= Solution
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 $1\leq r<\min(n,m)$, every line $tE_{ij}$ 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 $T_0X=k^{nm}$. Meanwhile a rank-$r$ chart with an invertible $r\times r$ block has the form
$$
\begin{pmatrix}B&C\\D&DB^{-1}C\end{pmatrix}.
$$
The 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 $r(n+m-r)<nm$, \b[zero is singular when $1\leq r<\min(n,m)$.]
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 \b[$\boxed{\dim X^{5,2,1}=6}$.] 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.
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