In a nonzero subvector space \(V\) of \(\RNrSpc{n}\) every orthonormal subset \(S\) of \(V\) can be complemented to an orthonormal basis of \(V\).
In a nonzero subvector space \(V\) of \(\RNrSpc{n}\) every orthonormal subset \(S\) of \(V\) can be complemented to an orthonormal basis of \(V\).
We know already that \(S\) is linearly independent and, hence, can be complemented to a basis of \(V\). So, if \(S=(\Vect{b}_1,\dots ,\Vect{b}_s)\), choose an ordered set of linearly independent vectors \(T=(\Vect{c}_1,\dots ,\Vect{c}_t)\) such that
\[\EuScript{A}\DefEq (\Vect{b}_1,\dots ,\Vect{b}_s,\Vect{c}_1,\dots ,\Vect{c}_t)\]is an ordered basis of \(V\). Then apply the Gram-Schmidt process to \(\EuScript{A}\) so as to obtain an ONB \(\EuScript{B}\) of \(V\). It only remains to observe that the vectors from \(S\) remain unchanged as they are already orthonormal, and this completes the proof.