Set of orthogonal vectors in $\mathbb{R}^n$ How can we show that a set of pairwise orthogonal vectors in $\mathbb{R}^n$ has size at most $n$? I know it seems very intuitive, but not sure what the formal proof would look like (whether "elementary" or not). 
 A: Take a linear combination of any set of $k$ vectors with this pairwise-inner-product property. Suppose this linear combination  happens to equal zero. Dot it with each of them in turn to show that each coefficient is zero. Hence the vectors are linearly independent. 
In an $n$-dimensional space, a linearly independent set has at most $n$ elements, hence $k \le n$. 
Note though, that the "each coeff is zero" step requires that the elements of your set all be nonzero. Otherwise the set $\{ (1, 0), (0, 1), (0, 0) \}$, all of which are pairwise orthogonal (at least by the "orthogonal means dot-product-is-zero" definition), would violate your conclusion. 
A: You can show that any set $\{x_1, ..., x_m \}$ of orthogonal vectors is a linearly independent set by noticing the following:
$(\sum_{j = 1}^m a_jx_j, x_k) = a_k$ for any $k$, so that if $\sum_{j = 1}^m a_jx_j = 0$ then $a_k = 0$ for all $k$, where $( , )$ is the inner product you are using in $\mathbb{R}^n$.
Then you just have to use the fact from linear algebra that you can find in any book that $n$ is the highest cardinality of any linearly independent set in $\mathbb{R}^n$.
A: Consider $n$ orthogonal, or however independent, vectors, then the corresponding $n$ x $n$ matrix has a non null determinant.
Thus a $(n+1)$th (non-null) vector can always be expressed as a linear (non-null) combination of the others, i.e. you cannot find a $(n+1)$th independent vector.
