Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 24G (Algebraic Geometry)

We use algebraic dimension and tangent spaces, equivalently after extending scalars to an algebraic closure; these are not dimensions of the finite set of rational points when $k$ is finite. The condition rank at most $r$ is equivalent to the vanishing of all $(r+1)\times (r+1)$ matrix minors. These are polynomials in the $nm$ entries, so define an affine determinantal variety $X^{n,m,r}$. The same equations make it a Zariski closed subset of $X^{n,m,r+1}$.

For $1\le r<\min (n,m)$, every line $t{E}_{ij}$ lies in $X$ because its matrices have rank at most one. Its tangent direction $E_{ij}$ therefore belongs to the Zariski tangent space at zero. The matrix units span all $nm$ directions, so $T_{0}X=k^{nm}$. Meanwhile a rank-$r$ chart with an invertible $r\times r$ block has the formThe free blocks have $r^2+r(m-r)+(n-r)r=r(n+m-r)$ entries. Rank-$r$ matrices are dense in $X$: the factorization $UV$ parametrizes all rank-at-most-$r$ matrices, and the full-rank factor pairs form a dense open subset. This also proves irreducibility, so the local dimension at zero equals the dimension of $X$. Since $r(n+m-r)<nm$, zero is singular when $1\le r<\min (n,m)$.

There is a necessary qualification to the printed claim: for $r=0$, $X=\{0\}$ is a smooth point, with zero-dimensional tangent space. Thus the claim is false if zero is allowed as a rank bound.

Finally, for $n=5,m=2,r=1$, the formula gives $\boxed{\dim X^{5,2,1}=6}$. Concretely, on a chart where an entry of the first column is nonzero, choose that column freely and make the second an arbitrary scalar multiple of it, giving five plus one parameters.