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So if I have a vector p and I project it onto itself I get

$$\operatorname{proj} \large~\vec{p}_{ ~\vec{p}} = \frac{\vec{p} \cdot \vec{p}}{\|\vec{p}\|^2}\vec{p} \rightarrow \frac{\vec{p} \cdot \vec{p}}{\\{p} \cdot \vec{p}}\vec{p}$$

So this leaves the final answer of $ \large~\vec{p}$

My question is if $ \large~\vec{p}$ is a zero vector will the resulting answer be undefined or will $ \large~\vec{p}$ be the zero vector?

Thanks

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  • $\begingroup$ You CANNOT "project" anything onto the zero vector. $\endgroup$ – user247327 Oct 19 '15 at 3:07
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    $\begingroup$ With that definition of the projector, it is undefined. With the geometric definition, the projection of anything onto the zero vector will be the zero vector. $\endgroup$ – Ian Oct 19 '15 at 3:10
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It depends on how you define projection. Naively the formula $$\mathrm{proj}_v p = \frac{p\cdot v}{v\cdot v}v$$ is not defined when $p=v=0$. However the limit $$\lim_{v,p\to 0} \frac{p\cdot v}{v\cdot v}v = 0$$ is well-defined and so there is nothing morally wrong with defining $\mathrm{proj}_00=0$ if this is somehow convenient for some calculation.

Finally following Ian's hint you can turn to the more general definition of projection of a point onto any set $S$:

$$\mathrm{proj}_S p = \arg\min_{q\in S} \|q-p\|$$ which is well-defined for any convex set, and in particular any linear subspace, and in particular $S=\{0\}$. You will get $\mathrm{proj}_{\{0\}} p = 0$ for any $p$.

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