Existence and Uniqueness of the Determinant

We proceed in several steps: first we derive a key consequence of the alternating property (antisymmetry under transposition, and more generally under any permutation of arguments), then we use multilinearity to expand any $D \in \Lambda^n(V)$ in coordinates relative to $\mathcal{B}$, showing this expansion is forced to have a specific form (proving uniqueness), and finally we check directly that the resulting candidate formula does define an alternating form (proving existence). This mirrors the classical technique of building up the Leibniz Permutation Formula from first principles.

Step 1: Alternating $\implies$ antisymmetric. Let $D \in \Lambda^n(V)$ (not yet assumed normalized). Fix $i \neq j$ and vectors $v_1,\dots,v_n \in V$. We show swapping $v_i$ and $v_j$ negates $D$. Consider the tuple with $v_i, v_j$ both replaced by $v_i + v_j$ in positions $i,j$; by the alternating property this tuple, having a repeated entry, gives $D=0$:

with $v_i+v_j$ in both positions $i$ and $j$. Expand using multilinearity in position $i$ first, then in position $j$, treating the other argument in that slot as fixed:

The first and last terms vanish since they have a repeated argument ($v_i$ in both slots $i,j$; respectively $v_j$ in both). Hence

i.e. swapping the arguments in positions $i,j$ negates $D$. Since every permutation $\sigma \in S_n$ is a product of transpositions, iterating this shows

for every $\sigma \in S_n$, where $\mathrm{sgn}(\sigma) = (-1)^k$ if $\sigma$ is a product of $k$ transpositions (this sign is well-defined, a standard fact about permutation parity).

Step 2: Coordinate expansion forces uniqueness. Let $D \in \Lambda^n(V)$ be arbitrary. Take any $v_1,\dots,v_n \in V$ and write each in coordinates relative to $\mathcal{B}$:

so $A = (a_{ik})$ is the matrix whose $i$-th row records the coordinates of $v_i$. Expand $D(v_1,\dots,v_n)$ by multilinearity in each of the $n$ arguments successively:

This is a completely general consequence of multilinearity: expanding one slot at a time in the basis $\mathcal{B}$ produces $n^n$ terms indexed by all functions $k : \{1,\dots,n\} \to \{1,\dots,n\}$. Now use the alternating property: if $k$ is not injective, i.e. $k_p = k_q$ for some $p \neq q$, then $D(e_{k_1},\dots,e_{k_n})$ has a repeated argument and is zero. Since $\{1,\dots,n\}$ is finite, an injective function $k:\{1,\dots,n\}\to\{1,\dots,n\}$ is automatically a bijection, i.e. a permutation $\sigma \in S_n$ with $k_i = \sigma(i)$. So only permutation terms survive:

By Step 1, $D(e_{\sigma(1)},\dots,e_{\sigma(n)}) = \mathrm{sgn}(\sigma)\, D(e_1,\dots,e_n)$. Substituting,

Formula $(\ast)$ shows that $D$ is completely determined by the single scalar $D(e_1,\dots,e_n)$: every value of $D$ on every tuple of arguments is that scalar times a fixed expression in the coordinates $a_{ik}$ (the Leibniz sum, independent of $D$). Hence the linear map

is injective: if $D(e_1,\dots,e_n) = 0$ then $(\ast)$ forces $D \equiv 0$ on all of $V^n$. This map is visibly linear (evaluation at a fixed point of a vector space of functions is linear). An injective linear map from $\Lambda^n(V)$ into the $1$-dimensional space $\mathbb{F}$ shows $\dim \Lambda^n(V) \le 1$. In particular there is at most one normalized form, proving uniqueness in part 2, modulo showing $\Lambda^n(V) \neq \{0\}$, which is existence.

Step 3: Existence — the candidate defined by $(\ast)$ works. Define $D_0 : V^n \to \mathbb{F}$ by, for $v_i = \sum_k a_{ik} e_k$,

We must verify $D_0$ is multilinear, alternating, and normalized. Multilinearity. Fix all arguments except $v_i$, and consider varying $v_i = \sum_k a_{ik} e_k$ only. Each summand of $D_0$ is $\mathrm{sgn}(\sigma) \prod_{\ell} a_{\ell \sigma(\ell)}$, and the only factor depending on the $i$-th argument is $a_{i\sigma(i)}$, which is exactly the $\sigma(i)$-coordinate of $v_i$, a linear functional of $v_i$. So each summand is linear in $v_i$, and a finite sum of functions linear in $v_i$ is linear in $v_i$. This holds for each $i$, so $D_0$ is multilinear. Alternating. Suppose $v_p = v_q$ for some $p \neq q$, i.e. $a_{pk} = a_{qk}$ for all $k$. Let $\tau = (p\,q)$ be the transposition swapping $p,q$. Partition $S_n$ into pairs $\{\sigma, \sigma\tau\}$ (this is a valid pairing since $\tau$ has order $2$ and $\sigma \mapsto \sigma\tau$ is a fixed-point-free involution on $S_n$, because $\sigma \tau = \sigma$ would force $\tau = e$, false). We compare the two terms in a pair:

Now $\mathrm{sgn}(\sigma\tau) = -\mathrm{sgn}(\sigma)$ since $\tau$ is a transposition. For the products: $(\sigma\tau)(\ell) = \sigma(\ell)$ for $\ell \neq p,q$, and $(\sigma\tau)(p) = \sigma(q)$, $(\sigma\tau)(q) = \sigma(p)$. So

Since $a_{pk}=a_{qk}$ for all $k$, we have $a_{p\sigma(q)} = a_{q\sigma(q)}$ and $a_{q\sigma(p)} = a_{p\sigma(p)}$, so

which equals the corresponding factor in $\prod_\ell a_{\ell \sigma(\ell)}$. Hence $\prod_\ell a_{\ell(\sigma\tau)(\ell)} = \prod_\ell a_{\ell\sigma(\ell)}$, and the two terms in the pair are

which cancel. Summing over all pairs gives $D_0(v_1,\dots,v_n) = 0$. So $D_0$ is alternating. Normalization. At $v_i = e_i$ for all $i$, the coordinate matrix is the identity: $a_{ik} = 1$ if $i=k$ and $0$ otherwise. Then $\prod_\ell a_{\ell\sigma(\ell)} = 0$ unless $\sigma(\ell) = \ell$ for every $\ell$, i.e. unless $\sigma = e$ (identity permutation), in which case the product is $1$. So

Thus $D_0 \in \Lambda^n(V)$ and $D_0(e_1,\dots,e_n) = 1$, establishing existence. Combined with Step 2 ($\dim \Lambda^n(V)\le 1$) and $D_0 \ne 0$ (since $D_0(e_1,\dots,e_n)=1\ne0$), we conclude $\dim \Lambda^n(V) = 1$ exactly, and $D_0$ is the unique normalized alternating form. $\blacksquare$