[proofplan] We prove the equivalence by a cyclic chain of implications $(1) \implies (2) \implies (3) \implies (1)$. For $(1) \implies (2)$, we take a nontrivial dependence relation and solve for the vector with the largest index carrying a nonzero coefficient. For $(2) \implies (3)$, any linear combination involving $v_j$ can be rewritten using the expression for $v_j$ in terms of the remaining vectors. For $(3) \implies (1)$, removing a vector that lies in the span of the others directly yields a nontrivial dependence relation. [/proofplan] [step:$(1) \implies (2)$: Extract a vector from a nontrivial dependence relation] Since the list $(v_1, \ldots, v_k)$ is linearly dependent, there exist scalars $a_1, \ldots, a_k \in F$, not all zero, such that \begin{align*} a_1 v_1 + a_2 v_2 + \cdots + a_k v_k = 0. \end{align*} Let $j = \max\{i \in \{1, \ldots, k\} : a_i \neq 0\}$. Since not all $a_i$ are zero, this maximum exists and $a_j \neq 0$. Solving for $v_j$: \begin{align*} v_j = -a_j^{-1}\bigl(a_1 v_1 + \cdots + a_{j-1} v_{j-1}\bigr). \end{align*} Here $a_j^{-1}$ exists because $a_j \neq 0$ and $F$ is a field. The right-hand side is a linear combination of vectors from the list $(v_1, \ldots, \hat{v}_j, \ldots, v_k)$, so $v_j \in \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. [guided] The goal is to show that one of the vectors in the list can be written as a linear combination of the others. We start from the definition of linear dependence: there exist scalars $a_1, \ldots, a_k \in F$, not all zero, such that \begin{align*} a_1 v_1 + a_2 v_2 + \cdots + a_k v_k = 0. \end{align*} To isolate a particular vector, we need a nonzero coefficient. Let $j = \max\{i \in \{1, \ldots, k\} : a_i \neq 0\}$. Why take the maximum index? Any index with a nonzero coefficient would work for this implication; the choice of the largest index is a convention that becomes useful in applications (for instance, in the proof that every spanning list can be reduced to a basis, where one removes vectors from the end of the list to preserve a prefix). Since $a_j \neq 0$ and $F$ is a field, the inverse $a_j^{-1} \in F$ exists. Moreover, $a_i = 0$ for all $i > j$ by the definition of $j$, so the dependence relation reduces to $a_1 v_1 + \cdots + a_j v_j = 0$. Rearranging: \begin{align*} v_j = -a_j^{-1}\bigl(a_1 v_1 + \cdots + a_{j-1} v_{j-1}\bigr). \end{align*} The right-hand side is a linear combination of vectors from the sub-list $(v_1, \ldots, v_{j-1})$, which is contained in $(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. Therefore $v_j \in \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. [/guided] [/step] [step:$(2) \implies (3)$: Rewrite any linear combination to avoid $v_j$] Suppose $v_j \in \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$ for some $j \in \{1, \ldots, k\}$. Then there exist scalars $b_i \in F$ (for $i \neq j$) such that \begin{align*} v_j = \sum_{\substack{i=1 \\ i \neq j}}^{k} b_i v_i. \end{align*} We claim that $\operatorname{span}(v_1, \ldots, v_k) = \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. The inclusion $\operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k) \subset \operatorname{span}(v_1, \ldots, v_k)$ holds because the shorter list is a sub-list of the longer one. For the reverse inclusion, let $v \in \operatorname{span}(v_1, \ldots, v_k)$. Then there exist scalars $c_1, \ldots, c_k \in F$ with $v = \sum_{i=1}^{k} c_i v_i$. Substituting the expression for $v_j$: \begin{align*} v = \sum_{\substack{i=1 \\ i \neq j}}^{k} c_i v_i + c_j v_j = \sum_{\substack{i=1 \\ i \neq j}}^{k} c_i v_i + c_j \sum_{\substack{i=1 \\ i \neq j}}^{k} b_i v_i = \sum_{\substack{i=1 \\ i \neq j}}^{k} (c_i + c_j b_i) v_i. \end{align*} This expresses $v$ as a linear combination of the list with $v_j$ removed, so $v \in \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. Since $v$ was arbitrary, $\operatorname{span}(v_1, \ldots, v_k) \subset \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. Combined with the other inclusion, removing $v_j$ does not change the span. [guided] We want to show that after removing $v_j$, the remaining vectors still span everything that the original list spanned. The idea is simple: any linear combination that uses $v_j$ can be rewritten by substituting the expression for $v_j$ in terms of the other vectors. Since $v_j \in \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$, there exist scalars $b_i \in F$ (for $i \neq j$) such that \begin{align*} v_j = \sum_{\substack{i=1 \\ i \neq j}}^{k} b_i v_i. \end{align*} We need to show $\operatorname{span}(v_1, \ldots, v_k) = \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. One inclusion is immediate: the shorter list is a sub-list of the longer one, so $\operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k) \subset \operatorname{span}(v_1, \ldots, v_k)$. For the reverse, take any $v \in \operatorname{span}(v_1, \ldots, v_k)$, so $v = \sum_{i=1}^{k} c_i v_i$ for some $c_1, \ldots, c_k \in F$. Replace $v_j$ by its expression: \begin{align*} v = \sum_{\substack{i=1 \\ i \neq j}}^{k} c_i v_i + c_j \sum_{\substack{i=1 \\ i \neq j}}^{k} b_i v_i = \sum_{\substack{i=1 \\ i \neq j}}^{k} (c_i + c_j b_i) v_i. \end{align*} This is a linear combination of the $(k-1)$ vectors $(v_1, \ldots, \hat{v}_j, \ldots, v_k)$, so $v \in \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. The key point is that the substitution works for any $v$ in the span, regardless of the coefficient $c_j$ — even if $c_j = 0$, the expression is valid. [/guided] [/step] [step:$(3) \implies (1)$: Construct a nontrivial dependence relation from a redundant vector] Suppose removing $v_j$ does not change the span for some $j \in \{1, \ldots, k\}$, i.e., $\operatorname{span}(v_1, \ldots, v_k) = \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. Since $v_j \in \operatorname{span}(v_1, \ldots, v_k) = \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$, there exist scalars $b_i \in F$ (for $i \neq j$) such that \begin{align*} v_j = \sum_{\substack{i=1 \\ i \neq j}}^{k} b_i v_i. \end{align*} Rearranging: \begin{align*} \sum_{\substack{i=1 \\ i \neq j}}^{k} b_i v_i + (-1) v_j = 0. \end{align*} This is a linear combination of $(v_1, \ldots, v_k)$ equaling zero in which the coefficient of $v_j$ is $-1 \neq 0$, so it is a nontrivial dependence relation. Therefore the list $(v_1, \ldots, v_k)$ is linearly dependent. [guided] We assume that removing some vector $v_j$ from the list does not change the span, i.e., $\operatorname{span}(v_1, \ldots, v_k) = \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. Our goal is to produce a nontrivial linear combination of $(v_1, \ldots, v_k)$ that equals zero. The key observation is that $v_j$ itself belongs to $\operatorname{span}(v_1, \ldots, v_k)$ (it is one of the vectors in the list), and since the span does not change upon removing $v_j$, we have $v_j \in \operatorname{span}(v_1, \ldots, \hat{v}_j, \ldots, v_k)$. This means there exist scalars $b_i \in F$ (for $i \neq j$) such that \begin{align*} v_j = \sum_{\substack{i=1 \\ i \neq j}}^{k} b_i v_i. \end{align*} Now we rearrange this equation by moving $v_j$ to the right-hand side: \begin{align*} \sum_{\substack{i=1 \\ i \neq j}}^{k} b_i v_i + (-1) v_j = 0. \end{align*} This is a linear combination of all $k$ vectors $(v_1, \ldots, v_k)$ that sums to the zero vector. Is it nontrivial? Yes: the coefficient of $v_j$ is $-1 \neq 0$, so not all coefficients are zero. By the definition of linear dependence, the list $(v_1, \ldots, v_k)$ is linearly dependent. Note that the logical structure here mirrors the reverse of implication $(1) \implies (2)$: there, we started with a dependence relation and isolated a vector; here, we start with an isolated vector and build a dependence relation. The two constructions are essentially inverses of each other, which is why the equivalence holds. [/guided] [/step]