Alternating Multilinear Forms

What single property of area and volume makes them so useful, and so rigid? Consider the area of the parallelogram spanned by two vectors in the plane. If the two vectors coincide, the parallelogram degenerates to a line segment. So the area is zero. If we swap the two vectors, the parallelogram keeps the same shape. But the orientation flips, and the signed area changes sign. These two innocuous observations — vanishing on repeats, and sign-flipping under a swap — turn out to force area to be a specific bilinear function of the coordinates. The same phenomenon in higher dimensions forces volume to be the determinant. This article isolates the algebraic skeleton behind that phenomenon: alternating multilinear forms. These are nn-linear scalar-valued functions of nn vector arguments. They vanish whenever two arguments are equal. We will show, from first principles, that this single vanishing condition already forces full antisymmetry. That means sign change under any transposition of two arguments. More generally, it means the sign of the permutation appears under any reordering. This is a genuinely nontrivial implication, not a definitional coincidence. Its proof is the engine behind everything that follows in the theory of determinants. The importance of this idea reaches far beyond bookkeeping. Alternating forms are the correct algebraic language for signed volume, orientation, and independence testing. They underlie the change-of-variables formula in multivariable calculus, the cross product in R3\mathbb{R}^3, the Jacobian criterion for local invertibility, and the exterior algebra used throughout differential geometry and physics. Understanding why alternation implies antisymmetry — and precisely when the converse fails — is what separates a working command of determinants from a memorized formula. This article builds the definition of an alternating multilinear form from the ground up. It proves the central theorem connecting vanishing-on-repeats to sign-reversal under permutations. It works through two concrete low-dimensional examples with genuinely different numbers at each step. It shows exactly where the implication breaks down if a hypothesis (like characteristic ≠2\ne 2) is dropped. It sets the stage — without yet proving it — for a further fact. On an nn-dimensional space, the alternating nn-linear forms constitute a one-dimensional space: the space of determinants.

Key takeaways
  • An alternating multilinear form ff on a vector space VV takes nn vector inputs, is linear in each one separately, and equals 00 whenever two inputs coincide.
  • Vanishing on repeated arguments forces antisymmetry: swapping any two arguments multiplies the value by −1-1 (assuming the scalar field has characteristic ≠2\ne 2).
  • More generally, permuting all nn arguments by σ∈Sn\sigma \in S_n multiplies the value by sgn⁡(σ)\operatorname{sgn}(\sigma).
  • Alternating forms vanish on any linearly dependent list of vectors, which is why determinants detect singular matrices.
  • In characteristic 22, "vanishes on repeats" and "antisymmetric" genuinely diverge — the implication one way survives, but the converse can fail.
  • Key idea: requiring a multilinear form to vanish on repeated inputs is a compact hypothesis that secretly encodes the entire sign-permutation structure of SnS_n.

Where this lives today

Alternating multilinear forms are the backbone of the definition of the determinant, characterized (as the next stage of this theory shows) as the essentially unique alternating nn-linear form on Fn\mathbb{F}^n normalized to send the standard basis to 11. This single idea radiates outward into many areas. In multivariable calculus, the Jacobian determinant measuring how a smooth map distorts volume is exactly the value of an alternating nn-form (the standard volume form pulled back by the derivative) evaluated on the images of basis vectors; this underlies the change of variables formula for multiple integrals. In differential geometry and physics, alternating multilinear forms generalize to differential forms, the language of Stokes' theorem, electromagnetism (the field strength tensor is an alternating 22-form), and general relativity. In algebraic topology, alternating forms and their generalizations (via exterior algebra) compute intersection numbers and orientation classes. In numerical linear algebra and statistics, the fact that alternating forms vanish on dependent vectors is exploited directly: testing whether a determinant is (near) zero is a standard numerical test for whether a set of vectors is (nearly) linearly dependent, with applications to detecting multicollinearity in regression design matrices. In theoretical computer science and cryptography, computing or approximating determinants and permanents (a non-alternating cousin) connects to complexity classes and to algorithms for counting perfect matchings in bipartite graphs. Finally, the sign representation of SnS_n appearing in Theorem (b) is the same sign character used throughout representation theory of the Symmetric Group and in the combinatorics of Young tableaux.

Ask the tutor: Why does requiring an nn-linear form to vanish only on pairs of equal arguments already force full sign-reversal under every permutation, not just transpositions?

DefinitionAlternating Multilinear Form

Let VV be a vector space over a field F\mathbb{F}, and let n≥1n \ge 1 be an integer. A function

f:Vn=V×V×⋯×V⏟n times→Ff : V^n = \underbrace{V \times V \times \cdots \times V}_{n \text{ times}} \to \mathbb{F}

is called an nn-linear form (or multilinear form) on VV if for every index k∈{1,…,n}k \in \{1, \dots, n\} and every fixed choice of vectors v1,…,vk−1,vk+1,…,vn∈Vv_1, \dots, v_{k-1}, v_{k+1}, \dots, v_n \in V, the map

v↦f(v1,…,vk−1,v,vk+1,…,vn)v \mapsto f(v_1, \dots, v_{k-1}, v, v_{k+1}, \dots, v_n)

is a linear map V→FV \to \mathbb{F}. That is, ff is linear in each argument separately, holding the others fixed. An nn-linear form ff is called alternating if

f(v1,…,vn)=0whenever vi=vj for some i≠j.f(v_1, \dots, v_n) = 0 \quad \text{whenever } v_i = v_j \text{ for some } i \ne j.

Equivalently (this equivalence is proved below as part of the central theorem), ff is alternating if and only if it is antisymmetric: for every transposition (i j)(i\ j),

f(v1,…,vi,…,vj,…,vn)=−f(v1,…,vj,…,vi,…,vn).f(v_1, \dots, v_i, \dots, v_j, \dots, v_n) = -f(v_1, \dots, v_j, \dots, v_i, \dots, v_n).

We write Λn(V)\Lambda^n(V) for the set of all alternating nn-linear forms on VV; it is itself a vector space over F\mathbb{F} under pointwise addition and scalar multiplication.

Intuition

Think of ff as a machine that eats an ordered list of nn vectors and outputs a number, linearly in each slot. Multilinearity alone is a very weak condition — it is satisfied by, for instance, the function that just multiplies together the first coordinates of all nn vectors. The alternating condition is what injects geometry: it says the machine reports zero whenever the list of vectors is "degenerate" in the specific sense of having a repeated entry. The picture to hold in mind is signed area in R2\mathbb{R}^2. Fix vectors u=(a,b)u = (a,b) and w=(c,d)w = (c,d). The signed area functional

f(u,w)=ad−bcf(u, w) = ad - bc

is bilinear in (u,w)(u,w), vanishes when u=wu = w (a "parallelogram" with two equal sides has zero area), and changes sign when we swap the two arguments: f(w,u)=cb−da=−(ad−bc)=−f(u,w)f(w,u) = cb - da = -(ad-bc) = -f(u,w). The alternating property is not an extra assumption bolted onto area — it is a direct algebraic reflection of the fact that a parallelogram degenerates into a segment when its two sides coincide, and reflects the fact that swapping the labels of the two sides reverses orientation. The theorem below shows this sign-flip is not special to area: it is a completely general consequence of the vanishing-on-repeats condition, for any nn and any alternating nn-linear form, as long as we can divide by 22 in the field of scalars.

Figure
Signed area of a parallelogram is alternating: swapping the two vectors flips the sign
u = (3,0) w = (1,2) f(u,w) = 3·2 − 0·1 = 6 → swap → w = (1,2) u = (3,0) f(w,u) = 1·0 − 2·3 = −6
TheoremAlternation forces total antisymmetry

Let VV be a vector space over a field F\mathbb{F} with char⁡(F)≠2\operatorname{char}(\mathbb{F}) \ne 2, and let f:Vn→Ff : V^n \to \mathbb{F} be an nn-linear form. Suppose ff is alternating, i.e.

f(v1,…,vn)=0 whenever vi=vj for some i≠j.f(v_1,\dots,v_n) = 0 \text{ whenever } v_i = v_j \text{ for some } i \ne j.

Then:

(a) Transposition antisymmetry. For any 1≤i<j≤n1 \le i < j \le n and any v1,…,vn∈Vv_1,\dots,v_n \in V, if vi′v_i' denotes the list with the entries in positions ii and jj swapped, then

f(v1,…,vi,…,vj,…,vn)=−f(v1,…,vj,…,vi,…,vn).f(v_1,\dots,v_i,\dots,v_j,\dots,v_n) = -f(v_1,\dots,v_j,\dots,v_i,\dots,v_n).

(b) Permutation formula. For every permutation σ∈Sn\sigma \in S_n,

f(vσ(1),…,vσ(n))=sgn⁡(σ) f(v1,…,vn).f(v_{\sigma(1)}, \dots, v_{\sigma(n)}) = \operatorname{sgn}(\sigma)\, f(v_1, \dots, v_n).

(c) Vanishing on dependent lists. If v1,…,vnv_1,\dots,v_n are linearly dependent, then f(v1,…,vn)=0f(v_1,\dots,v_n) = 0. Conversely, if char⁡(F)≠2\operatorname{char}(\mathbb{F}) \ne 2 and ff satisfies the antisymmetry condition in (a) for all i≠ji \ne j, then ff is alternating.

Proof

We prove each part in turn. Throughout, fix an nn-linear form ff and use only multilinearity and the alternating hypothesis (vanishing on repeats) unless stated otherwise. The technique for (a) is the standard bilinear expansion trick: substitute a sum into one slot and expand using linearity, then use alternation to kill the appropriate terms.

Proof of (a). Fix indices i<ji < j and fix all vectors vkv_k for k≠i,jk \ne i,j. To lighten notation, write g(x,y):=f(v1,…,x,…,y,…,vn)g(x,y) := f(v_1,\dots,x,\dots,y,\dots,v_n), where xx occupies slot ii and yy occupies slot jj, and all other slots are the fixed vkv_k. Multilinearity of ff in slots ii and jj means gg is bilinear: linear in xx with yy fixed, and linear in yy with xx fixed. The alternating hypothesis applied to the list with vi=vjv_i = v_j (both equal to some vector uu) says precisely

g(u,u)=0for every u∈V.(∗)g(u,u) = 0 \quad \text{for every } u \in V. \tag{$\ast$}

Now let u,w∈Vu, w \in V be arbitrary (these will become our vi,vjv_i, v_j). Apply (∗)(\ast) to u+wu + w:

0=g(u+w,u+w).0 = g(u+w, u+w).

Expand using bilinearity of gg (linearity in the first slot, then linearity in the second slot of each resulting term):

g(u+w,u+w)=g(u,u+w)+g(w,u+w)=[g(u,u)+g(u,w)]+[g(w,u)+g(w,w)].g(u+w,u+w) = g(u,u+w) + g(w,u+w) = \big[g(u,u) + g(u,w)\big] + \big[g(w,u) + g(w,w)\big].

By (∗)(\ast), g(u,u)=0g(u,u) = 0 and g(w,w)=0g(w,w) = 0, so this collapses to

0=g(u,w)+g(w,u),0 = g(u,w) + g(w,u),

i.e.

g(u,w)=−g(w,u).g(u,w) = -g(w,u).

Unwinding the definition of gg, this says exactly

f(v1,…,u,…,w,…,vn)=−f(v1,…,w,…,u,…,vn),f(v_1,\dots,u,\dots,w,\dots,v_n) = -f(v_1,\dots,w,\dots,u,\dots,v_n),

with uu in slot ii and ww in slot jj on the left. Since u,w,v1,…,vnu,w,v_1,\dots,v_n (other than positions i,ji,j) were arbitrary, this proves (a). Note that no division by 22 was used in this step — the identity g(u,w)=−g(w,u)g(u,w) = -g(w,u) follows from (∗)(\ast) alone, for any field. (The characteristic-≠2\ne 2 hypothesis will only be needed for the converse direction.)

Proof of (b). This is an induction on the structure of SnS_n using the fact, proved in the theory of Permutations, that every permutation σ∈Sn\sigma \in S_n can be written as a product of transpositions, say

σ=τ1τ2⋯τm,\sigma = \tau_1 \tau_2 \cdots \tau_m,

and that sgn⁡(σ)=(−1)m\operatorname{sgn}(\sigma) = (-1)^m is well-defined independent of the chosen factorization (this is the standard sign homomorphism sgn⁡:Sn→{±1}\operatorname{sgn}: S_n \to \{\pm 1\}). We first check the claim for a single transposition τ=(i j)\tau = (i\ j): applying part (a) with the roles of "swap slots i,ji,j" gives exactly

f(vτ(1),…,vτ(n))=−f(v1,…,vn)=sgn⁡(τ) f(v1,…,vn),f(v_{\tau(1)}, \dots, v_{\tau(n)}) = -f(v_1,\dots,v_n) = \operatorname{sgn}(\tau)\, f(v_1,\dots,v_n),

since sgn⁡(τ)=−1\operatorname{sgn}(\tau) = -1 for a transposition. Now suppose the claim holds for all permutations expressible as a product of fewer than mm transpositions, and let σ=τ1⋯τm\sigma = \tau_1 \cdots \tau_m. Write σ′=τ2⋯τm\sigma' = \tau_2 \cdots \tau_m, so σ=τ1σ′\sigma = \tau_1 \sigma' and σ\sigma is obtained from σ′\sigma' by an additional transposition. Applying the single-transposition case to the list (vσ′(1),…,vσ′(n))(v_{\sigma'(1)}, \dots, v_{\sigma'(n)}) with the transposition τ1\tau_1 applied to positions, we get

f(vτ1(σ′(1)),…,vτ1(σ′(n)))=−f(vσ′(1),…,vσ′(n)).f(v_{\tau_1(\sigma'(1))}, \dots, v_{\tau_1(\sigma'(n))}) = -f(v_{\sigma'(1)}, \dots, v_{\sigma'(n)}).

The left side is f(vσ(1),…,vσ(n))f(v_{\sigma(1)}, \dots, v_{\sigma(n)}) since σ=τ1∘σ′\sigma = \tau_1 \circ \sigma'. By the inductive hypothesis, f(vσ′(1),…,vσ′(n))=sgn⁡(σ′)f(v1,…,vn)f(v_{\sigma'(1)},\dots,v_{\sigma'(n)}) = \operatorname{sgn}(\sigma') f(v_1,\dots,v_n), so

f(vσ(1),…,vσ(n))=−sgn⁡(σ′) f(v1,…,vn)=sgn⁡(τ1)sgn⁡(σ′) f(v1,…,vn)=sgn⁡(σ) f(v1,…,vn),f(v_{\sigma(1)},\dots,v_{\sigma(n)}) = -\operatorname{sgn}(\sigma')\, f(v_1,\dots,v_n) = \operatorname{sgn}(\tau_1)\operatorname{sgn}(\sigma')\, f(v_1,\dots,v_n) = \operatorname{sgn}(\sigma)\, f(v_1,\dots,v_n),

using sgn⁡(τ1)=−1\operatorname{sgn}(\tau_1) = -1 and multiplicativity of sgn⁡\operatorname{sgn}. This completes the induction, proving (b) for all σ∈Sn\sigma \in S_n.

Proof of (c). Suppose v1,…,vnv_1,\dots,v_n are linearly dependent. Then some vkv_k is a linear combination of the others:

vk=∑m≠kcmvmv_k = \sum_{m \ne k} c_m v_m

for scalars cm∈Fc_m \in \mathbb{F} (if instead all viv_i are zero except possibly using a nontrivial relation with a different index solved for, the same argument applies to whichever index has a nonzero coefficient in a dependence relation; such an index exists by definition of linear dependence). By multilinearity in slot kk,

f(v1,…,vk−1,vk,vk+1,…,vn)=∑m≠kcm f(v1,…,vk−1,vm,vk+1,…,vn).f(v_1,\dots,v_{k-1}, v_k, v_{k+1},\dots,v_n) = \sum_{m \ne k} c_m\, f(v_1,\dots,v_{k-1}, v_m, v_{k+1},\dots,v_n).

In each term on the right, the vector vmv_m (with m≠km \ne k) appears in slot kk and also appears in its original slot mm elsewhere in the list — that is, the list (v1,…,vk−1,vm,vk+1,…,vn)(v_1,\dots,v_{k-1},v_m,v_{k+1},\dots,v_n) has vmv_m repeated (once in position mm, once in position kk). By the alternating hypothesis, every term on the right vanishes. Hence the left side is 00, i.e. f(v1,…,vn)=0f(v_1,\dots,v_n)=0, proving (c).

Proof of the converse. Assume char⁡(F)≠2\operatorname{char}(\mathbb{F}) \ne 2 and assume ff is antisymmetric under every transposition of two slots (property (a), taken now as hypothesis). We must show ff vanishes whenever two arguments coincide. Suppose vi=vj=uv_i = v_j = u for some i≠ji \ne j. Apply antisymmetry to the transposition swapping slots ii and jj:

f(v1,…,u,…,u,…,vn)=−f(v1,…,u,…,u,…,vn),f(v_1,\dots,u,\dots,u,\dots,v_n) = -f(v_1,\dots,u,\dots,u,\dots,v_n),

since swapping two equal entries changes nothing about the list itself, yet the antisymmetry hypothesis asserts the value equals its own negative. Writing c=f(v1,…,u,…,u,…,vn)c = f(v_1,\dots,u,\dots,u,\dots,v_n), we obtain c=−cc = -c, i.e. 2c=02c = 0. Since char⁡(F)≠2\operatorname{char}(\mathbb{F}) \ne 2, the scalar 2=1+12 = 1+1 is invertible in F\mathbb{F}, so we may multiply both sides by 2−12^{-1} to conclude c=0c = 0. This is exactly the alternating condition, completing the converse. ■\blacksquare

∎

Worked examples

Example 1: A trilinear alternating form on R3\mathbb{R}^3 built from the standard determinant, evaluated on genuinely asymmetric vectors. Let V=R3V = \mathbb{R}^3 and define f:V3→Rf : V^3 \to \mathbb{R} by

f(u,v,w)=det⁡(u1v1w1u2v2w2u3v3w3)=u1(v2w3−v3w2)−v1(u2w3−u3w2)+w1(u2v3−u3v2).f(u,v,w) = \det \begin{pmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{pmatrix} = u_1(v_2 w_3 - v_3 w_2) - v_1(u_2 w_3 - u_3 w_2) + w_1(u_2 v_3 - u_3 v_2).

(We take the standard formula for a 3×33\times 3 determinant as given here for illustration; its full derivation as the alternating form is the subject of a later theorem.) Take three concrete, deliberately asymmetric vectors:

u=(2,−1,3),v=(0,4,1),w=(5,2,−2).u = (2, -1, 3), \qquad v = (0, 4, 1), \qquad w = (5, 2, -2).

Compute f(u,v,w)f(u,v,w) directly:

f(u,v,w)=2(4⋅(−2)−1⋅2)−0(…)+5((−1)⋅1−3⋅4)f(u,v,w) = 2\big(4\cdot(-2) - 1\cdot 2\big) - 0\big(\dots\big) + 5\big((-1)\cdot 1 - 3\cdot 4\big)
=2(−8−2)−0+5(−1−12)=2(−10)+5(−13)=−20−65=−85.= 2(-8-2) - 0 + 5(-1-12) = 2(-10) + 5(-13) = -20 - 65 = -85.

Now swap uu and ww (a transposition of the first and third arguments) and recompute f(w,v,u)f(w,v,u) directly, as a check on Theorem (a):

f(w,v,u)=det⁡(50224−1−213)=5(4⋅3−(−1)⋅1)−0(…)+2(2⋅1−4⋅(−2))f(w,v,u) = \det\begin{pmatrix} 5 & 0 & 2 \\ 2 & 4 & -1 \\ -2 & 1 & 3\end{pmatrix} = 5(4\cdot 3 - (-1)\cdot 1) - 0\big(\dots\big) + 2\big(2\cdot 1 - 4\cdot(-2)\big)
=5(12+1)+2(2+8)=5(13)+2(10)=65+20=85.= 5(12+1) + 2(2+8) = 5(13) + 2(10) = 65 + 20 = 85.

Indeed f(w,v,u)=85=−(−85)=−f(u,v,w)f(w,v,u) = 85 = -(-85) = -f(u,v,w), exactly as Theorem (a) predicts for the transposition swapping slots 11 and 33. Next, check the full permutation formula (Theorem (b)) with the cyclic permutation σ=(1 2 3)\sigma = (1\,2\,3) (sending slot 1→2,2→3,3→11 \to 2, 2\to 3, 3 \to 1, so that σ\sigma has sgn⁡(σ)=+1\operatorname{sgn}(\sigma) = +1 since it is a 33-cycle, a product of two transpositions). We should get f(v,w,u)=f(u,v,w)=−85f(v,w,u) = f(u,v,w) = -85:

f(v,w,u)=det⁡(05242−11−23)=0(…)−5(4⋅3−(−1)⋅1)+2(4⋅(−2)−2⋅1)f(v,w,u) = \det\begin{pmatrix} 0 & 5 & 2 \\ 4 & 2 & -1 \\ 1 & -2 & 3\end{pmatrix} = 0\big(\dots\big) - 5\big(4\cdot 3 - (-1)\cdot 1\big) + 2\big(4\cdot(-2) - 2\cdot 1\big)
=−5(12+1)+2(−8−2)=−5(13)+2(−10)=−65−20=−85.= -5(12+1) + 2(-8-2) = -5(13) + 2(-10) = -65 - 20 = -85.

This matches sgn⁡(σ)f(u,v,w)=(+1)(−85)=−85\operatorname{sgn}(\sigma) f(u,v,w) = (+1)(-85) = -85 exactly, confirming Theorem (b) numerically. Finally, illustrate part (c): since u=(2,−1,3)u = (2,-1,3), v=(0,4,1)v = (0,4,1), w=(5,2,−2)w = (5,2,-2) are (we should check) linearly independent (their determinant −85≠0-85 \ne 0 confirms this), f(u,v,w)≠0f(u,v,w) \ne 0 is consistent — Theorem (c) only forces vanishing for dependent lists, and here they are independent, so nonvanishing is exactly what we expect, not a contradiction. To see (c) at work, replace ww with w′=3u−v=(6,−7,8)w' = 3u - v = (6,-7,8), a vector explicitly dependent on u,vu,v:

f(u,v,w′)=f(u,v,3u−v)=3f(u,v,u)−f(u,v,v)=3⋅0−0=0,f(u,v,w') = f(u,v,3u-v) = 3f(u,v,u) - f(u,v,v) = 3\cdot 0 - 0 = 0,

using multilinearity in the third slot and the alternating property applied to the two resulting lists (each of which has a repeated vector). This matches Theorem (c) directly, without needing to expand the 3×33\times 3 determinant at all.

Example 2: A bilinear alternating form on a 44-dimensional space, computed via the antisymmetrization trick from Theorem (a)'s proof. Let V=R4V = \mathbb{R}^4 with standard basis e1,e2,e3,e4e_1,e_2,e_3,e_4, and suppose f:V×V→Rf: V \times V \to \mathbb{R} is any alternating bilinear form (we do not yet know a formula — we will reconstruct enough of it from a few given values using only the theorem). Suppose we are told:

f(e1,e2)=5,f(e1,e3)=−2,f(e1,e4)=0,f(e2,e3)=7,f(e2,e4)=1,f(e3,e4)=−4.f(e_1,e_2) = 5, \quad f(e_1,e_3) = -2, \quad f(e_1,e_4)=0, \quad f(e_2,e_3) = 7, \quad f(e_2,e_4) = 1, \quad f(e_3,e_4)=-4.

(This is exactly the data needed to determine ff completely on basis vectors, once we know ff is alternating and hence antisymmetric — a fact we now exploit.) By Theorem (a), f(ej,ei)=−f(ei,ej)f(e_j,e_i) = -f(e_i,e_j) for i<ji<j, and by the alternating hypothesis itself, f(ei,ei)=0f(e_i,e_i)=0 for each ii. So the full 4×44\times 4 array of values f(ei,ej)f(e_i,e_j) is the antisymmetric matrix

A=(05−20−50712−70−40−140).A = \begin{pmatrix} 0 & 5 & -2 & 0 \\ -5 & 0 & 7 & 1 \\ 2 & -7 & 0 & -4 \\ 0 & -1 & 4 & 0 \end{pmatrix}.

Now use bilinearity to compute ff on two genuinely asymmetric, non-basis vectors:

x=2e1−e2+3e4,y=e1+4e2−e3.x = 2e_1 - e_2 + 3e_4, \qquad y = e_1 + 4e_2 - e_3.

Expand by bilinearity:

f(x,y)=f(2e1−e2+3e4, e1+4e2−e3).f(x,y) = f(2e_1 - e_2 + 3e_4,\ e_1 + 4e_2 - e_3).

Distribute over the first slot, then the second, picking out the 4×3=124 \times 3 = 12 terms with nonzero coefficient product (dropping any term where the same basis vector appears in both slots, since those vanish by the alternating property):

f(x,y)=2f(e1,e1)+8f(e1,e2)−2f(e1,e3)−f(e2,e1)−4f(e2,e2)+f(e2,e3)+3f(e4,e1)+12f(e4,e2)−3f(e4,e3).f(x,y) = 2f(e_1,e_1)+8f(e_1,e_2) - 2f(e_1,e_3) - f(e_2,e_1) - 4f(e_2,e_2) + f(e_2,e_3) + 3f(e_4,e_1) + 12 f(e_4,e_2) - 3f(e_4,e_3).

Now substitute known values, using f(e1,e1)=f(e2,e2)=0f(e_1,e_1) = f(e_2,e_2) = 0, f(e2,e1)=−f(e1,e2)=−5f(e_2,e_1) = -f(e_1,e_2) = -5, f(e4,e1)=−f(e1,e4)=0f(e_4,e_1) = -f(e_1,e_4) = 0, f(e4,e2)=−f(e2,e4)=−1f(e_4,e_2) = -f(e_2,e_4) = -1, f(e4,e3)=−f(e3,e4)=4f(e_4,e_3) = -f(e_3,e_4) = 4:

f(x,y)=2(0)+8(5)−2(−2)−(−5)−4(0)+7+3(0)+12(−1)−3(4)f(x,y) = 2(0) + 8(5) - 2(-2) - (-5) - 4(0) + 7 + 3(0) + 12(-1) - 3(4)
=0+40+4+5−0+7+0−12−12=32.= 0 + 40 + 4 + 5 - 0 + 7 + 0 - 12 - 12 = 32.

So f(x,y)=32f(x,y) = 32. As a consistency check using Theorem (a) directly (rather than recomputing from scratch), we should have f(y,x)=−32f(y,x) = -32. This is a genuine, independently checkable numerical prediction of the theorem, illustrating that once the antisymmetric array of basis values is fixed, the value of ff on any pair of vectors — expressed in coordinates — is completely and consistently determined by multilinearity plus alternation, without ever needing a separate formula for ff.

Counterexamplewhy characteristic ≠ 2 is needed

The converse direction of the theorem — "antisymmetric implies alternating" — genuinely requires char⁡(F)≠2\operatorname{char}(\mathbb{F}) \ne 2, and the forward direction's proof of the converse breaks down precisely at the step "divide by 22." Here is a concrete bilinear form over the field F2={0,1}\mathbb{F}_2 = \{0,1\} (with 1+1=01+1=0) that is antisymmetric in the naive sense but not alternating. Let V=F22V = \mathbb{F}_2^2 with basis e1,e2e_1, e_2, and define the bilinear form ff by

f(e1,e1)=1,f(e1,e2)=0,f(e2,e1)=0,f(e2,e2)=1,f(e_1,e_1) = 1, \qquad f(e_1,e_2) = 0, \qquad f(e_2,e_1)=0, \qquad f(e_2,e_2)=1,

extended bilinearly to all of V×VV \times V (so f(u,v)=u1v1+u2v2f(u,v) = u_1v_1 + u_2v_2 in coordinates, the standard "dot product" over F2\mathbb{F}_2).

Checking antisymmetry. Over F2\mathbb{F}_2, every element equals its own negative, since −1=1-1 = 1 (because 1+1=01 + 1 = 0 means −1=1-1 = 1 in F2\mathbb{F}_2). So the antisymmetry condition f(u,v)=−f(v,u)f(u,v) = -f(v,u) is equivalent to the symmetry condition f(u,v)=f(v,u)f(u,v) = f(v,u) over this field. Our form f(u,v)=u1v1+u2v2f(u,v) = u_1v_1+u_2v_2 is visibly symmetric (f(u,v)=f(v,u)f(u,v)=f(v,u) trivially from the formula), hence it is also "antisymmetric" in the (vacuous, in this field) sense f(u,v)=−f(v,u)f(u,v) = -f(v,u).

But ff is not alternating. Take u=v=e1u = v = e_1. Then

f(e1,e1)=1≠0.f(e_1,e_1) = 1 \ne 0.

So ff fails to vanish on the repeated pair (e1,e1)(e_1,e_1), even though it satisfies the antisymmetry identity f(u,v)=−f(v,u)f(u,v)=-f(v,u) for every pair u,vu,v (this identity holds trivially, as noted, because −1=1-1=1 over F2\mathbb{F}_2).

Diagnosing the failure. Retracing the converse proof: from f(u,u)=−f(u,u)f(u,u) = -f(u,u) we correctly deduce 2f(u,u)=02f(u,u) = 0. But over F2\mathbb{F}_2, the scalar 2=1+1=02 = 1+1 = 0, so the equation 2f(u,u)=02f(u,u)=0 becomes the vacuous statement 0=00 = 0, which forces no constraint on f(u,u)f(u,u) whatsoever. The step "divide both sides by 22" is precisely the place where char⁡≠2\operatorname{char} \ne 2 was used, and here 22 is simply not invertible (it is not even nonzero) — division is impossible, and indeed the conclusion fails. This is exactly why the standard convention in characteristic 22 is to define "alternating" by the vanishing-on-repeats condition directly, rather than via antisymmetry, since the two notions are no longer equivalent. The forward direction of the theorem (alternating ⇒\Rightarrow antisymmetric) still holds in every characteristic, as our proof of part (a) never used division by 22; only the converse needs the hypothesis.

Scope, edge cases & common traps

SituationWhat holdsWhat can fail
char⁡(F)≠2\operatorname{char}(\mathbb{F}) \ne 2 (e.g. R,C,Q\mathbb{R}, \mathbb{C}, \mathbb{Q})Alternating   ⟺  \iff antisymmetric; full permutation formula (b)—
char⁡(F)=2\operatorname{char}(\mathbb{F}) = 2Alternating ⇒\Rightarrow antisymmetric still holdsAntisymmetric ⇏\not\Rightarrow alternating (see counterexample)
nn vectors linearly dependentf(v1,…,vn)=0f(v_1,\dots,v_n)=0 automatically, by Theorem (c)Do not assume the converse: f=0f=0 does not imply dependence unless dim⁡Λn(V)\dim\Lambda^n(V)-type uniqueness results are separately invoked
n>dim⁡Vn > \dim VEvery list is automatically dependent, so every alternating nn-form is identically zeroEasy to forget — leads to confusing 00-form with "no alternating forms exist" (they exist, they're just trivial)
Only linearity assumed, alternation not checkedMultilinear ≠\ne alternating in generalThe product-of-first-coordinates example is multilinear but not alternating

The mathematician behind it

Origins · 1 minHermann Grassmann and how this was discovered
HG
Hermann Grassmann
1809–1877 · 19th-century Germany, linear algebra and exterior algebra

Hermann Grassmann was a German polymath — schoolteacher, linguist, and mathematician — who in his 1844 work Die lineale Ausdehnungslehre ("The Theory of Linear Extension") introduced what we now call exterior algebra, the natural home for alternating multilinear forms. Working largely in isolation from the mainstream mathematical community of his time, Grassmann developed an algebra of "extensive quantities" in which the product of a quantity with itself vanishes — precisely the alternating condition at the heart of this article, generalized to an entire algebraic structure. His ideas were slow to be appreciated; much of his mathematical work only gained recognition decades later, after his death, when mathematicians such as Giuseppe Peano and later differential geometers recognized exterior algebra as the natural setting for determinants, volume forms, and eventually differential forms in the calculus on manifolds.

History
- 1750s-1770s: Cramer and Vandermonde study determinant-like expressions for solving linear systems, without a general alternating-form framework.
- 1812: Cauchy gives one of the first systematic treatments of the determinant, including the multiplication theorem and connections to permutations.
- 1844: Grassmann publishes the Ausdehnungslehre, founding exterior algebra and the general theory of alternating products.
- 1880s-1900s: Weierstrass and Kronecker develop axiomatic treatments of determinants via multilinear, alternating, normalized functions of the columns.
- 20th century: The modern axiomatic formulation — alternating multilinear forms as the defining structure behind the determinant — becomes standard in linear algebra textbooks following Bourbaki-style treatments.
Is every antisymmetric multilinear form automatically alternating?

Not in every field. If the scalar field has characteristic different from 22 (which includes R,C,Q\mathbb{R}, \mathbb{C}, \mathbb{Q}, and any field where 1+1≠01+1 \ne 0), then yes: antisymmetric and alternating are equivalent, as the theorem proves in both directions. However, over a field of characteristic 22 (such as F2\mathbb{F}_2), the equivalence can fail. The forward implication (alternating ⇒\Rightarrow antisymmetric) is always true in any characteristic, but the converse relies on dividing by 22, which is impossible when 2=02 = 0 in the field. The counterexample in this article exhibits an explicit bilinear form over F2\mathbb{F}_2 that is symmetric (hence trivially "antisymmetric" since −1=1-1=1) but does not vanish on the diagonal.

Why does the alternating property force a form to vanish on linearly dependent inputs?

If v1,…,vnv_1,\dots,v_n are linearly dependent, one vector — say vkv_k — can be written as a linear combination of the others. Multilinearity lets us expand f(v1,…,vn)f(v_1,\dots,v_n) in the kk-th slot as a sum of terms, each obtained by replacing vkv_k with one of the other vectors vmv_m (times its coefficient). But in each such term, vmv_m now appears twice in the list — once in its original slot, once in slot kk — so the alternating hypothesis forces every term in the expansion to vanish. Summing zero terms gives zero, so f(v1,…,vn)=0f(v_1,\dots,v_n)=0. This is exactly Theorem (c), proven rigorously in the proof block above.

What is the difference between multilinear and alternating multilinear?

Multilinearity is a much weaker condition: it only says the form is linear in each argument separately, with the others held fixed. Many multilinear forms are not alternating — for example, the form that multiplies together the first coordinates of nn vectors is multilinear but assigns a nonzero value even when all the input vectors are equal (as long as their first coordinate is nonzero), so it is not alternating. Alternating multilinear forms are a special, much more rigid subclass singled out by the extra vanishing-on-repeats condition, and this rigidity is exactly what makes them useful for detecting linear dependence and encoding orientation.

Does the alternating property depend on the order in which we list the vectors?

The value of ff certainly depends on the order — that is the whole point of Theorem (b), the permutation formula, which tells you precisely how the value changes (by a factor of sgn⁡(σ)\operatorname{sgn}(\sigma)) when you reorder the inputs by a permutation σ\sigma. What does not depend on order is whether ff vanishes: since sgn⁡(σ)=±1≠0\operatorname{sgn}(\sigma) = \pm 1 \ne 0 in any field where ±1\pm 1 makes sense, the equation f(vσ(1),…,vσ(n))=sgn⁡(σ)f(v1,…,vn)f(v_{\sigma(1)},\dots,v_{\sigma(n)}) = \operatorname{sgn}(\sigma) f(v_1,\dots,v_n) shows one side is zero exactly when the other is. So "the form vanishes on this unordered set of vectors" is a well-defined notion, even though the precise nonzero value it takes is order-sensitive.

Can an alternating multilinear form be nonzero on an nn-dimensional space with nn arguments?

Yes, and this is exactly the situation of the determinant: on an nn-dimensional space VV, alternating nn-linear forms exist and are nonzero exactly when applied to a linearly independent (hence basis-forming) list of nn vectors, by the contrapositive of Theorem (c). If instead you feed more than nn vectors into an alternating form on an nn-dimensional space (say n+1n+1 vectors), the list is automatically linearly dependent (since dim⁡V=n\dim V = n), so the value must be zero by Theorem (c). This is why alternating nn-linear forms on an nn-dimensional space are the natural objects of study — they are the maximal case before forced vanishing sets in.

How does this relate to the determinant of a matrix?

The determinant of an n×nn \times n matrix AA can be characterized as the unique alternating nn-linear form of the columns (or rows) of AA that assigns the value 11 to the identity matrix. This article develops exactly the algebraic machinery — vanishing on repeats, forced antisymmetry, the permutation formula — that makes such a characterization possible and well-defined; the existence and uniqueness of this special normalized alternating form, and the explicit Leibniz formula it produces via the permutation sum, are treated as the natural continuation of this material.

Is the space of all alternating nn-linear forms on an nn-dimensional space large?

No — it turns out to be exactly one-dimensional. Every alternating nn-linear form on an nn-dimensional vector space is a scalar multiple of any single nonzero one (such as the determinant with respect to a chosen basis). This uniqueness-up-to-scalar fact is a deeper theorem building directly on the permutation formula proven here, since expanding any alternating form in coordinates relative to a basis forces the Leibniz-style permutation sum, up to an overall scalar (the value on the basis itself). That theorem is the natural next step after establishing the antisymmetry properties in this article.

Where this leads