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Is every hyperplane in $\mathbb{R}^n$ determined by a unique normal vector? And why?

I analysed for $\mathbb{R}$, a hiperplane in $\mathbb{R}$ is a point, so the hyperplane is $PX= \alpha$, with $\alpha \in \mathbb{R}$. So $X=\alpha / P$.

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    $\begingroup$ The normal vector isn't unique. But the span of any normal vector will be the same one-dimensional subspace. To prove this you use that theorem that says that you can always complete a basis of a subspace to a basis of the ambient space. Then you just prove that any such extension for a basis of your hyperplane will be made by adding a vector determined up to scalar multiplication (you can do this part by proof by contradiction). $\endgroup$
    – user137731
    Nov 20 '15 at 0:30
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    $\begingroup$ If you choose a unit normal vector it's almost unique-choosing the sign requires some additional information, namely, an orientation. $\endgroup$ Nov 20 '15 at 0:44
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No. Consider $\mathbb{R}^3$, and the (hyper)planes $x = 0$ and $x = 1$. Both have $[c, 0, 0]$ as normal vectors for all $c \neq 0$. So they are neither determined by their normal vectors, nor is their normal vector unique.

Note that these vectors comprise all their normal vectors, so there does not exist a normal vector which can distinguish these two planes.

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    $\begingroup$ I think what OP is asking is a bit different from what you have answered. OP wants: Given a hyperplane does there exist a unique vector to identify that hyperplane. Your answer: given a vector you can have many planes associated with it. They are not the same thing. $\endgroup$
    – Anurag A
    Nov 20 '15 at 0:44
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    $\begingroup$ @AnuragA Good call, I've updated my answer accordingly. $\endgroup$ Nov 20 '15 at 0:47
  • $\begingroup$ @JohnColanduoni $\endgroup$
    – dcg
    Nov 20 '15 at 1:09

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