What is the dimension of the orthogonal complement of a hyperplane? I'm studying from my textbook, and from what I understand a hyperplane is a set of vectors with dimension $n-1$ that is the orthogonal complement of a set of vectors with dimension $n$. However there is this bit in the textbook that is really confusing me:
Since hyperplanes through the origin of $\Bbb{R}^n$ are the subspaces of dimension $n - 1$, their orthogonal complements are the subspaces of dimension 1, which are lines through the origin of $\Bbb{R}^n$.
I'm confused on one aspect. Shouldn't the dimension of the orthogonal complement of a hyperplane be $n$? Why does it say the orthogonal complement of a hyperplane are of dimension $1$, when it says earlier that the dimension of a hyperplane is $n-1$? I must not be understanding something correctly.
 A: Any $k$-dimensional linear subspace of an $n$-dimensional vector space intersects its orthogonal complement trivially. This is because the only vector that is perpendicular to itself is the $0$-vector. Hence the orthogonal complement is at most $(n-k)$-dimensional. In particular this means the orthogonal complement of a hyperplane is at most $1$-dimensional, and certainly not $n$-dimensional.
That the orthogonal complement is in fact precisely $(n-k)$-dimensional can be seen from the fact that any orthogonal basis for the $k$-dimensional subspace can be extended to an orthogonal basis for the $n$-dimensional vector space itself, by the basis extension theorem and Gram-Schmidt.
A: In $\mathbb R^n$, the orthogonal complement of a $k$ dimensional subspace is $n-k$ dimensional. Therefore, the orthogonal complement of the $n-1$ dimensional hyperplane is $n-(n-1)=1$ dimensional. There is no reason that I can think of to expect that the orthogonal complement of an $n-1$ dimensional subspace should be $n$ dimensional.
A: A hyperplane means just a subspace of dimension $n-1$, if the vector space has dimension $n$. 
The subspace plus its orthogonal complement must add up to the whole space, so the orthogonal complement of the hyperplane has dimension $1$, it is the line along its normal vector.
A: For intuition, think about $\mathbb{R}^3$. Here, a hyperplane of dimension $n-1$ is simply a plane of dimension 2 (passing through the origin). The orthogonal complement of this plane is the set of vectors perpendicular to it (i.e. the span of a normal vector to the plane). This forms a line through the origin. 
For further generality, the orthogonal complement of a $k$ dimensional subspace of $\mathbb{R}^n$ has dimension $n-k$. For your case, $k=n-1$.
