[proofplan] The forward direction uses the spanning property of a basis to get existence of a representation and linear independence to force uniqueness. The reverse direction observes that the existence of a representation for every vector gives spanning, while uniqueness of the trivial representation of $\mathbf{0}$ gives independence. [/proofplan] [step:Deduce existence of a representation from the spanning property ($\Rightarrow$, existence)] Suppose $\{e_1, \ldots, e_n\}$ is a basis for $V$. Since the set spans $V$, for every $v \in V$ there exist scalars $\lambda_1, \ldots, \lambda_n \in \mathbb{F}$ with \begin{align*} v = \sum_{i=1}^{n}\lambda_i e_i. \end{align*} [/step] [step:Deduce uniqueness of the representation from linear independence ($\Rightarrow$, uniqueness)] Suppose $v = \sum_{i=1}^{n}\lambda_i e_i = \sum_{i=1}^{n}\mu_i e_i$ are two representations. Subtracting: \begin{align*} \sum_{i=1}^{n}(\lambda_i - \mu_i)e_i = \mathbf{0}. \end{align*} Since $\{e_1, \ldots, e_n\}$ is linearly independent, $\lambda_i - \mu_i = 0$ for all $i \in \{1, \ldots, n\}$. Hence $\lambda_i = \mu_i$ for all $i$, establishing uniqueness. [/step] [step:Recover spanning and independence from unique representation ($\Leftarrow$)] Suppose every $v \in V$ has a unique representation $v = \sum_{i=1}^{n}\lambda_i e_i$. **Spanning:** The existence of at least one representation for every $v \in V$ means $\{e_1, \ldots, e_n\}$ spans $V$. **Linear independence:** Suppose $\sum_{i=1}^{n}\lambda_i e_i = \mathbf{0}$. The zero vector also has the representation $\mathbf{0} = \sum_{i=1}^{n} 0 \cdot e_i$. By uniqueness, $\lambda_i = 0$ for all $i$. Hence the set is linearly independent, and $\{e_1, \ldots, e_n\}$ is a basis. [/step]