Determinants and Multilinear Forms

Definition Let $V$ be a vector space over $\mathbb{F}$ and let $k \ge 1$. A function $D \colon V^k \to \mathbb{F}$ is a multilinear form, or $k$-linear form, if it is linear in each argument when the other arguments are held fixed. For every slot, all vectors and all scalars $a, c \in \mathbb{F}$,

where $u$, $w$ and $au + cw$ sit in the same slot and the other arguments are unchanged. In this lesson $V = \mathbb{F}^n$ and $k = n$, so $D$ takes the $n$ columns of a square matrix. For $A \in \mathbb{F}^{n \times n}$ with columns $a_1, \dots, a_n$ we write $D(A) = D(a_1, \dots, a_n)$.

Definition A multilinear form $D$ on $V^k$ is alternating if $D(v_1, \dots, v_k) = 0$ whenever $v_i = v_j$ for some $i \ne j$. Three rules follow from multilinearity and this condition. First, exchanging two arguments changes the sign. Put $u + v$ into both of those slots, keep the other slots fixed, and expand,

The repeated terms vanish, so $D(v, u) = -D(u, v)$. Second, adding a multiple of one argument to another does not change the value, because $D(u + cv, v) = D(u, v) + c\,D(v, v) = D(u, v)$. Third, if $v_1, \dots, v_k$ are linearly dependent, then $D(v_1, \dots, v_k) = 0$. Solve a nontrivial relation for one of the vectors, substitute the result into its slot and expand, and every term has a repeated argument. Over $\mathbb{R}$ or $\mathbb{C}$ the sign rule also implies the alternating condition, since $D(v, v) = -D(v, v)$ forces $2D(v, v) = 0$.

Definition A permutation of $\{1, \dots, n\}$ is a bijection $\sigma$ from this set to itself, recorded by the list $(\sigma(1), \dots, \sigma(n))$. The set of all $n!$ permutations is $S_n$. An inversion of $\sigma$ is a pair of positions $i < j$ with $\sigma(i) > \sigma(j)$. If $N(\sigma)$ is the number of inversions, the sign of $\sigma$ is

Exchanging two neighboring entries of the list changes $N(\sigma)$ by exactly one, so it flips the sign. Exchanging the entries in positions $i < j$ can be done with $2(j - i) - 1$ neighboring exchanges, an odd number, so any exchange of two entries flips the sign. A permutation and its inverse have the same number of inversions, hence the same sign. Sorting the list by exchanging inverted neighbors lowers $N(\sigma)$ by one each time, so it takes exactly $N(\sigma)$ exchanges. For an alternating multilinear form $D$ on $(\mathbb{F}^n)^n$, each exchange flips the sign of $D$, which gives

Definition Let $n \ge 1$. A normalized determinant on $\mathbb{F}^{n \times n}$ is a function $D \colon \mathbb{F}^{n \times n} \to \mathbb{F}$ that is multilinear and alternating in the columns and satisfies $D(I) = D(e_1, \dots, e_n) = 1$. One such function exists. For $A = [a_{ij}]$ define

Each term takes exactly one entry from each column, so the sum is linear in each column. If columns $p \ne q$ are equal, pair each $\sigma$ with the permutation $\sigma'$ that agrees with $\sigma$ except that $\sigma'(p) = \sigma(q)$ and $\sigma'(q) = \sigma(p)$. The two products are equal because the columns coincide, and the signs are opposite because the two lists differ by one exchange, so the terms cancel in pairs and the sum is $0$. At the identity matrix only the identity permutation gives a nonzero product, and it contributes $1$.

Definition Let $n \ge 1$ and let $D \colon \mathbb{F}^{n \times n} \to \mathbb{F}$ be multilinear and alternating in the columns. The uniqueness theorem says that for every $A \in \mathbb{F}^{n \times n}$

In particular $\det$ is the only normalized determinant. To prove it, write column $j$ as $a_j = \sum_i a_{ij} e_i$ and expand every slot by multilinearity. This gives a sum over all choices of one row index $i_j$ for each column $j$,

A choice that repeats an index contributes zero because $D$ is alternating. The remaining choices use every index once, so they are the permutations $\sigma$ with $i_j = \sigma(j)$, and $D(e_{\sigma(1)}, \dots, e_{\sigma(n)}) = \operatorname{sgn}(\sigma) D(I)$. What is left is $D(I)$ times the permutation formula. In that formula the term of $\det A^{\mathsf{T}}$ for $\sigma$ equals the term of $\det A$ for $\sigma^{-1}$, with the same sign and the same factors in a different order, which proves $\det A^{\mathsf{T}} = \det A$. So the rules for columns also hold for rows, and a matrix with linearly dependent columns or rows has determinant $0$.

Definition Let $n \ge 1$. For all $A, B \in \mathbb{F}^{n \times n}$

To prove it, fix $A$ and set $D(v_1, \dots, v_n) = \det(Av_1, \dots, Av_n)$. Because $v \mapsto Av$ is linear, $D$ is multilinear and alternating, and $D(I) = \det A$. By uniqueness $D(B) = \det A \det B$ for every $B$. The columns of $AB$ are $Ab_1, \dots, Ab_n$, so $D(B) = \det(AB)$. The argument never divides by $\det A$, so it covers singular $A$ too. Two consequences follow. A square matrix $A$ is invertible exactly when $\det A \ne 0$, since $\det A \det A^{-1} = \det I = 1$ in one direction and dependent columns give $0$ in the other. When $A$ is invertible, $\det(A^{-1}) = 1/\det A$.