# Hyperplane and convexity

A hyperplane $H$ in $\mathbb{R}^n$ is defined by $H=\left\{x:p\cdot x=\alpha,x \in \mathbb{R}^n\right\}$, where $p\in\mathbb{R}^n$ and $\alpha\in\mathbb{R}$. The vector $p$ is the normal vector of the hyperplane.

Consider the hyperplane where $p = (2,4)$ and $\alpha = 1$

a) this hyperplane represents a linear subspace of $\mathbb{R}^2$?

b) is this hyperplane a convex set? Is it compact?

my notes:

The hyperplane in the case of $\mathbb{R^2}$ will be a line, which is a convex set, but I believe it will not be compact because it is not bounded. Am I right?

I'm not sure how to formally demonstrate this.

And a doubt: $p.x$ would not have to be equal to zero? Because the normal vector is not always orthogonal to $x$?

• You are right about convexity and compactness. To you doubt: I do not really understand why you think this is important here. Can you explain? Also what is your opinion to a)? Mar 27, 2018 at 12:39
• my doubt: how can $p\cdot x=\alpha$ and $\alpha \neq 0$? a) A straight line is not always a linear subspace of R? since it is closed for sum and multiplication by scalar. Mar 27, 2018 at 12:55
• You are right about a). I still don't get it. $p\cdot x$ gives a number, why not $\alpha$? Mar 27, 2018 at 13:11

Such a hyperplane is also called an affine hyperplane, where an affine subspace of a vector space $$V$$ is a translation of a subspace, i.e. is of the form $$U+a$$ for an $$U\le V$$ and $$a\in V$$. A hyperplane is a one codimension subspace. In an inner product space (such as $$\Bbb R^n$$ with the dot product) it's indeed equivalent to being the orthogonal complement of a one dimensional subspace (one vector).
Note also that $$H=\{x\mid p\cdot x=\alpha\}$$ is indeed a translation of $$p^\perp:=\{x\mid p\cdot x=0\}$$, namely take an arbitrary $$a$$ such that $$p\cdot a=\alpha$$, then $$H=p^\perp+a$$.
a) Since the given $$\alpha$$ is nonzero, $$H$$ will not be a linear subspace (it doesn't even contain $$0$$ as you observed it).
b) Correct, it's convex: show that every affine subspace $$A$$ is closed under convex (moreover, affine) combinations: $$v_i\in A, \, \sum\alpha_i=1\implies\sum\alpha_iv_i\in A$$.
And, indeed it's not bounded, hence not compact. (Consider e.g. the open cover $$B_n:=\{x\mid x\cdot x < n^2\}$$ and observe that any nontrivial (affine) subspace contains a vector of arbitrarily long length.)