> [!definition] > > Let $R$ be a [[Ring|ring]] and $E$ and $F$ be $R$-[[Module|modules]], and $\lambda \in L^k(E, F)$ be a [[Multilinear Map|multilinear map]], then the following are equivalent: > 1. $\lambda \circ \sigma = \sgn(\sigma) \cdot \lambda$ for any $\sigma \in S_k$. > 2. For any $v \in E^k$, if there exists $(i, j)$ such that $i \ne j$ and $v_i = v_j$, then $\lambda v = 0$. > 3. For any $v \in E^k$, if there exists $1 \le i < k$ such that $v_i = v_{i+1}$, then $\lambda v =0$. > > If the above holds, then $\lambda$ is an **alternating** multilinear map. The set $L^k_a(E, F) \subset L^k(E, F)$ is the space of alternating multilinear maps from $E$ to $F$. > > *Proof*. $(1) \Rightarrow (2) \Rightarrow (3)$ directly. > > $(3) \Rightarrow (1)$: Assume that $k = 2$ for simplicity, then > $ > \begin{align*} > \lambda(x + y, x + y) &= \lambda(x, x + y) + \lambda(y, x + y) \\ > &= \lambda(x, x) + \lambda(x, y) + \lambda(y, x) + \lambda(y, y) \\ > &= \lambda(x, y) + \lambda(y, x) = 0 > \end{align*} > $ > [!theorem] > > Let $\lambda: E^n \to F$ be an alternating linear map, $\bracsn{e^j}_1^n \subset E$, $\bracs{a_{i, j}} \subset R$, and $w_i = a_{i, j}e^j$[^1], then > $ > \lambda(w_1, \cdots, w_n) = \lambda(e^1, \cdots, e^n)\sum_{\sigma \in S_n}\sgn{(\sigma)}\paren{\prod_{i = 1}^na_{i, \sigma(i)}} > $ > *Proof*. Inductively expand > $ > \lambda(w_1, \cdots, w_n) = \sum_{\sigma: [n] \to [n]}\paren{\prod_{i = 1}^na_{i, \sigma(i)}}\lambda(e^{\sigma(1)}, \cdots, e^{\sigma(n)}) > $ > If $\sigma: [n] \to [n]$ is not bijective, then $\lambda$ vanishes on $(e^{\sigma(1)}, \cdots, e^{\sigma(n)})$. > [!theorem] Cramer's Rule[^2] > > Let $\lambda: E^n \to F$ be an alternating linear map, $\bracsn{e^j}_1^n \subset E$, $\seqf{\alpha_j} \subset R$, and $v = \alpha_je^j$, then > $ > \alpha_j\lambda(e^1, \cdots, e^n) = \lambda(e^1, \cdots, e^{j-1}, v, e^{j+1}, \cdots, e^n) > $ > *Proof*. Assume without loss of generality that $j = 1$, then > $ > \begin{align*} > \lambda(e^1, \cdots, e^{j-1}, v, e^{j+1}, \cdots, e^n) &= \alpha_k\lambda(e^1, \cdots, e^{j-1}, e^{k}, e^{j+1}, \cdots, e^n) \\ > &= \alpha_j\lambda(e^1, \cdots, e^n) > \end{align*} > $ > since $\lambda(e^1, \cdots, e^{j-1}, e^{k}, e^{j+1}, \cdots, e^n) = 0$ whenever $k \ne j$. [^1]: Einstein summation. [^2]: Lang, Theorem XIII.4.4.