Determinant as the Unique Alternating Multilinear Form

Derivation

Determinant as the Unique Alternating Multilinear Form

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Statement

On the space of \(n\times n\) matrices over a field \(K\), there is exactly one function \(D:(K^{n})^{n}\to K\) of the ordered columns \((\mathbf{a}_1,\dots,\mathbf{a}_n)\) that is (i) multilinear — linear in each column separately, (ii) alternating — \(D=0\) whenever two columns coincide, and (iii) normalized — \(D(\mathbf{e}_1,\dots,\mathbf{e}_n)=1\) on the standard basis. That unique function is the determinant \(\det\), it equals the Leibniz sum \(\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i}A_{\sigma(i)\,i}\), and from the characterization alone it obeys the product law \(\det(AB)=\det(A)\det(B)\).

Why it matters

The determinant is usually introduced by a formula — cofactor expansion or the permutation sum \(\sum_{\sigma}\operatorname{sgn}(\sigma)\prod_i A_{i\sigma(i)}\) — that looks arbitrary and grows factorially. The characterization inverts this: three structural axioms pin down a single scalar, so every determinant identity becomes a two-line consequence of multilinearity and antisymmetry rather than a combinatorial accident. Multiplicativity \(\det(AB)=\det(A)\det(B)\) — the fact that makes the determinant a group homomorphism \(GL_n(K)\to K^{\times}\) — falls out immediately.

Physically the alternating multilinear form is the algebra of oriented \(n\)-volume: \(\det\) is the signed volume of the parallelepiped spanned by the columns, the Jacobian factor \(d^{\,n}\mathbf{y}=|\det J|\,d^{\,n}\mathbf{x}\) in every change of variables, and the object whose sign encodes orientation and whose vanishing signals linear dependence. Uniqueness is why "volume scaling under a linear map" and "the determinant" are literally the same number.

Assumptions

The scalars form a field (or at least a commutative ring with \(1\)).Linearity in each column needs scalar multiplication to distribute and commute; over a non-commutative ring "linear in each column" splits into left/right versions, the Leibniz sum loses permutation-invariance, and no single scalar-valued form satisfies all three axioms. Multilinearity holds in each column with the others held fixed.Drop it and the signed-sum-of-products structure collapses: there is no reason \(\det(c\,\mathbf{a}_j,\dots)=c\det(\dots)\), and volume scaling is lost. The number of column-slots equals the ambient dimension \(n\) (square matrices).For \(m\neq n\) vectors in \(K^n\) an alternating multilinear form either vanishes identically (\(m>n\)) or is not top-degree (\(m<n\)); the one-dimensional uniqueness result is special to the top exterior power \(\Lambda^{n}K^{n}\). Alternating is used in the strong form "\(D=0\) when two arguments are equal," not merely "\(D\) changes sign under a swap."In characteristic \(2\), \(+1=-1\), so sign-antisymmetry is vacuous (\(2D=0\) automatically) and does not force \(D\) to vanish on repeated columns; the equal-columns axiom is the correct hypothesis in all characteristics and is what the proof actually needs. The dimension of \(K^n\) is a well-defined invariant \(n\).Uniqueness is stated relative to a normalization on an \(n\)-element basis; without basis-size invariance (prior result vector-space-basis-dimension-invariance) the phrase "normalized to \(1\) on the standard basis" would be ambiguous, and "top degree" would not mean the same thing in every basis.

Derivation

\[ D(\dots,\mathbf{u},\dots,\mathbf{v},\dots)=-\,D(\dots,\mathbf{v},\dots,\mathbf{u},\dots). \]

Antisymmetry from the axiom: expand \(0=D(\dots,\mathbf{u}+\mathbf{v},\dots,\mathbf{u}+\mathbf{v},\dots)\) by multilinearity in both repeated slots; the two equal-argument terms vanish, leaving \(D(\mathbf{u},\mathbf{v})+D(\mathbf{v},\mathbf{u})=0\). So swapping two columns flips the sign. A

\[ \mathbf{a}_j=\sum_{i=1}^{n}A_{ij}\,\mathbf{e}_i,\qquad j=1,\dots,n. \]

Expand each column in the standard basis; \(A_{ij}\) is the \(i\)-th coordinate of column \(j\). Pure bookkeeping, valid because \(\{\mathbf{e}_i\}\) is a basis. A

\[ D(\mathbf{a}_1,\dots,\mathbf{a}_n)=\sum_{i_1=1}^{n}\!\cdots\!\sum_{i_n=1}^{n} A_{i_1 1}A_{i_2 2}\cdots A_{i_n n}\; D(\mathbf{e}_{i_1},\dots,\mathbf{e}_{i_n}). \]

Apply multilinearity: pull each column's sum out one slot at a time, the scalars \(A_{ij}\) coming through linearity. The sum runs over all \(n^{n}\) index tuples \((i_1,\dots,i_n)\). B

\[ D(\mathbf{e}_{i_1},\dots,\mathbf{e}_{i_n})=0\quad\text{whenever }i_p=i_q\text{ for some }p\neq q. \]

Alternating: a repeated basis vector means two equal arguments, so the form vanishes. Only tuples with all indices distinct survive — i.e. \((i_1,\dots,i_n)=(\sigma(1),\dots,\sigma(n))\) for a permutation \(\sigma\in S_n\). A

\[ D(\mathbf{e}_{\sigma(1)},\dots,\mathbf{e}_{\sigma(n)})=\operatorname{sgn}(\sigma)\,D(\mathbf{e}_1,\dots,\mathbf{e}_n). \]

Sort the permuted basis columns back to standard order by transpositions; each swap contributes a factor \(-1\) (Step 1), and \(\operatorname{sgn}(\sigma)=(-1)^{\#\text{transpositions}}\) is well-defined because the parity of a permutation is a fixed invariant. This is the one nontrivial ingredient. C

\[ D(\mathbf{a}_1,\dots,\mathbf{a}_n)=D(\mathbf{e}_1,\dots,\mathbf{e}_n)\;\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\,\prod_{i=1}^{n}A_{\sigma(i)\,i}. \]

Substitute Steps 4–5 into Step 3: only permutation tuples survive, and each contributes \(\operatorname{sgn}(\sigma)\) times the constant \(D(\mathbf{e}_1,\dots,\mathbf{e}_n)\). The value on any matrix is fixed once that single number is chosen — so the space of alternating multilinear forms is one-dimensional. B

\[ \det A:=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n}A_{\sigma(i)\,i},\qquad D=D(\mathbf{e}_1,\dots,\mathbf{e}_n)\,\det. \]

Impose normalization \(D(\mathbf{e}_1,\dots,\mathbf{e}_n)=1\); this selects the unique member, the Leibniz determinant. Existence — that this sum really is multilinear and alternating — is checked separately (pairing \(\sigma\) with \(\sigma\cdot(\text{swap})\) cancels equal-column terms), so exactly one normalized alternating multilinear form exists. B

\[ D_B(\mathbf{x}_1,\dots,\mathbf{x}_n):=\det\!\big(B[\,\mathbf{x}_1\ \cdots\ \mathbf{x}_n\,]\big). \]

Fix a matrix \(B\) and define a new form by left-multiplying the column matrix by \(B\) before taking \(\det\). The map \(\mathbf{x}\mapsto B\mathbf{x}\) is linear. C

\[ D_B\ \text{is multilinear and alternating in the columns }\mathbf{x}_j. \]

Because \(\mathbf{x}\mapsto B\mathbf{x}\) is linear, \(D_B\) is linear in each \(\mathbf{x}_j\); because equal columns \(\mathbf{x}\) give equal columns \(B\mathbf{x}\), the inner \(\det\) vanishes, so \(D_B\) is alternating. Hence \(D_B\) is an alternating multilinear form. B

\[ D_B=D_B(\mathbf{e}_1,\dots,\mathbf{e}_n)\,\det=\det(B)\,\det, \]

By Step 6 every alternating multilinear form is its standard-basis value times \(\det\); here \(D_B(\mathbf{e}_1,\dots,\mathbf{e}_n)=\det(B\,I)=\det B\). This is the crux — multiplicativity is uniqueness in disguise. C

\[ \det(BA)=D_B(\mathbf{a}_1,\dots,\mathbf{a}_n)=\det(B)\,\det(A)\;\Longrightarrow\;\det(AB)=\det(A)\det(B). \]

Evaluate \(D_B\) on the columns of \(A\): \(B[\mathbf{a}_1\cdots\mathbf{a}_n]=BA\), and Step 10 at those columns gives the product law. Renaming (or using \(\det A=\det A^{\top}\)) gives the symmetric statement. B

Result

\[ \det=\text{unique alternating multilinear form with }\det I=1,\qquad \boxed{\;\det(AB)=\det(A)\,\det(B)\;} \]

Reading. There is one and only one scalar function of a square matrix that scales linearly with each column, flips sign when two columns swap (hence vanishes on dependent columns), and equals \(1\) on the identity. Because "apply \(B\), then take that unique form" is itself such a form, it must be a constant multiple of \(\det\), and the constant is \(\det B\) — giving the multiplicative law for free.

Units check. As a pure algebraic identity \(\det\) is dimensionless, but if column \(j\) carries physical dimension \([x_j]\) then multilinearity forces \([\det A]=\prod_j[x_j]\). For the volume reading with all columns lengths \([L]\), \([\det]=[L]^n\), the dimension of \(n\)-volume; both sides of \(\det(AB)=\det(A)\det(B)\) then carry \([a]^n[b]^n\) and balance, and for a Jacobian \(\partial(\mathbf{y})/\partial(\mathbf{x})\), \([\det J]=[y]^n/[x]^n\), exactly the factor \(d^{\,n}\mathbf{y}=|\det J|\,d^{\,n}\mathbf{x}\) demands.

Limiting cases

\(n=1\): \(\det[a]=a\); multilinearity is ordinary linearity, alternation is vacuous, and the product law is the field multiplication \((ab)=a\,b\).
\(A=I\): normalization gives \(\det I=1\); then \(\det B=\det(B I)=\det(B)\det(I)\) is consistent.
Triangular \(A\): any non-identity \(\sigma\) forces an off-diagonal zero factor, so only \(\sigma=\mathrm{id}\) survives and \(\det A=\prod_i A_{ii}\).
Singular \(A\) (dependent columns): one column is a combination of the others, so alternation gives \(\det A=0\), and \(\det(AB)=0\) for every \(B\).
\(B=A^{-1}\): \(\det(A)\det(A^{-1})=\det I=1\), so \(\det(A^{-1})=1/\det A\) — forcing \(\det A\neq0\) for invertibility.
Orthogonal / rotation \(A\): \(\det A=\pm1\), volume-preserving; \(+1\) for proper rotations, \(-1\) for reflections — the sign is the orientation the axioms track.

Breaks when

Characteristic \(2\) (or any field with \(1=-1\)), using only the weak axiom. "Antisymmetric" (\(D\to-D\) on a swap) and "alternating" (\(D=0\) on repeated columns) stop being equivalent: antisymmetry gives \(2D=0\), automatic and informationless. With only antisymmetry the space of forms need not be one-dimensional and uniqueness — hence \(\det(AB)=\det A\det B\) via Step 10 — can fail. The equal-columns axiom must be used.
Non-square arrays (\(m\neq n\)). The characterization presumes as many columns as the dimension; with fewer or more there is no scalar-valued normalized top form (some basis index must repeat when \(m>n\)), and one falls back to minors / the Cauchy–Binet formula.
Non-commutative scalars (e.g. quaternionic entries). "Linear in each column" becomes ambiguous (left vs right), the Leibniz sum is not permutation-invariant, and \(\det(AB)=\det(A)\det(B)\) fails; only surrogates survive, e.g. the Dieudonné determinant valued in \(K^{\times}/[K^{\times},K^{\times}]\).
Infinite-dimensional operators. There is no top exterior power, no finite Leibniz sum, and no normalization on a finite standard basis; only restricted classes (identity-plus-trace-class, giving the Fredholm determinant) admit a determinant with a product law, and only on that class.

Failure modes

Confusing alternating with symmetric: dropping \(\operatorname{sgn}\) gives the permanent \(\operatorname{per}(A)=\sum_\sigma\prod_i A_{\sigma(i)i}\) and expecting \(\operatorname{per}(AB)=\operatorname{per}(A)\operatorname{per}(B)\). The product law is bought by alternation, not by multilinearity alone; the permanent has no product rule.
Confusing alternating with antisymmetric: proving only sign-flip-under-swap and thinking one is done. The vanishing-on-equal-columns statement is strictly stronger in characteristic \(2\) and is the axiom the uniqueness proof requires.
Assuming \(\det\) is additive: writing \(\det(A+B)=\det A+\det B\). Multilinearity is linearity in one column at a time with the others fixed, not additivity in the whole matrix; \(\det\) is degree \(n\) in the entries.
Mis-scaling: writing \(\det(cA)=c\det A\). Scaling all \(n\) columns pulls one factor of \(c\) from each, so \(\det(cA)=c^{\,n}\det A\).
Row/column index slip in Leibniz: summing \(\prod_i A_{i\,\sigma(i)}\) versus \(\prod_i A_{\sigma(i)\,i}\) — both are correct and equal (\(\det A=\det A^{\top}\)), but mixing them mid-proof produces sign errors.
Assuming parity is obvious: using \(\operatorname{sgn}(\sigma)\) without having established that \(\#\text{transpositions}\bmod 2\) is well-defined; the whole uniqueness argument rests on that lemma.

Discussion

The deep content is that the space of alternating multilinear forms of top degree \(n\) on an \(n\)-dimensional space is one-dimensional, \(\Lambda^n(K^{n})^{*}\cong K\). Step 6 is exactly this: any such form is determined by its single value on a basis, and normalization picks a distinguished generator. This is why the determinant is not one formula among many but a canonical object — it is the coordinate of the top exterior power \(\Lambda^{n}(K^{n})\cong K\), and a linear map \(A\) acts on that one-dimensional space by multiplication by the scalar \(\det A\).

The product rule then becomes almost tautological: composition of maps corresponds to composition of their actions on \(\Lambda^{n}\), and scalars multiply, so \(\Lambda^n(AB)=\Lambda^n(A)\Lambda^n(B)\) is \(\det(AB)=\det(A)\det(B)\). The same one line gives \(\det(A^{-1})=(\det A)^{-1}\) and shows \(\det:GL_n(K)\to K^{\times}\) is a group homomorphism whose kernel \(SL_n(K)\) is the volume-and-orientation-preserving maps. Because the characterization is basis-free, \(\det\) descends to a well-defined invariant of a linear operator — a change of basis by \(P\) multiplies it by \(\det P\,\det P^{-1}=1\) — a fact that leans on dimension-invariance so that "top degree" means the same thing in every basis.

Physically \(|\det A|\) is the factor by which \(A\) scales \(n\)-volume, and its sign records whether orientation is preserved. In continuum mechanics \(\det\) of the deformation gradient \(F\) is the local volume ratio \(J=\det F\) (with \(J=1\) for incompressible flow); in the change-of-variables theorem \(|\det J|\) is the Jacobian factor; in quantum many-body theory the antisymmetry of fermionic wavefunctions is literally the alternating property, so a Slater determinant is the unique (up to normalization) totally antisymmetric product state.

The one subtle logical dependency is the well-definedness of \(\operatorname{sgn}(\sigma)\) (Step 5). One proves independently — via inversion counts, or via the action on \(\prod_{i<j}(x_i-x_j)\) — that the parity of the number of transpositions expressing \(\sigma\) is invariant; only then is \(\operatorname{sgn}:S_n\to\{\pm1\}\) a homomorphism, and only then does the Leibniz formula give a consistent value. Existence (that the Leibniz sum actually is alternating) is separate from uniqueness and quietly uses \(\operatorname{sgn}\) being a homomorphism; the characterization theorem is the conjunction of both.

Common misconceptions. The determinant is not defined by cofactor expansion — that is one computational scheme, justified after the fact by uniqueness. "Multilinear" does not mean linear in the matrix as a whole. "Alternating" is subtly stronger than "antisymmetric" off characteristic \(0\) — always axiomatize with "vanishes on repeated columns." And "unique" needs all three axioms: dropping normalization leaves a one-parameter family, dropping alternation admits the permanent, dropping multilinearity admits essentially anything.

Worked examples

\[ A=\begin{pmatrix}2&1\\ 0&3\end{pmatrix},\qquad B=\begin{pmatrix}1&4\\ 2&1\end{pmatrix}. \]

Verify \(\det(AB)=\det A\,\det B\) at \(n=2\). Symbol first: \(\det\begin{pmatrix}p&q\\ r&s\end{pmatrix}=ps-qr\), the two-term Leibniz sum (identity minus swap). A

\[ \det A=(2)(3)-(1)(0)=6,\qquad \det B=(1)(1)-(4)(2)=1-8=-7. \]

Insert numbers; \(\det A\,\det B=(6)(-7)=-42\). A

\[ AB=\begin{pmatrix}2\cdot1+1\cdot2 & 2\cdot4+1\cdot1\\ 0\cdot1+3\cdot2 & 0\cdot4+3\cdot1\end{pmatrix}=\begin{pmatrix}4&9\\ 6&3\end{pmatrix},\quad \det(AB)=(4)(3)-(9)(6)=12-54=-42. \]

Ordinary matrix product, then the same \(2\times2\) formula. A

\[ \det(AB)=-42=(6)(-7)=\det A\,\det B\ \checkmark \]

Reading. Two independent routes agree exactly, as the theorem forces: the composite scales signed area by \(-42\), the product of the individual factors. Units check. All entries dimensionless here, so \(\det\) is a pure number; if columns were lengths \([L]\), each \(\det\) would be an area \([L]^2\) and \(\det(AB)\) an \([L]^4\) product, consistently.

\[ M=\begin{pmatrix}1&2&3\\ 0&1&4\\ 5&6&0\end{pmatrix},\qquad \text{find }\det M\text{ and }\det(2M). \]

Use the axioms directly: cofactor expansion (a special case of the Leibniz sum) for \(\det M\), then the multilinear scaling law \(\det(cM)=c^{\,n}\det M\). B

\[ \det M=1\,(1\cdot0-4\cdot6)-2\,(0\cdot0-4\cdot5)+3\,(0\cdot6-1\cdot5). \]

Expand along the first row; each cofactor is a \(2\times2\) determinant. Symbols arranged before arithmetic. A

\[ \det M=1(-24)-2(-20)+3(-5)=-24+40-15=1. \]

Arithmetic. A

\[ \det(2M)=2^{\,n}\det M=2^{3}\cdot1=8,\qquad n=3. \]

Multilinearity pulls a factor \(2\) out of each of the three columns, i.e. \(c^n\) with \(c=2\); equivalently \(\det(2I_3)\det(M)=8\cdot1\) by the product law. Numbers last. B

\[ \det M=1,\qquad \det(2M)=8=\det(2I_3)\,\det M\ \checkmark \]

Reading. Scaling every column by \(2\) multiplies a \(3\times3\) determinant by \(2^3=8\), not by \(2\) — the volume reading: doubling every axis multiplies \(3\)-volume by eight. Units check. \(\det M\) has dimension \([L]^3\) (a volume) if columns are lengths; \(c=2\) is dimensionless, so \(\det(2M)\) stays \([L]^3\), scaled by the pure number \(8\).

Problems

Compute \(\det A\) and \(\det B\) for \(A=\begin{pmatrix}3&0\\ 1&2\end{pmatrix}\) and \(B=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}\), then verify \(\det(AB)=\det A\,\det B\) by direct multiplication.
Solution
\(\det A=3\cdot2-0\cdot1=6\); \(\det B=0\cdot0-(-1)\cdot1=1\). \(AB=\begin{pmatrix}3\cdot0+0\cdot1 & 3\cdot(-1)+0\cdot0\\ 1\cdot0+2\cdot1 & 1\cdot(-1)+2\cdot0\end{pmatrix}=\begin{pmatrix}0&-3\\ 2&-1\end{pmatrix}\), so \(\det(AB)=0\cdot(-1)-(-3)\cdot2=6\). And \(\det A\,\det B=6\cdot1=\boxed{6}\). (\(B\) is a \(90^\circ\) rotation, \(\det=1\), orientation-preserving.)
An alternating multilinear form \(D\) on \(2\times2\) matrices has \(D(\mathbf{e}_1,\mathbf{e}_2)=\lambda\). Express \(D\begin{pmatrix}a&b\\ c&d\end{pmatrix}\) in terms of \(\lambda\), and identify which \(\lambda\) gives the determinant.
Solution
Columns are \(\mathbf{a}_1=a\mathbf{e}_1+c\mathbf{e}_2\), \(\mathbf{a}_2=b\mathbf{e}_1+d\mathbf{e}_2\). Expand by multilinearity: \(D=ab\,D(\mathbf{e}_1,\mathbf{e}_1)+ad\,D(\mathbf{e}_1,\mathbf{e}_2)+cb\,D(\mathbf{e}_2,\mathbf{e}_1)+cd\,D(\mathbf{e}_2,\mathbf{e}_2)\). Diagonal terms vanish (alternation) and \(D(\mathbf{e}_2,\mathbf{e}_1)=-\lambda\), so \(D=\lambda(ad-bc)\). The determinant is \(\lambda=1\); this exhibits the one-dimensionality of the space of forms.
Prove from the axioms that adding a multiple of one column to another leaves \(\det\) unchanged: \(\det(\dots,\mathbf{a}_j+c\,\mathbf{a}_k,\dots)=\det(\dots,\mathbf{a}_j,\dots)\) for \(j\neq k\).
Solution
By multilinearity in slot \(j\), \(\det(\dots,\mathbf{a}_j+c\,\mathbf{a}_k,\dots)=\det(\dots,\mathbf{a}_j,\dots)+c\,\det(\dots,\mathbf{a}_k,\dots)\), where in the second term column \(j\) has been replaced by \(\mathbf{a}_k\). That determinant has \(\mathbf{a}_k\) in both slot \(j\) and slot \(k\) — two equal columns — so by alternation it is \(0\). Hence the determinant is unchanged. \(\blacksquare\) This justifies the elementary column operation of Gaussian elimination.
Given \(\det A=5\) for a \(4\times4\) matrix, compute \(\det(3A)\), \(\det(A^{-1})\), and \(\det(A^{2})\).
Solution
\(\det(3A)=3^{4}\det A=81\cdot5=405\) (one factor \(3\) per column). \(\det(A^{-1})=1/\det A=1/5\), from \(\det(AA^{-1})=\det I=1\). \(\det(A^{2})=(\det A)^{2}=25\), by the product law with \(B=A\). Answers: \(\boxed{405,\ 1/5,\ 25}\).
The permanent is \(\operatorname{per}(A)=\sum_{\sigma}\prod_i A_{\sigma(i)\,i}\) (no sign). For \(A=\begin{pmatrix}1&1\\ 1&1\end{pmatrix}\) and \(B=\begin{pmatrix}1&1\\ 1&1\end{pmatrix}\), compute \(\operatorname{per}A,\ \operatorname{per}B,\ \operatorname{per}(AB)\) and show \(\operatorname{per}(AB)\neq\operatorname{per}A\,\operatorname{per}B\), explaining which axiom fails.
Solution
\(\operatorname{per}A=A_{11}A_{22}+A_{12}A_{21}=1+1=2\), likewise \(\operatorname{per}B=2\). \(AB=\begin{pmatrix}2&2\\ 2&2\end{pmatrix}\), so \(\operatorname{per}(AB)=2\cdot2+2\cdot2=8\), while \(\operatorname{per}A\,\operatorname{per}B=2\cdot2=4\neq8\). The permanent is multilinear and normalized (\(\operatorname{per}I=1\)) but NOT alternating — no sign flip on a column swap. The product law is exactly the property alternation buys, so the permanent lacks it. \(\blacksquare\)