Since \(\Vect{x}\) belongs to \(\span(S)\), there exist pairwise distinct vectors \(\Vect{a}_1,\dots ,\Vect{a}_m\) in \(S\) and nonzero numbers \(t_1,\dots ,t_m\) such that

\(\Vect{x}\)

\(=\)

\(t_1 \Vect{a}_1 + \cdots + t_m \Vect{a}_m\)

It remains to show that these vectors and numbers are unique. We explain the idea, leaving a rigorous proof (induction on \(m\) and \(k\)) to the astute reader. So suppose

\(\Vect{x}\)

\(=\)

\(s_1 \Vect{b}_1 + \cdots + s_k \Vect{b}_k\)

for pairwise distinct vectors \(\Vect{b}_1,\dots ,\Vect{b}_k\) in \(S\) and nonzero numbers \(s_1,\dots ,s_k\). It follows that

\(t_1 \Vect{a}_1 + \cdots + t_m \Vect{a}_m\ -\ (s_1 \Vect{b}_1 + \cdots + s_k \Vect{b}_k)\)

\(=\)

\(\Vect{0}\)

We first show that each \(\Vect{a}_i\) occurs in the list of \(\Vect{b}\)-vectors. – Suppose not. Then linear independence of the set \(S\) forces \(t_i=0\), contradicting the assumption that \(t_1,\dots ,t_m\neq 0\). The same kind of argument shows that each \(\Vect{b}_j\) occurs in the list of \(\Vect{a}\)-vectors. Consequently, \(m=k\) and we may assume that \(\Vect{a}_1=\Vect{b}_1\), ... , \(\Vect{a}_m=\Vect{b}_m\). This proves that the vectors needed to express \(\Vect{x}\) as a linear combination of vectors in \(S\) are unique. It remains to show that \(s_1=t_1\), ... , \(s_m=t_m\).

The equation above is equivalent to

\((t_1-s_1)\Vect{a}_1 + \cdots + (t_m-s_m) \Vect{a}_m\)

\(=\)

\(\Vect{0}\)

Linear independence of \(S\) implies \((t_j-s_j)=0\), for \(1\leq i\leq m\); i.e. \(t_i=s_i\), for \(1\leq i\leq m\). – This completes the proof.