Given a vector $\mathbf{v} \in \mathbb{R}^n$, the set of vectors in $\mathbb{R}^n$ orthogonal to $\mathbf{v}$, namely $$\{\mathbf{u} \in \mathbb{R}^n: \mathbf{u} \cdot \mathbf{v}=0\},$$ forms a subspace. In fact, it is the null space of the $1 \times n$ matrix $$\left( \begin{matrix} v_1 & v_2 & \cdots & v_n \\ \end{matrix} \right)$$ if $\mathbf{v}=(v_1,v_2,\ldots,v_n)$.

Question: Is there a specific name or notation for this vector space?

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    $\begingroup$ It is called a hyperplane. $\endgroup$ – John Douma Aug 27 '13 at 22:53
  • $\begingroup$ The notation is $\langle \mathbf{v}\rangle^\perp$ and if $F$ is a subspace or even a set then it's orthogonal is denoted by $F^\perp$. $\endgroup$ – user63181 Aug 27 '13 at 22:56
  • $\begingroup$ @user69810 The tangent space of a point at the hyperplane, and the plane is defined by using $\mathbf{v}$ as normal, to be precise. $\endgroup$ – Shuhao Cao Aug 27 '13 at 23:00
  • $\begingroup$ @ShuhaoCao Don't you need to start with some kind of surface to get a tangent space? In this case we are given the vector but we are not told that it is the normal to a surface. $\endgroup$ – John Douma Aug 27 '13 at 23:08
  • $\begingroup$ @user69810: The set $\{(1,y):y\in\mathbb R\}$ is a hyperplane in $\mathbb R^2$ but is not the space of vectors orthogonal to any vector. $\endgroup$ – Rahul Aug 27 '13 at 23:13

We usually denote it "W perp", a W with an upside down T as a superscript. We call it the orthogonal complement.


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    $\begingroup$ Here $W$ is the line spanned by $v$. It's the ortogonal complement in the sense that $\Bbb R^n=W\oplus W^\perp$. $\endgroup$ – Andrea Mori Aug 27 '13 at 22:59
  • $\begingroup$ Although W need not be a line. It is more general. W perp is the set of all vectors orthogonal to the subspace W. $\endgroup$ – Demetri Pananos Aug 27 '13 at 23:01
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    $\begingroup$ Certainly. Yet the OP was asking for the orthogonal of a vector. $\endgroup$ – Andrea Mori Aug 27 '13 at 23:03
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    $\begingroup$ But the perpendicular applies to any subset of a vector space--it is always a subspace since $S^\perp=\text{span}(S)^\perp$. $\endgroup$ – Alex Youcis Aug 27 '13 at 23:11

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