Determinants of Matrices over a Commutative Ring

Entrywise ring addition gives its abelian-group laws. Substitution in the product formula gives $I_mA=A=AI_n$ and both distributive laws, since all off-diagonal identity terms are zero and the diagonal term is one.

For compatible $A,B,C$, the $(i,\ell)$ entry of $(AB)C$ is $\sum_k(\sum_j a_{ij}b_{jk})c_{k\ell}$, while that of $A(BC)$ is $\sum_j a_{ij}(\sum_k b_{jk}c_{k\ell})$. Distributivity and finite-sum Fubini turn both into $\sum_{j,k}a_{ij}b_{jk}c_{k\ell}$.

The $(k,i)$ entry of $(AB)^{\mathsf T}$ is $\sum_j a_{ij}b_{jk}=\sum_j b_{jk}a_{ij}$, the $(k,i)$ entry of $B^{\mathsf T}A^{\mathsf T}$; transposing twice restores each entry. Empty sums are zero, so every argument also covers zero-sized shapes.

By [L1], [L3] and [L6], both matrix carriers are the same function set. The formulas in [L2], [L4] and [L5] agree entry for entry, including empty sums and zero-sized shapes.

Since the products and identities agree, the two-sided inverse equations in [L7] and [L8] select the same matrices and the same unique inverses; then the conjugation formulas in [L7] and [L9] select the same similar pairs.

Put $u+v$ in both selected positions. Alternation gives $0=F(\ldots,u+v,\ldots,u+v,\ldots)$.

Expanding twice gives $0=F(\ldots,u,\ldots,v,\ldots)+F(\ldots,v,\ldots,u,\ldots)$ because the two equal-column terms vanish. Adding the inverse of the first term yields the antisymmetry formula, without dividing by $2$.

For $r<n$, let $e_r$ be the column with entry $1$ in row $r$ and $0$ elsewhere. Since column $i$ is $\sum_{r<n}a_{ri}e_r$, repeated multilinearity expands $F(A)$ as the finite sum over tuples $(r_i)_{i<n}$ of $\bigl(\prod_{i<n}a_{r_i i}\bigr)F(e_{r_0},\ldots,e_{r_{n-1}})$.

If a tuple repeats a row index, its $F$-value is zero by alternation. The surviving tuples use every element of $n$ once, so they are precisely the permutations $r_i=\sigma(i)$.

Factor $\sigma$ into transpositions. Repeated antisymmetry changes $F(I_n)$ by one minus sign per transposition. Each transposition has sign $-1$, so [L4] identifies the product of these signs with $\operatorname{sgn}(\sigma)$; equivalently [L3] and [L7] identify it with the factorisation-independent inversion parity. Substitution into step 2.1 gives the formula.

By [L2]–[L4], the displayed finite sum of signed monomials is an element $P_n$ of the real polynomial ring in the $n^2$ variables $x_{ri}$.

Iterate [L5] through the finite list of variables, sending each $x_{ri}$ to $a_{ri}$. The resulting homomorphism sends each monomial to $\prod_{i<n}a_{\sigma(i),i}$ and therefore sends $P_n$ to the Leibniz sum $\det(A)$ from [L1].

Hence determinant is the polynomial function given by this explicit finite sum of finite products in the matrix entries.

With every column but column $q$ fixed, each Leibniz monomial contains exactly one entry from column $q$. Distributing finite sums therefore proves additivity and scalar compatibility in that column, and $q$ was arbitrary.

Suppose columns $p,q$ are equal. Pair each $\sigma$ with $\sigma\circ(p\ q)$. Commutativity makes the paired monomials equal, while [L5] makes their signs opposite, so every pair sums to zero and the determinant vanishes.

At $I_n$, every nonidentity permutation selects an off-diagonal zero, while the identity term is $1$. Thus $\det(I_n)=1$. This also covers $n=1$ and the zero ring, where the equality reads $0=0$.

Fact [L2] proves existence of a function with the three properties.

If $F$ has them, normalization gives $F(I_n)=1$, so [L1] gives $F(A)=\det(A)$ for every $A$. Hence the function is unique.

If one column is zero, multilinearity gives $\det(A)=\det(A)+\det(A)$ by writing that column as $0+0$; cancellation in the additive group gives $\det(A)=0$.

If two columns are equal, alternation in [L1] gives $\det(A)=0$ directly, regardless of whether the columns are adjacent.

Substituting [L2] into [L1] gives $\det(A^{\mathsf T})=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i<n}a_{i,\sigma(i)}$.

Reindex the sum by $\tau=\sigma^{-1}$ and the product by $j=\sigma(i)$. Commutativity and [L3] turn the expression into $\sum_{\tau\in S_n}\operatorname{sgn}(\tau)\prod_{j<n}a_{\tau(j),j}=\det(A)$.

Transposition sends a selected row, a row sum, a row scaling or a row swap to the corresponding operation on a selected column, without changing determinant.

Apply the column multilinearity and alternation of [L2] to the transpose, then apply [L1] again. The resulting identities are exactly row multilinearity and row alternation for the original matrix.

Row multilinearity gives $\det(\ldots,cR_p,\ldots)=c\det(A)$ for every scalar $c$, including $c=0$ and zero divisors.

Row antisymmetry gives $\det(A')=-\det(A)$ when $A'$ is obtained by a row swap.

Replacing $R_p$ by $R_p+cR_q$ expands the new determinant as $\det(A)+cD$, where $D$ has row $R_q$ in both positions $p$ and $q$; alternation gives $D=0$, so the determinant is unchanged.

The derivations used only row multilinearity and alternation, not cancellation, matrix invertibility or scalar invertibility. Hence all three formulas include singular matrices and arbitrary scalars.

For an upper triangular matrix, a nonzero Leibniz term requires $\sigma(i)\le i$ for every $i$. A permutation with these inequalities is the identity: starting at $i=0$ and proceeding upward, bijectivity forces $\sigma(i)=i$. Thus every nonidentity term contains a zero entry.

The identity term has sign $1$ and equals $\prod_{i<n}a_{ii}$, so it is the whole determinant.

For a lower triangular matrix the same argument uses $\sigma(i)\ge i$ and proceeds downward, again forcing the identity. The argument includes $n=1$ and makes no cancellation assumption, so zero divisors and the zero ring cause no exception.

Define $F(B):=\det(AB)$. Each column of $AB$ is $A$ times the corresponding column of $B$, so distributivity makes $F$ column-multilinear; equal columns of $B$ give equal columns of $AB$, so $F$ is alternating.

Since $AI_n=A$, one has $F(I_n)=\det(A)$.

Apply [L2] to $F$: $\det(AB)=F(B)=F(I_n)\det(B)=\det(A)\det(B)$. Neither matrix was assumed invertible.

Apply [L1] to the two equations in [L4]: $\det(A)\det(A^{-1})=\det(I_n)=1$ and $\det(A^{-1})\det(A)=1$.

Thus $\det(A)$ is a unit with inverse $\det(A^{-1})$ by [L2]. In the zero ring, $0=1$ and the same two-sided equation still gives the unit conclusion.

If $A$ is invertible, [L1] and [L6] make $\det(A)$ a unit of $\mathbb R$, hence nonzero by [L7].

Conversely suppose $A$ is not invertible and row reduce it to its RREF $R$. By [L3], $R$ has fewer than $n$ pivots and hence a zero row. Its pivot in row $i$, when present, lies in a column at least $i$, so every entry below the diagonal is zero; thus $R$ is upper triangular.

The zero row forces a zero diagonal entry, so [L5] gives $\det(R)=0$.

Along the finite real row reduction, swaps and row additions preserve whether determinant is zero, and a scaling uses a nonzero scalar whose inverse gives the reverse implication. Hence $\det(A)=0$.

Steps 1.2–3.1 prove the contrapositive $\det(A)\ne0\Rightarrow A$ invertible; combining it with step 1.1 proves the equivalence.

By [L2], [L4] and [L5], $\det(A)\det(A^{-1})=\det(I_n)=1$, so $\det(A^{-1})$ is an inverse of the unit $\det(A)$.

Uniqueness of the inverse in [L3] gives $\det(A^{-1})=\det(A)^{-1}$.

By [L1], [L2] and associativity, $\det(B)=\det(P^{-1})\det(A)\det(P)$.

Substitute [L3] and commute the scalar factors in $R$: $\det(P)^{-1}\det(P)=1$, leaving $\det(B)=\det(A)$.