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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 $n-m$.

Thanks.

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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 $n-k$.

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  • $\begingroup$ Thanks, Jiri. What happens if the constraints are non-linear? $\endgroup$ – russell Sep 12 '11 at 8:57
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    $\begingroup$ I was already typing ... $\endgroup$ – Jiri Kriz Sep 12 '11 at 8:59

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