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 $x$?

  • $\begingroup$ 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


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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.