \in H^\*\_{\rm dilation}(V) \cong {\mathbb Z}[t]$, and the character of $t\cdot Id\_n$ on $V\_\lambda$ is 
$\dim(V\_\lambda) t^{|\lambda|}$.


(Of course, the character is a symmetric polynomial. Correspondingly, we should look at $[X\_r] \in H^\*\_{GL(n)}(Mat\_{n\times n}) = H^\*\_T(Mat\_{n\times n})^{S\_n}$.)


Next, let's go beyond square matrices to general $k\times n$.
But most satisfyingly, instead of asking that the whole thing have rank $r$, let $X\_{\vec n}$ be the $k \times \infty$ matrices where the left $n\_i$ columns has rank $\leq i$, for each $i=0,\ldots,k$. After column $n\_k$ there are no conditions, so including those columns has no effect on the degree. The case you're thinking about is $n\_{i} = n+max(0,i-r)$. The data of $\vec n$ is equivalent to a partition with $\leq k$ rows.


You might be most satisfied looking at
$$ [X\_{\vec n}] \in H^\*\_{GL(k)}(Mat\_{k\times \infty}) = H^\*\_{GL(k)}(\text{full rank matrices}) = H^\*(GL(k) \backslash \text{full rankers}) = H^\*(Gr\_k({\mathbb C}^\infty)) \cong Rep(End({\mathbb C}^k)) $$
where the last two rings-with-bases are well known to be isomorphic (see 
[Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian?](https://mathoverflow.net/questions/14219/why-do-littlewood-richardson-coefficients-describe-the-cohomology-of-the-grassma) ). In particular the degree of $X\_{\vec n}$ is the 
dimension of the corresponding irrep of $GL(k)$ (which extends to $End({\mathbb C}^k)$).


(Note that to go from all $k\times \infty$ matrices to the full rank ones, the locus removed is $\infty$-codimensional, so doesn't affect the cohomology.)


(Here's a Gröbner basis computation you might like. Replace each of the determinants defining $X\_{\vec n}$ by its diagonal-product summand. Decompose the resulting monomial scheme into components, which are coordinate spaces inside matrices.
We can draw a component on a $k\times \infty$ strip, where $m\_{ij}=0$ is drawn as a $j$ at position $(i,j)$. Now move those $j$s Northwest until they get stuck, and you will always get the same shape worth of numbers (determined by $\vec n$), with a semistandard Young tableau in it. From this one can show that the character is given by a sum over SSYT. This stuff is in my paper with Miller and Yong. For a first example, try $k=2$, $n\_0 = 1, n\_1 = 3, n\_2 = 4$. Then the equations are $m\_{11} = m\_{21} = \underline{m\_{13} m\_{24}} - m\_{14} m\_{23} = 0$. The drawings are
$$ \begin{pmatrix} 1 & 1 & &\cdots \\ 2 & & &\cdots \end{pmatrix},
 \begin{pmatrix} 1 & & & \cdots \\ 2 & &2 &\cdots \end{pmatrix}
$$
which march NW to be the two SSYT for the $V\_{2+1}$ irrep of $GL(2)$.)


