[proofplan]
We prove the equivalence by showing both directions. For $(1) \Rightarrow (2)$, the definition of a basis gives spanning (existence of a representation) and linear independence (uniqueness of that representation, established by showing any two representations must coincide). For $(2) \Rightarrow (1)$, unique representability directly implies both spanning and linear independence.
[/proofplan]
[step:Show that a basis provides existence of a representation]
Assume $\mathcal{B}$ is a basis of $V$. By definition, $\mathcal{B}$ is a linearly independent spanning set. Since $\mathcal{B}$ spans $V$, every $v \in V$ can be written as a linear combination
\begin{align*}
v = \sum_{i=1}^n \alpha_i b_i
\end{align*}
for some scalars $\alpha_1, \ldots, \alpha_n \in F$. This establishes the existence part of statement (2).
[guided]
Assume $\mathcal{B}$ is a basis of $V$. Recall that a basis is, by definition, a subset that is both linearly independent and spanning. The spanning property means that the subspace $\operatorname{span}(\mathcal{B}) = \left\{ \sum_{i=1}^n \alpha_i b_i : \alpha_i \in F \right\}$ equals all of $V$. Therefore, for any $v \in V$, there exist scalars $\alpha_1, \ldots, \alpha_n \in F$ such that
\begin{align*}
v = \sum_{i=1}^n \alpha_i b_i.
\end{align*}
This gives the existence of at least one representation. But we still need uniqueness -- that is where linear independence enters.
[/guided]
[/step]
[step:Deduce uniqueness of the representation from linear independence]
Suppose $v \in V$ admits two representations:
\begin{align*}
v = \sum_{i=1}^n \alpha_i b_i = \sum_{i=1}^n \beta_i b_i.
\end{align*}
Subtracting yields
\begin{align*}
\sum_{i=1}^n (\alpha_i - \beta_i)\, b_i = 0_V.
\end{align*}
Since $\mathcal{B}$ is linearly independent, the only solution to this equation is $\alpha_i - \beta_i = 0$ for every $i \in \{1, \ldots, n\}$. Hence $\alpha_i = \beta_i$ for all $i$, and the representation is unique. This completes the direction $(1) \Rightarrow (2)$.
[guided]
To establish uniqueness, suppose for contradiction that some $v \in V$ has two distinct representations:
\begin{align*}
v = \sum_{i=1}^n \alpha_i b_i \quad \text{and} \quad v = \sum_{i=1}^n \beta_i b_i,
\end{align*}
where $\alpha_j \neq \beta_j$ for at least one index $j$. Subtracting the second expression from the first gives
\begin{align*}
0_V = v - v = \sum_{i=1}^n (\alpha_i - \beta_i)\, b_i.
\end{align*}
This is a linear combination of $b_1, \ldots, b_n$ that equals the zero vector, with the coefficient $\alpha_j - \beta_j \neq 0$. But $\mathcal{B}$ is linearly independent, which means the only linear combination of its elements that produces $0_V$ is the one where all coefficients are zero. This contradicts $\alpha_j - \beta_j \neq 0$, so no two distinct representations can exist. Combined with the existence from the previous step, every $v \in V$ has exactly one representation as a linear combination of $\mathcal{B}$, completing the direction $(1) \Rightarrow (2)$.
[/guided]
[/step]
[step:Derive spanning and linear independence from unique representability]
Now assume statement (2): every $v \in V$ can be written uniquely as $v = \sum_{i=1}^n \alpha_i b_i$.
**Spanning.** Since every vector in $V$ admits such a representation, $\mathcal{B}$ spans $V$.
**Linear independence.** Suppose $\sum_{i=1}^n \gamma_i b_i = 0_V$ for some $\gamma_1, \ldots, \gamma_n \in F$. The zero vector also has the representation $0_V = \sum_{i=1}^n 0 \cdot b_i$. By uniqueness of the representation, $\gamma_i = 0$ for all $i \in \{1, \ldots, n\}$. Therefore $\mathcal{B}$ is linearly independent.
Since $\mathcal{B}$ is both linearly independent and spanning, $\mathcal{B}$ is a basis of $V$. This completes the direction $(2) \Rightarrow (1)$ and establishes the equivalence.
[guided]
Now assume statement (2): every $v \in V$ can be written uniquely as $v = \sum_{i=1}^n \alpha_i b_i$ with $\alpha_i \in F$. We must show that $\mathcal{B}$ is a basis, i.e., that $\mathcal{B}$ is both linearly independent and spanning.
**Spanning.** The hypothesis directly states that every $v \in V$ is a linear combination of $b_1, \ldots, b_n$. This is exactly the definition of $\operatorname{span}(\mathcal{B}) = V$, so $\mathcal{B}$ spans $V$.
**Linear independence.** Suppose $\sum_{i=1}^n \gamma_i b_i = 0_V$ for some scalars $\gamma_1, \ldots, \gamma_n \in F$. We need to show all $\gamma_i$ are zero. The key observation is that the zero vector itself has a representation as a linear combination of $\mathcal{B}$: the zero combination $0_V = \sum_{i=1}^n 0 \cdot b_i$. By hypothesis, the representation of every vector -- including $0_V$ -- is unique. Since we have two representations of $0_V$ (one with coefficients $\gamma_i$ and one with all coefficients zero), uniqueness forces $\gamma_i = 0$ for every $i \in \{1, \ldots, n\}$. This is exactly the definition of linear independence.
Since $\mathcal{B}$ is both spanning and linearly independent, $\mathcal{B}$ is a basis of $V$, completing the direction $(2) \Rightarrow (1)$ and the proof of the equivalence.
[/guided]
[/step]
