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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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1 Answer 1

up vote 2 down vote accepted

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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Thanks, Jiri. What happens if the constraints are non-linear? –  russell Sep 12 '11 at 8:57
    
I was already typing ... –  Jiri Sep 12 '11 at 8:59
    
Thanks again. :-) –  russell Sep 12 '11 at 9:05

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