Proposition 6

*Let E be a Hausdorff prehilbertian space, and let $(a_n)_{n \in I}$ (I an interval of $\mathbf{N}$ with origin 1) be a countable (finite or not) independent family of vectors of E. There exists an orthonormal family $(e_n)_{n \in I}$, and only one, in E, with the following properties:

1) for every integer $p \in I$, the vector subspace of E generated by $e_1, e_2, ..., e_p$ is identical with the vector subspace of E generated by $a_1, a_2, ..., a_p$; 2) for every integer $p \in I$, the number $\langle a_p|e_p \rangle$ is real and $> 0$.

In fact, let $V_n$ be the subspace (of dimension n) generated by $a_1, a_2, ..., a_n$. If $n + 1 \in I$ and $b_{n+1} = a_{n+1} - p_{V_n}(a_{n+1})$ (where $p_{V_n}$ is the orthoprojector onto the complete subspace $V_n$), the line $K b_{n+1}$ is the orthogonal of $V_n$ in $V_{n+1}$. If the $e_n$ satisfy condition 1) of the proposition, we must have $e_{n+1} = \lambda b_{n+1}$; the condition $\|e_{n+1}\| = 1$ then implies $|\lambda|^2 \|b_{n+1}\|^2 = 1$ and the condition $\langle a_{n+1}|e_{n+1} \rangle > 0$ implies $\lambda \langle a_{n+1}|b_{n+1} \rangle > 0$; this completely determines $\lambda$, and we have proved that we can determine, by induction, an orthonormal family $(e_n)_{n \in I}$, and only one, so as to satisfy conditions 1) and 2) of the proposition.

The sequence $(e_n)_{n \in I}$ is said to be obtained by orthonormalisation from the independent family $(a_n)_{n \in I}$. It is clear that the vector subspace generated by the family $(e_n)$ is identical with the vector subspace generated by the family $(a_n)$. In particular, if $(a_n)$ is a total sequence, so is $(e_n)$, which is then an orthonormal basis of E; hence we get: