Let $\left\{q_1, q_2, \dots, q_k\right\} \subset \mathbb{R}^n$, $k<n$, be an orthonormal collection. Let $\left\{e_1, e_2, \dots, e_n\right\} \subset \mathbb{R}^n$ be the standard (orthonormal) basis of $\mathbb{R}^n$. We wish to define $q_{k+1}, \dots, q_{n}$ so that $\left\{q_1, q_2, \dots, q_n\right\}$ is an orthonormal basis.
I have been able to experimentally validate the following 'algorithm' for doing this:
Set $s:= k$.
For $ i = 1, 2, \dots, n$:
- If $e_i - \sum_{k=1}^s (e_i^T q_k)q_k \ne 0$, then set $q_{s+1} := \frac{e_i - \sum_{k=1}^s (e_i^T q_k)q_k }{\Vert e_i - \sum_{k=1}^s (e_i^T q_k)q_k \Vert_2}$. Set $ s = s+1 $.
It is clear from the Gram-Schmidt orthonormalisation that if $$e_i - \sum_{k=1}^s (e_i^T q_k)q_k \ne 0,$$ then $\left\{q_1, \dots, q_s, q_{s+1}\right\}$ will be an orthonormal set.
However, what is unclear to me is, how can I be sure that in the end I indeed have precisely $n$ vectors $q_1, q_2, \dots, q_n$?
Hint: $s+1 \leq n$ becasue $q_i$'s are linearly idependent elments of $\mathbb R^{n}$. Now use your definition of $q_{i+1}$ to check that $span \{e_1,e_2,..,e_n\} \subseteq span \{q_1,q_2,..,q_{s+1}\}$. This gives $n \leq s+1$.