[proofplan] The four claims are proved in sequence. Linearity is immediate from the form of the defining sum: $\operatorname{Alt}(T)$ is a fixed $\mathbb{K}$-linear combination of evaluations of $T$. To show $\operatorname{Alt}(T)$ is alternating, we permute its arguments by a fixed $\tau \in S_k$ and reindex the sum by $\rho := \tau \circ \sigma$; the multiplicativity of $\operatorname{sgn}$ extracts a factor of $\operatorname{sgn}(\tau)$. If $T$ is already alternating, each of the $k!$ terms collapses to $\operatorname{sgn}(\sigma)^2\, T(v_1,\dots,v_k) = T(v_1,\dots,v_k)$, so the average reproduces $T$. Idempotence then follows by combining (1) and (3): $\operatorname{Alt}(T) \in \Lambda^k(V^*)$, hence $\operatorname{Alt}$ fixes it. [/proofplan]

[step:Verify that $\operatorname{Alt}(T)$ is a well-defined multilinear form on $V^k$] Fix $T \in \operatorname{Mult}^k(V)$. For each $\sigma \in S_k$, define \begin{align*} T_\sigma : V^k &\to \mathbb{K} \\ (v_1,\dots,v_k) &\mapsto T(v_{\sigma(1)},\dots,v_{\sigma(k)}). \end{align*} Since $T$ is multilinear and the map $(v_1,\dots,v_k) \mapsto (v_{\sigma(1)},\dots,v_{\sigma(k)})$ is a permutation of coordinates, $T_\sigma$ is multilinear in each argument $v_i$. The expression \begin{align*} \operatorname{Alt}(T) \;=\; \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma)\, T_\sigma \end{align*} is a finite $\mathbb{K}$-linear combination of multilinear forms (the division by $k! \neq 0$ in $\mathbb{K}$ uses $\operatorname{char}(\mathbb{K}) = 0$), and $\operatorname{Mult}^k(V)$ is a $\mathbb{K}$-[vector space](/page/Vector%20Space), so $\operatorname{Alt}(T) \in \operatorname{Mult}^k(V)$. [/step]

[step:Prove linearity of $\operatorname{Alt}$] Let $S, T \in \operatorname{Mult}^k(V)$ and $\lambda, \mu \in \mathbb{K}$. For every $(v_1,\dots,v_k) \in V^k$ and every $\sigma \in S_k$, \begin{align*} (\lambda S + \mu T)(v_{\sigma(1)},\dots,v_{\sigma(k)}) \;=\; \lambda\, S(v_{\sigma(1)},\dots,v_{\sigma(k)}) + \mu\, T(v_{\sigma(1)},\dots,v_{\sigma(k)}), \end{align*} by definition of the [vector space](/page/Vector%20Space) structure on $\operatorname{Mult}^k(V)$. Multiplying by $\operatorname{sgn}(\sigma) \in \{\pm 1\}$, summing over $\sigma \in S_k$, and dividing by $k!$, we obtain \begin{align*} \operatorname{Alt}(\lambda S + \mu T)(v_1,\dots,v_k) &= \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma)\bigl[\lambda\, S(v_{\sigma(1)},\dots,v_{\sigma(k)}) + \mu\, T(v_{\sigma(1)},\dots,v_{\sigma(k)})\bigr] \\ &= \lambda \cdot \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma)\, S(v_{\sigma(1)},\dots,v_{\sigma(k)}) + \mu \cdot \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma)\, T(v_{\sigma(1)},\dots,v_{\sigma(k)}) \\ &= \lambda\, \operatorname{Alt}(S)(v_1,\dots,v_k) + \mu\, \operatorname{Alt}(T)(v_1,\dots,v_k). \end{align*} Since this holds for every $(v_1,\dots,v_k) \in V^k$, $\operatorname{Alt}(\lambda S + \mu T) = \lambda \operatorname{Alt}(S) + \mu \operatorname{Alt}(T)$. [/step]

[step:Show $\operatorname{Alt}(T)$ is alternating by reindexing the symmetric group sum]Fix $T \in \operatorname{Mult}^k(V)$, $\tau \in S_k$, and $(v_1,\dots,v_k) \in V^k$. We compute the effect of permuting the arguments of $\operatorname{Alt}(T)$ by $\tau$: \begin{align*} \operatorname{Alt}(T)(v_{\tau(1)},\dots,v_{\tau(k)}) &= \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma)\, T(v_{\tau(\sigma(1))},\dots,v_{\tau(\sigma(k))}) \\ &= \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma)\, T(v_{(\tau\sigma)(1)},\dots,v_{(\tau\sigma)(k))}). \end{align*} We reindex by $\rho := \tau\sigma \in S_k$. Since left multiplication by $\tau$ is a bijection $S_k \to S_k$, the map $\sigma \mapsto \rho$ is a bijection, and $\sigma = \tau^{-1}\rho$. By the [Sign Homomorphism](/theorems/778), $\operatorname{sgn}: S_k \to \{\pm 1\}$ is a group homomorphism, so \begin{align*} \operatorname{sgn}(\sigma) = \operatorname{sgn}(\tau^{-1}\rho) = \operatorname{sgn}(\tau^{-1})\operatorname{sgn}(\rho) = \operatorname{sgn}(\tau)\operatorname{sgn}(\rho), \end{align*} where the last equality uses $\operatorname{sgn}(\tau^{-1}) = \operatorname{sgn}(\tau)^{-1} = \operatorname{sgn}(\tau)$ (since $\operatorname{sgn}(\tau) \in \{\pm 1\}$). Substituting, \begin{align*} \operatorname{Alt}(T)(v_{\tau(1)},\dots,v_{\tau(k)}) &= \frac{1}{k!}\sum_{\rho \in S_k} \operatorname{sgn}(\tau)\operatorname{sgn}(\rho)\, T(v_{\rho(1)},\dots,v_{\rho(k)}) \\ &= \operatorname{sgn}(\tau) \cdot \frac{1}{k!}\sum_{\rho \in S_k} \operatorname{sgn}(\rho)\, T(v_{\rho(1)},\dots,v_{\rho(k)}) \\ &= \operatorname{sgn}(\tau)\, \operatorname{Alt}(T)(v_1,\dots,v_k). \end{align*} Since $\tau \in S_k$ and $(v_1,\dots,v_k) \in V^k$ were arbitrary, $\operatorname{Alt}(T)$ satisfies the defining transformation law of an alternating multilinear form. Therefore $\operatorname{Alt}(T) \in \Lambda^k(V^*)$, which establishes claim (1).[/step]

[guided]We want to show that the form $\operatorname{Alt}(T)$ — which we built by averaging signed permutations of $T$ — is itself alternating. By definition, this means: for every $\tau \in S_k$ and every $(v_1,\dots,v_k) \in V^k$, \begin{align*} \operatorname{Alt}(T)(v_{\tau(1)},\dots,v_{\tau(k)}) = \operatorname{sgn}(\tau)\,\operatorname{Alt}(T)(v_1,\dots,v_k). \end{align*} The strategy is to write out the left-hand side from the definition, then perform a change of summation variable that converts the permuted arguments back into the standard form $(v_1,\dots,v_k)$, paying a factor of $\operatorname{sgn}(\tau)$ along the way. Fix $\tau \in S_k$ and $(v_1,\dots,v_k) \in V^k$. By definition of $\operatorname{Alt}$ applied to the permuted tuple, \begin{align*} \operatorname{Alt}(T)(v_{\tau(1)},\dots,v_{\tau(k)}) &= \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma)\, T(w_{\sigma(1)},\dots,w_{\sigma(k)}), \end{align*} where we have set $w_i := v_{\tau(i)}$ for clarity. Now $w_{\sigma(i)} = v_{\tau(\sigma(i))} = v_{(\tau\sigma)(i)}$, so the inner term is $T(v_{(\tau\sigma)(1)},\dots,v_{(\tau\sigma)(k)})$. The natural change of variable is $\rho := \tau\sigma$. *Why is this a valid reindexing?* Because left multiplication by $\tau$ is a bijection $S_k \to S_k$ — its inverse is left multiplication by $\tau^{-1}$. So as $\sigma$ ranges over all of $S_k$, $\rho = \tau\sigma$ also ranges over all of $S_k$ exactly once. Under this reindexing, $\sigma = \tau^{-1}\rho$, so we need to express $\operatorname{sgn}(\sigma)$ in terms of $\operatorname{sgn}(\rho)$. This is exactly what the [Sign Homomorphism](/theorems/778) provides: $\operatorname{sgn}: S_k \to \{\pm 1\}$ is a group homomorphism, so \begin{align*} \operatorname{sgn}(\sigma) = \operatorname{sgn}(\tau^{-1}\rho) = \operatorname{sgn}(\tau^{-1})\,\operatorname{sgn}(\rho). \end{align*} Furthermore $\operatorname{sgn}(\tau^{-1}) = \operatorname{sgn}(\tau)^{-1}$ (homomorphism), and since $\operatorname{sgn}(\tau) \in \{\pm 1\}$, its inverse equals itself: $\operatorname{sgn}(\tau^{-1}) = \operatorname{sgn}(\tau)$. Thus $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\tau)\operatorname{sgn}(\rho)$. Substituting, \begin{align*} \operatorname{Alt}(T)(v_{\tau(1)},\dots,v_{\tau(k)}) &= \frac{1}{k!}\sum_{\rho \in S_k} \operatorname{sgn}(\tau)\operatorname{sgn}(\rho)\, T(v_{\rho(1)},\dots,v_{\rho(k)}) \\ &= \operatorname{sgn}(\tau)\cdot \frac{1}{k!}\sum_{\rho \in S_k} \operatorname{sgn}(\rho)\, T(v_{\rho(1)},\dots,v_{\rho(k)}) \\ &= \operatorname{sgn}(\tau)\, \operatorname{Alt}(T)(v_1,\dots,v_k), \end{align*} where we pulled the constant $\operatorname{sgn}(\tau)$ out of the sum and recognised the remaining expression as the definition of $\operatorname{Alt}(T)(v_1,\dots,v_k)$. Since $\tau$ and $(v_1,\dots,v_k)$ were arbitrary, $\operatorname{Alt}(T)$ satisfies the alternating transformation law, so $\operatorname{Alt}(T) \in \Lambda^k(V^*)$. This proves claim (1): $\operatorname{Alt}$ takes values in $\Lambda^k(V^*)$.[/guided]

[step:Show $\operatorname{Alt}$ fixes already-alternating forms] Let $\alpha \in \Lambda^k(V^*)$ and $(v_1,\dots,v_k) \in V^k$. By the defining alternating property, for every $\sigma \in S_k$, \begin{align*} \alpha(v_{\sigma(1)},\dots,v_{\sigma(k)}) = \operatorname{sgn}(\sigma)\, \alpha(v_1,\dots,v_k). \end{align*} Substituting into the definition of $\operatorname{Alt}$ and using $\operatorname{sgn}(\sigma)^2 = 1$, \begin{align*} \operatorname{Alt}(\alpha)(v_1,\dots,v_k) &= \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma)\, \alpha(v_{\sigma(1)},\dots,v_{\sigma(k)}) \\ &= \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma)^2\, \alpha(v_1,\dots,v_k) \\ &= \frac{1}{k!}\sum_{\sigma \in S_k} \alpha(v_1,\dots,v_k) \\ &= \frac{|S_k|}{k!}\,\alpha(v_1,\dots,v_k) \\ &= \alpha(v_1,\dots,v_k), \end{align*} since $|S_k| = k!$. As $(v_1,\dots,v_k)$ was arbitrary, $\operatorname{Alt}(\alpha) = \alpha$, proving claim (3). [/step]

[step:Deduce idempotence by combining the range and fixed-point properties] Let $T \in \operatorname{Mult}^k(V)$. By the previous steps, $\operatorname{Alt}(T) \in \Lambda^k(V^*)$ (claim (1)), and $\operatorname{Alt}$ fixes every element of $\Lambda^k(V^*)$ (claim (3)). Applying claim (3) with $\alpha := \operatorname{Alt}(T)$, \begin{align*} \operatorname{Alt}(\operatorname{Alt}(T)) = \operatorname{Alt}(T). \end{align*} This proves claim (4). Combined with linearity (claim (2)) and surjectivity onto $\Lambda^k(V^*)$ — which follows from claims (1) and (3), since $\operatorname{Alt}(\alpha) = \alpha$ for $\alpha \in \Lambda^k(V^*)$ — the map $\operatorname{Alt}: \operatorname{Mult}^k(V) \to \Lambda^k(V^*)$ is a linear projection onto $\Lambda^k(V^*)$, completing the proof. [/step]