A list ${latex.inlinev_{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.inlinev = a_{1}v_{1} + ... + a_{n}v_{n}} where ${latex.inlinea_{1}, ..., a_{n} \in F}.
${latex.inline\Rightarrow} Suppose that ${latex.inlinev_{1}, ..., v_{n}} is a basis of V. Let ${latex.inlinev \in V}. Because ${latex.inlinev_{1}, ..., v_{n}} spans V, there exists ${latex.inlinea_{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.inlinec_{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.inline0 = (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} Suppose every ${latex.inlinev \in V} can be written uniquely in the form ${latex.inlinev = 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.