Question

let S be a finite set in a vector space V with the property that every x in V has a unique representation as a linear combination of S. Show that S is a basis of V.

Answer 1

For the sake of the proof, suppose that \(S=\{\mathbf{v}_1,\ldots,\mathbf{v}_k\}\). By definition, we know that \(\mathrm{span}\,(S)\subseteq V\). Since every \(\mathbf{x}\in V\) can be written uniquely in the form \(\mathbf{x}=\alpha_1\mathbf{v}_1+\cdots+\alpha_k\mathbf{v}_k\), then every \(\mathbf{x}\in \mathrm{span}\,(S)\) which implies that \(V\subseteq\mathrm{span}\,(S)\). Therefore \(\mathrm{span}\,(S) = V\).

Last thing we need to show is that this spanning set is linearly independent. Suppose that \(\mathbf{0}\in S\). Then if \(\mathbf{v}_i=\mathbf{0}\) for some \(i\), then

\(\begin{aligned} \mathbf{x} &=\alpha_1\mathbf{v}_1+\cdots+\alpha_{i-1}\mathbf{v}_{i-1}+\alpha_i\mathbf{v}_i+\alpha_{i+1}\mathbf{v}_{i+1}+\cdots + \alpha_k\mathbf{v}_k\\ &= \alpha_1\mathbf{v}_1+\cdots+\alpha_{i-1}\mathbf{v}_{i-1}+\alpha_i\mathbf{0}+\alpha_{i+1}\mathbf{v}_{i+1}+\cdots + \alpha_k\mathbf{v}_k\end{aligned}\)

would no longer be a unique representation of vectors in \(S\) (all because \(\alpha_i\) could be any real number we want it to be and it wouldn't change \(\mathbf{x}\)). This contradicts the assumption that \(\mathbf{x}\) can be represented uniquely by vectors in \(S\). Therefore, \(\mathbf{0}\notin S\).

Now suppose there are \(\alpha_i\) such that \(\alpha_1\mathbf{v}_1 +\cdots +\alpha_k\mathbf{v}_k = \mathbf{0}\). By our assumptions, \(\mathbf{0}\) must be uniquely represented by vectors in \(S\). Since \(\mathbf{0}\notin S\), it is impossible to express any vector \(\mathbf{v}_i\in S\) in terms of any of the other vectors in \(S\). Therefore, the only way we can uniquely have \(\alpha_1\mathbf{v}_1 +\cdots +\alpha_k\mathbf{v}_k = \mathbf{0}\) is when \(\alpha_1=\cdots=\alpha_k=0\). Therefore, \(S\) is a linearly independent set of vectors.

Since \(S\) is a linearly independent set of vectors that spans all of \(V\), it now follows that \(S\) is a basis for \(V\).

I hope this makes sense! :-)