A set of $n$ orthonormal vectors provides a basis for $\mathbb R^n$ Let $\{v_1,...,v_n\} \subseteq \mathbb R^n$ be a set of mutually orthogonal, non-zero vectors of length $1$.
How does one prove that this set is a basis of $\mathbb R^n$?
I have already proved that they are linearly independent, but I still need to prove that they span $\mathbb R^n$. I have also read that a set of $n$ linearly independent vectors in $\mathbb R^n$   is a basis, but I couldn't find a proof of this theorem. 
 A: Assume they don't span $\mathbb{R}^n$. Then you can keep adding vectors $u_1,..u_k$ so that at each step, $v_1,...,v_n, u_1,...,u_k$ is free. Since $\mathbb{R}^n$ is finite dimensional, this process stops eventually, so that $v_1,...,v_n,u_1,...u_k$ is free and spans $\mathbb{R}^n$, so it's a basis
But a well-known theorem asserts that any two bases of a (finite dimensional - or with the axiom of choice, any) vector space have the same cardinality, so that $n+k = n$, $k=0$ : $v_1,...,v_n$ spans $\mathbb{R}^n$.
For a more detailed proof you can look up the incomplete basis theorem, which asserts that in a finite dimensional vector space, any free family can be expanded to a basis (actually this is true of any vector space if you add the axiom of choice)
A: I don't know if this would count as extremely rigorous, but I was just working on this question today, and this is how I would answer it.
We have the set of vectors which are orthogonal and of unit length. Therefore, they form some sort of basis for the space (since the vectors are linearly independent). We then note that this basis is given by the span of $(v_1,...,v_n)$, which means the dimension of this space is the number of vector elements that we used. The number of vectors is $n$, so the dimension is $n$. This is the same dimension as $\mathbb R^n$. Since the dimension of $\mathbb R^n$ is $n$, we know that any linearly independent set of $n$ vectors will span this space, which means our set is a basis for $\mathbb R^n$.
A: Any linearly independent set with cardinality $n$ forms a basis for  $\mathbb{R}^n$. Even it is true for any $n$ dimensional vector space.
