Grateful if someone could tell me whether my rationale is correct:
If I impose $m$ constraints on $\mathbb R^n$ (where $n<\infty$), then the set has $n-m$ degrees of freedom. Hence this subspace has dimension $n-m$.
If you impose $m$ linear and linearly independent constraints on $\mathbb R^n$, then the set has the dimension $n-m$. The vector space of these $m$ constraints has the dimension $m$.
If only $k, k \le m$ (and not more) of these constraints are linearly independent, then the subspace of the constraints has the dimension $k$ and the orthogonal subspace has the dimension $n-k$.