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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 $x$?

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  • $\begingroup$ 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)? $\endgroup$
    – M. Winter
    Mar 27, 2018 at 12:39
  • $\begingroup$ 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. $\endgroup$ Mar 27, 2018 at 12:55
  • $\begingroup$ You are right about a). I still don't get it. $p\cdot x$ gives a number, why not $\alpha$? $\endgroup$
    – M. Winter
    Mar 27, 2018 at 13:11

1 Answer 1

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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 $H=p^\perp+a$.

Now to the questions:

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.)

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