[proofplan] The proof has two directions. For the forward direction, we first establish that an alternating form flips sign under any transposition of two of its arguments by expanding $\alpha$ on a tuple with the sum $v_i + v_j$ inserted at positions $i$ and $j$; multilinearity produces four terms, two of which vanish by alternation, and the surviving cross terms cancel against each other to give the sign flip. We then propagate this from transpositions to arbitrary permutations by using [Transposition Decomposition](/theorems/777) — every $\sigma \in S_k$ is a product of transpositions whose parity matches $\operatorname{sgn}(\sigma)$. For the converse, applying the permutation formula to a transposition that swaps two equal arguments gives $\alpha = -\alpha$, and since $\operatorname{char} \mathbb{F} \neq 2$ this forces $\alpha = 0$ on tuples with a repeated entry. [/proofplan] [step:Assume $\alpha$ alternating and prove transposition sign-flip] Suppose $\alpha \in \operatorname{Mult}^k(V)$ is alternating. Fix indices $1 \le i < j \le k$ and fix vectors $v_1, \dots, v_k \in V$. We will show that swapping the entries at positions $i$ and $j$ flips the sign of $\alpha$: \begin{align*} \alpha(v_1, \dots, \underbrace{v_j}_{i}, \dots, \underbrace{v_i}_{j}, \dots, v_k) = -\alpha(v_1, \dots, \underbrace{v_i}_{i}, \dots, \underbrace{v_j}_{j}, \dots, v_k). \end{align*} Consider the tuple obtained from $(v_1, \dots, v_k)$ by replacing both the $i$-th and $j$-th entries by $v_i + v_j$. Since the resulting tuple has a repeated entry (namely $v_i + v_j$ at both positions $i$ and $j$), the alternating hypothesis gives \begin{align*} \alpha(\dots, \underbrace{v_i + v_j}_{i}, \dots, \underbrace{v_i + v_j}_{j}, \dots) = 0. \end{align*} Expanding the left-hand side using multilinearity in the $i$-th and $j$-th arguments (with all other entries held fixed), we obtain a sum of four terms: \begin{align*} &\alpha(\dots, \underbrace{v_i}_{i}, \dots, \underbrace{v_i}_{j}, \dots) + \alpha(\dots, \underbrace{v_i}_{i}, \dots, \underbrace{v_j}_{j}, \dots) \\ &\qquad + \alpha(\dots, \underbrace{v_j}_{i}, \dots, \underbrace{v_i}_{j}, \dots) + \alpha(\dots, \underbrace{v_j}_{i}, \dots, \underbrace{v_j}_{j}, \dots) = 0. \end{align*} The first term has $v_i$ at both positions $i$ and $j$, and the last term has $v_j$ at both positions; both vanish by the alternating hypothesis. The two surviving cross terms therefore satisfy \begin{align*} \alpha(\dots, \underbrace{v_i}_{i}, \dots, \underbrace{v_j}_{j}, \dots) + \alpha(\dots, \underbrace{v_j}_{i}, \dots, \underbrace{v_i}_{j}, \dots) = 0, \end{align*} which is the claimed transposition sign-flip. [guided] We want to prove that an alternating form flips sign under a transposition of two of its arguments. The clever device is to use the vanishing condition not on the original tuple, but on a deformed tuple where we have **forced** a repetition: replace both the $i$-th and $j$-th entries by $v_i + v_j$. Why this particular replacement? Because expanding $\alpha$ on such a tuple by multilinearity produces exactly four terms — two "diagonal" terms with a repeated entry (which vanish by alternation) and two "cross" terms that are precisely the configuration we want to compare. Explicitly, the tuple in which positions $i$ and $j$ both equal $v_i + v_j$ has a repeated entry, so by the alternating hypothesis, \begin{align*} \alpha(\dots, \underbrace{v_i + v_j}_{i}, \dots, \underbrace{v_i + v_j}_{j}, \dots) = 0. \end{align*} By multilinearity (applied first in the $i$-th slot, then in the $j$-th slot, with all other arguments held fixed), \begin{align*} 0 &= \alpha(\dots, v_i, \dots, v_i + v_j, \dots) + \alpha(\dots, v_j, \dots, v_i + v_j, \dots) \\ &= \alpha(\dots, v_i, \dots, v_i, \dots) + \alpha(\dots, v_i, \dots, v_j, \dots) \\ &\quad + \alpha(\dots, v_j, \dots, v_i, \dots) + \alpha(\dots, v_j, \dots, v_j, \dots). \end{align*} The first and fourth terms each have a repeated argument and hence vanish by alternation. We are left with \begin{align*} \alpha(\dots, v_i, \dots, v_j, \dots) + \alpha(\dots, v_j, \dots, v_i, \dots) = 0, \end{align*} which rearranges to the transposition sign-flip \begin{align*} \alpha(\dots, v_j, \dots, v_i, \dots) = -\alpha(\dots, v_i, \dots, v_j, \dots). \end{align*} This is a structural device worth remembering: to extract anti-symmetry from a vanishing condition on coincident arguments, plug in the **sum** at the coincident positions and let multilinearity do the bookkeeping. [/guided] [/step] [step:Propagate sign-flip from transpositions to arbitrary permutations] Let $\sigma \in S_k$ be an arbitrary permutation. By [Transposition Decomposition](/theorems/777), we may write \begin{align*} \sigma = \tau_1 \tau_2 \cdots \tau_m \end{align*} as a product of $m$ transpositions $\tau_1, \dots, \tau_m \in S_k$. The sign is multiplicative, $\operatorname{sgn}(\tau_1 \cdots \tau_m) = \prod_{l=1}^m \operatorname{sgn}(\tau_l) = (-1)^m$, so $\operatorname{sgn}(\sigma) = (-1)^m$. Define, for each $l \in \{0, 1, \dots, m\}$, the permutation $\rho_l := \tau_{m-l+1} \tau_{m-l+2} \cdots \tau_m$ (the empty product giving $\rho_0 = \operatorname{id}$), so that $\rho_m = \sigma$. We show by induction on $l$ that \begin{align*} \alpha\bigl(v_{\rho_l(1)}, \dots, v_{\rho_l(k)}\bigr) = (-1)^l\, \alpha(v_1, \dots, v_k). \tag{$\ast$} \end{align*} The base case $l = 0$ is trivial since $\rho_0 = \operatorname{id}$. For the inductive step, suppose ($\ast$) holds for some $l < m$ and write $\tau_{m-l} = (a\ b)$ with $a < b$. Then $\rho_{l+1} = \tau_{m-l} \rho_l$, so the tuple $(v_{\rho_{l+1}(1)}, \dots, v_{\rho_{l+1}(k)})$ is obtained from $(v_{\rho_l(1)}, \dots, v_{\rho_l(k)})$ by swapping the entries at positions $a$ and $b$. By the transposition sign-flip established in the previous step, \begin{align*} \alpha\bigl(v_{\rho_{l+1}(1)}, \dots, v_{\rho_{l+1}(k)}\bigr) = -\alpha\bigl(v_{\rho_l(1)}, \dots, v_{\rho_l(k)}\bigr) = -(-1)^l\, \alpha(v_1, \dots, v_k) = (-1)^{l+1}\, \alpha(v_1, \dots, v_k), \end{align*} where the middle equality uses the inductive hypothesis. This completes the induction. Setting $l = m$ in ($\ast$) gives \begin{align*} \alpha\bigl(v_{\sigma(1)}, \dots, v_{\sigma(k)}\bigr) = (-1)^m\, \alpha(v_1, \dots, v_k) = \operatorname{sgn}(\sigma)\, \alpha(v_1, \dots, v_k), \end{align*} which is the claimed identity. [guided] We have proved that swapping two arguments flips the sign of $\alpha$. We now want the analogous statement for an **arbitrary** permutation $\sigma \in S_k$. The bridge is the fact that every permutation is a product of transpositions, and the parity of the number of transpositions is precisely $\operatorname{sgn}(\sigma)$. Concretely, by [Transposition Decomposition](/theorems/777), write $\sigma = \tau_1 \tau_2 \cdots \tau_m$ for some $m$ and transpositions $\tau_l$. Since the [sign homomorphism](/theorems/778) satisfies $\operatorname{sgn}(\tau) = -1$ for every transposition $\tau$ and is multiplicative, $\operatorname{sgn}(\sigma) = (-1)^m$. We must show that permuting the arguments by $\sigma$ multiplies $\alpha$ by $(-1)^m$. The cleanest way is induction on the number of transpositions already applied. Define the partial products $\rho_l := \tau_{m-l+1} \cdots \tau_m$ (so $\rho_0 = \operatorname{id}$ and $\rho_m = \sigma$). We claim \begin{align*} \alpha\bigl(v_{\rho_l(1)}, \dots, v_{\rho_l(k)}\bigr) = (-1)^l\, \alpha(v_1, \dots, v_k). \end{align*} The base case $l = 0$ is the identity statement $\alpha(v_1, \dots, v_k) = \alpha(v_1, \dots, v_k)$. For the inductive step, write $\tau_{m-l} = (a\ b)$. The permutation $\rho_{l+1} = \tau_{m-l} \rho_l$ acts on tuples by first applying $\rho_l$ and then swapping the entries at positions $a$ and $b$. Why? Because if we write $w_i := v_{\rho_l(i)}$, then $v_{\rho_{l+1}(i)} = w_{\tau_{m-l}(i)}$, and $\tau_{m-l}$ swaps positions $a$ and $b$. By the transposition sign-flip from Step 1, applied to the tuple $(w_1, \dots, w_k)$, \begin{align*} \alpha\bigl(v_{\rho_{l+1}(1)}, \dots, v_{\rho_{l+1}(k)}\bigr) = -\alpha(w_1, \dots, w_k) = -\alpha\bigl(v_{\rho_l(1)}, \dots, v_{\rho_l(k)}\bigr). \end{align*} Combined with the inductive hypothesis, this gives $(-1)^{l+1}\, \alpha(v_1, \dots, v_k)$. Taking $l = m$ yields the desired formula with $\operatorname{sgn}(\sigma) = (-1)^m$. A subtlety: the decomposition $\sigma = \tau_1 \cdots \tau_m$ is not unique — the number $m$ depends on the choice of decomposition. What is well-defined is the **parity** of $m$, which equals $\operatorname{sgn}(\sigma)$. Our argument computes $(-1)^m$ along a specific decomposition, but the final value $\operatorname{sgn}(\sigma)\, \alpha(v_1, \dots, v_k)$ is intrinsic to $\sigma$. [/guided] [/step] [step:Prove the converse via swapping equal arguments] Conversely, suppose the permutation formula \begin{align*} \alpha\bigl(v_{\sigma(1)}, \dots, v_{\sigma(k)}\bigr) = \operatorname{sgn}(\sigma)\, \alpha(v_1, \dots, v_k) \tag{$\dagger$} \end{align*} holds for every $\sigma \in S_k$ and every tuple $(v_1, \dots, v_k) \in V^k$. Fix indices $i < j$ and a tuple $(v_1, \dots, v_k)$ with $v_i = v_j$; we must show $\alpha(v_1, \dots, v_k) = 0$. Apply ($\dagger$) with $\sigma$ equal to the transposition $\tau = (i\ j) \in S_k$. Since $\operatorname{sgn}(\tau) = -1$, \begin{align*} \alpha\bigl(v_{\tau(1)}, \dots, v_{\tau(k)}\bigr) = -\alpha(v_1, \dots, v_k). \end{align*} But $\tau$ swaps only positions $i$ and $j$, so $v_{\tau(l)} = v_l$ for $l \notin \{i, j\}$, while $v_{\tau(i)} = v_j = v_i$ and $v_{\tau(j)} = v_i = v_j$ (using $v_i = v_j$). Hence the permuted tuple is identical to the original tuple, and the left-hand side equals $\alpha(v_1, \dots, v_k)$. Therefore \begin{align*} \alpha(v_1, \dots, v_k) = -\alpha(v_1, \dots, v_k), \end{align*} which gives $2\, \alpha(v_1, \dots, v_k) = 0$. Since $\operatorname{char} \mathbb{F} \neq 2$, the scalar $2 \in \mathbb{F}$ is invertible, and we conclude $\alpha(v_1, \dots, v_k) = 0$. This shows $\alpha$ is alternating and completes the proof. [guided] We assume the permutation formula and want to deduce alternation, i.e., that $\alpha$ vanishes on any tuple with a repeated entry. Suppose $v_i = v_j$ for some $i < j$ in the tuple $(v_1, \dots, v_k)$. The idea is to pick the permutation formula at the **single transposition** $\tau = (i\ j)$, which swaps positions $i$ and $j$. On the one hand, the formula ($\dagger$) gives \begin{align*} \alpha\bigl(v_{\tau(1)}, \dots, v_{\tau(k)}\bigr) = \operatorname{sgn}(\tau)\, \alpha(v_1, \dots, v_k) = -\alpha(v_1, \dots, v_k), \end{align*} since transpositions have sign $-1$. On the other hand, the permuted tuple $(v_{\tau(1)}, \dots, v_{\tau(k)})$ is what we get from $(v_1, \dots, v_k)$ by interchanging the $i$-th and $j$-th entries — and these two entries are equal! So the permuted tuple is **literally identical** to the original tuple, and the left-hand side simplifies to $\alpha(v_1, \dots, v_k)$. Combining the two evaluations yields \begin{align*} \alpha(v_1, \dots, v_k) = -\alpha(v_1, \dots, v_k), \qquad \text{i.e.,} \qquad 2\, \alpha(v_1, \dots, v_k) = 0. \end{align*} Here is where the assumption $\operatorname{char} \mathbb{F} \neq 2$ enters: the scalar $2$ is invertible in $\mathbb{F}$, so we may divide by it and conclude $\alpha(v_1, \dots, v_k) = 0$. **Failure mode if $\operatorname{char} \mathbb{F} = 2$:** in characteristic $2$, the equation $2\, \alpha = 0$ is automatic and gives no information. In that setting the two conditions ("vanishes on repeated arguments" vs. "transforms by $\operatorname{sgn}$") split apart: the latter reduces to symmetry ($\operatorname{sgn} = 1$ in $\mathbb{F}_2$), which is strictly weaker than the former. This is the standard reason multilinear algebra texts adopt characteristic $\neq 2$ when discussing alternating forms via the permutation rule. [/guided] [/step]