[proofplan] We prove the equivalence by showing $(1) \implies (2)$, $(2) \implies (3)$, and $(3) \implies (1)$. The implication $(1) \implies (2)$ is immediate from the definition of a basis. For $(2) \implies (3)$, we use the [Extension of Independent Sets to Bases](/theorems/3270) to extend the independent set to a basis, then argue by cardinality that no extension is needed. For $(3) \implies (1)$, we use the [Extraction of Bases from Spanning Sets](/theorems/3271) to extract a basis, then again argue by cardinality that no extraction is needed. [/proofplan]

[step:$(1) \implies (2)$: a basis is linearly independent by definition] By definition, a basis for $V$ is a linearly independent subset that spans $V$. In particular, if $S$ is a basis, then $S$ is linearly independent. [/step]

[step:$(2) \implies (3)$: a linearly independent set of size $n = \dim V$ must span $V$]Assume $S = \{v_1, \ldots, v_n\}$ is linearly independent. By the [Extension of Independent Sets to Bases](/theorems/3270), there exist vectors $v_{n+1}, \ldots, v_n \in V$ such that $\{v_1, \ldots, v_n, v_{n+1}, \ldots, v_n\}$ is a basis for $V$. Since $S$ already has $n = \dim V$ elements and every basis for $V$ has exactly $n$ elements, the extension adds no new vectors: the set of appended vectors is empty. Therefore $S$ itself is a basis, and in particular $S$ spans $V$.[/step]

[guided]We want to show that $S = \{v_1, \ldots, v_n\}$, which is linearly independent by hypothesis, also spans $V$. The strategy is to extend $S$ to a basis using a known theorem, then use cardinality to conclude the extension is vacuous. The [Extension of Independent Sets to Bases](/theorems/3270) states: if $\{u_1, \ldots, u_k\}$ is a linearly independent subset of a vector space of dimension $n$ with $k \le n$, then there exist vectors $v_{k+1}, \ldots, v_n$ such that $\{u_1, \ldots, u_k, v_{k+1}, \ldots, v_n\}$ is a basis for $V$. We apply this theorem with $k = n$. The theorem guarantees the existence of vectors $v_{n+1}, \ldots, v_n$ completing $S$ to a basis. But the index range from $n+1$ to $n$ is empty, so no new vectors are appended. Therefore $S = \{v_1, \ldots, v_n\}$ is itself a basis for $V$. In particular, since every basis spans $V$, the set $S$ spans $V$. The underlying principle is that in a vector space of dimension $n$, a linearly independent set cannot exceed $n$ elements. When such a set has exactly $n$ elements, the extension theorem forces the extension to be trivial, leaving the original set as a basis.[/guided]

[step:$(3) \implies (1)$: a spanning set of size $n = \dim V$ must be a basis]Assume $S = \{v_1, \ldots, v_n\}$ spans $V$. By the [Extraction of Bases from Spanning Sets](/theorems/3271), some subset $B \subset S$ is a basis for $V$. Every basis for $V$ has exactly $n$ elements, so $|B| = n$. Since $B \subset S$ and $|S| = n$, we conclude $B = S$. Therefore $S$ is a basis for $V$.[/step]

[guided]We know $S = \{v_1, \ldots, v_n\}$ spans $V$ and want to show it is a basis. The strategy mirrors the previous step: apply a theorem that extracts a basis from a spanning set, then use cardinality to show that no vectors were actually removed. The [Extraction of Bases from Spanning Sets](/theorems/3271) states that from any finite spanning set, one can extract a subset that forms a basis. Applying this to $S$, we obtain a subset $B \subset S$ that is a basis for $V$. Since $B$ is a basis for $V$, it has exactly $\dim V = n$ elements, so $|B| = n$. On the other hand, $B \subset S$ and $|S| = n$, which forces $|B| \le |S| = n$. Since $|B| = n = |S|$ and $B \subset S$, we conclude $B = S$. Therefore $S$ is a basis for $V$. In particular, $S$ is linearly independent (as every basis is), which combined with the spanning hypothesis confirms that $S$ satisfies the definition of a basis. The cardinality argument works because a basis for an $n$-dimensional space has exactly $n$ elements -- a spanning set of size $n$ cannot lose any elements during extraction.[/guided]