Orthogonal complement of a single vector Is there a quick way to show that if v is an element of an n dimensional real inner product space space, the orthogonal complement of v is n-1 dimensional? I can do this by using gram-schmidt and showing that V is a direct sum, however the way a question is worded suggests there is a quicker way if we are considering just the orthogonal complement of one vector?
I'm pretty sure I'm missing something very obvious, but its very annoying!
Thanks
 A: Hint: Let $\text{Span}(v) = U\subset V$, and consider the linear map $T\colon V\to U$ given by orthogonal projection. What is $\ker(T)$?
A: By the way, your statement is incorrect. The dimension of the orthogonal complement is $n$ if $v$ is the $0$ vector. So, exclude that case from the rest of the discussion.
If $V$ is finite-dimensional, then you can argue using a column vector $x$. Every $0$ in the column vector gives you a standard basis vector in the orthogonal complement. For the remaining column vector entries, choose any two, say $x_{j}$ and $x_{k}$. Create a new column vector with all zeros except at $j$ and $k$, where you have swapped the $j$ and $k$ entries of $x$ and have negated one of them. (Negative-reciprocal slope is the idea.) Just fix the $j$, and allow $k$ to vary over all the other non-zero entries and you can count that you've found $K-1$ vectors in the orthogonal complement if there are $K$ non-zero entries in $x$. You can see that you end up with $n-1$ linearly-independent vectors in the orthogonal complement because of the strategic choice of vectors. And, you can't have dimension $n$ for the orthogonal complement because, otherwise, $v$ would be orthogonal to itself, making $v=0$.
