A list \({latex.inline[v_{1}, ..., v_{n}](v_{1}, ..., v_{n})} of vectors in V is a basis of V if and only if every \){latex.inlinev \in V} can be written uniquely in the form \({latex.inline[v = a_{1}v_{1} + ... + a_{n}v_{n}](v = a_{1}v_{1} + ... + a_{n}v_{n})} where \){latex.inlinea{1}, ..., a{n} \in F}.
\({latex.inline[\Rightarrow](\Rightarrow)} Suppose that \){latex.inlinev{1}, ..., v{n}} is a basis of V. Let \({latex.inline[v \in V](v \in V)}. Because \){latex.inlinev{1}, ..., v{n}} spans V, there exists \({latex.inline[a_{1}, ..., a_{n} \in F](a_{1}, ..., a_{n} \in F)} such that \){latex.inlinev = a{1}v{1} + ... + a{n}v{n}}. We now have to show that the choice of a’s is unique. Suppose not. Then there exists \({latex.inline[c_{1}, ..., c_{n}](c_{1}, ..., c_{n})} are scalars such that we also have \){latex.inlinev = c{1}v{1} + ... + c{n}v{n}}. We can subtract the two equations and get that \({latex.inline[0 = (a_{1} - c_{1})v_{1} + ... + (a_{n} - c_{n})v_{n}](0 = (a_{1} - c_{1})v_{1} + ... + (a_{n} - c_{n})v_{n})}. This implies that each \){latex.inlinea{k} - c{k} = 0 \Rightarrow a{k} = c{k}}. But then we have that each a is unique.
\({latex.inline[\Leftarrow](\Leftarrow)} Suppose every \){latex.inlinev \in V} can be written uniquely in the form \({latex.inline[v = a_{1}v_{1} + ... + a_{n}v_{n}](v = a_{1}v_{1} + ... + a_{n}v_{n})}. This implies that the list spans V. To show that \){latex.inlinev{1}, ..., v{n}} is linearly independent, suppose ${latex.inline0 = a{1}v{1} + ... + a{n}v{n}}. Well, by the uniqueness of the that representation, we have that all a’s = 0. This means the list is linearly independent. Thus, we have a linearly independent list that also spans V which is a basis of V.