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$V \cong \mathbb{R}^{n}$ is a vector space

$H = \{v \in V \mid \alpha \cdot v = 0\}$ is a linear hyperplane where $\alpha$ is a fixed nonzero vector in $V$ and $\alpha \cdot v$ is the usual dot product.

I know a plane is defined to be the set of vectors $v$ which are orthogonal to a normal vector $\alpha$. The vectors $v$ and $\alpha$ are, to my understanding, position vectors. So vectors which begin from the origin and end at a point in space.

I thought a plane was defined by $\alpha \cdot (v_1 - v_2) = 0$. That is, the difference of the vectors and the dot product with $\alpha$ makes a plane. But $H$ doesn't say to take a difference of the vectors. So how does $H$ define a hyperplane?

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  • $\begingroup$ $v_1-v_2=v$ for some $v\in V$. $\endgroup$ – Dave Sep 14 '17 at 16:46
  • $\begingroup$ Well, $v = (v - 0)$. To be less glib - in your last paragraph, if we hold $v_2$ constant, the set of $v_1$s that satisfy the equation gives you a hyperplane. Note that $v_2$ lies in this set. Thus, this is the hyperplane with normal vector $\alpha$ that passes through $v_2$. $H$ as you've defined it is the same thing with $v_2 = 0$. $\endgroup$ – stochasticboy321 Sep 14 '17 at 16:49
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The condition you stated is also satisfied by the elements in your $H$ above: $\alpha \cdot v_1=0, \alpha \cdot v_2=0$ imply $\alpha \cdot (v_1-v_2)=0,$ so such an object is then called linear.

Your so-called plane may be defined by something like $ \alpha \cdot v=k$ for some constant $k$ then any two vectors on it will also satisfy $\alpha \cdot (v_1-v_2)=0,$ but it will not be a linear subspace unless $k=0.$

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  • $\begingroup$ Tommy, why does $\alpha \cdot v = k$ denote a translation of the hyperplane $H$ ? $\endgroup$ – Taln Sep 14 '17 at 17:39
  • $\begingroup$ I think it can be viewed as $\alpha \cdot (v-c)=0$ for any constant vector such that $\alpha \cdot c + k=0$ if you want to see it as a translation of a hyperplane subspace. $\endgroup$ – tommy xu3 Sep 14 '17 at 17:46

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