Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans

For $r=0$ there is nothing to construct, and the successive-span assertion at $k=0$ is equality of zero subspaces.

Suppose $e_0,\dots ,e_{k-1}$ have been constructed orthonormally with the required span equalities. Define $u_k$ by the displayed formula. For $i<k$, linearity and orthonormality give $⟨u_k,e_i⟩=⟨v_k,e_i⟩-⟨v_k,e_i⟩=0$.

If $u_k=0$, then $v_k$ lies in $\mathrm{span}(e_0,\dots ,e_{k-1})=\mathrm{span}(v_0,\dots ,v_{k-1})$, say $v_k={\sum }_{i<k}{\mu }_i{v}_i$; then the scalars ${\lambda }_i={\mu }_i$ for $i<k$, ${\lambda }_k=-1_F$ and ${\lambda }_i=0_F$ for $i>k$ satisfy ${\sum }_i{\lambda }_i{v}_i=0_V$ with ${\lambda }_k\mathrm{\ne }{0}_F$, contradicting the independence of $v$ through [L2]. Hence $u_k\mathrm{\ne }0$, and [L3] makes $e_k=u_k\mathrm{/}\parallel u_k\parallel$ a unit vector orthogonal to its predecessors.

The formula for $u_k$ shows $e_k$ lies in $\mathrm{span}(v_0,\dots ,v_k)$, while its rearrangement shows $v_k$ lies in $\mathrm{span}(e_0,\dots ,e_k)$. Together with the induction hypothesis these give both inclusions in the span equality at $k+1$.

Induction constructs the stated list and proves every successive-span equality. Its vectors are nonzero and orthogonal, so [L4] also confirms their independence; their unit norms make the list orthonormal.