[proofplan] The construction is inductive. At each step $n$, we subtract from $v_n$ its orthogonal projection onto $\operatorname{span}\{e_1, \ldots, e_{n-1}\}$ and normalise. Orthonormality is verified directly, span preservation follows because each $e_n$ is a linear combination of $v_1, \ldots, v_n$ and vice versa, and non-degeneracy of the intermediate vector follows from linear independence. [/proofplan] [step:Handle the base case $n = 1$] Since $v_1 \ne 0$ (linear independence), set $e_1 = v_1 / \|v_1\|_H$. Then $\|e_1\|_H = 1$ and $\operatorname{span}\{e_1\} = \operatorname{span}\{v_1\}$. [/step] [step:Perform the inductive step: subtract the projection and normalise] Suppose $e_1, \ldots, e_{n-1}$ are orthonormal with $\operatorname{span}\{e_1, \ldots, e_j\} = \operatorname{span}\{v_1, \ldots, v_j\}$ for $j \leq n-1$. Define \begin{align*} \tilde{e}_n := v_n - \sum_{k=1}^{n-1} (v_n, e_k)_H\, e_k. \end{align*} [claim:Non-Degeneracy] $\tilde{e}_n \ne 0$. [/claim] [proof] If $\tilde{e}_n = 0$, then $v_n \in \operatorname{span}\{e_1, \ldots, e_{n-1}\} = \operatorname{span}\{v_1, \ldots, v_{n-1}\}$, contradicting linear independence. [/proof] Set $e_n = \tilde{e}_n / \|\tilde{e}_n\|_H$. [/step] [step:Verify orthonormality and span preservation] [claim:Orthonormality] $(e_n, e_j)_H = 0$ for $j \leq n-1$ and $\|e_n\|_H = 1$. [/claim] [proof] For $j \leq n - 1$: \begin{align*} (\tilde{e}_n, e_j)_H = (v_n, e_j)_H - \sum_{k=1}^{n-1} (v_n, e_k)_H\, (e_k, e_j)_H = (v_n, e_j)_H - (v_n, e_j)_H = 0, \end{align*} using $(e_k, e_j)_H = \delta_{kj}$. [/proof] [claim:Span Preservation] $\operatorname{span}\{e_1, \ldots, e_n\} = \operatorname{span}\{v_1, \ldots, v_n\}$. [/claim] [proof] By construction, $e_n$ is a linear combination of $v_n$ and $e_1, \ldots, e_{n-1}$ (hence of $v_1, \ldots, v_n$). Conversely, $v_n = \|\tilde{e}_n\|_H e_n + \sum_{k=1}^{n-1} (v_n, e_k)_H e_k \in \operatorname{span}\{e_1, \ldots, e_n\}$. [/proof] [/step] [step:Extend to the infinite case] When $\{v_k\}_{k=1}^\infty$ is countably infinite, the induction produces $\{e_k\}_{k=1}^\infty$ with the stated properties for every $n$. Taking closures: $\overline{\operatorname{span}\{e_k\}} = \overline{\operatorname{span}\{v_k\}}$. [/step]