A function $\delta :{\text{M}}_{n\times n}(F)\to F$ is called an n-linear function if it is a linear function of each row of an $n\times n$ matrix when the remaining $n-1$ rows are held fixed, that is, $\delta$ is n-linear if, for every $r=1,2,\dots ,n,$ we have

whenever k is a scalar and u, v, and each $a_i$ are vectors in ${\text{F}}^n$.

An n-linear function $\delta :{\text{M}}_{n\times n}(F)\to F$ is called alternating if, for each $A\in {\text{M}}_{n\times n}(F)$ we have $\delta (A)=0$ whenever two adjacent rows of A are identical.

(a) Let $A\in {\text{M}}_{n\times n}(F),$ and let B be the matrix obtained from A by interchanging rows r and s, where $r<s$. We first establish the result in the case that $s=r+1$. Because $\delta :{\text{M}}_{n\times n}(F)\to F$ is an n-linear function that is alternating, we have

Thus $\delta (B)=-\delta (A)$.

Next suppose that $s>r+1,$ and let the rows of A be $a_1,a_2,\dots ,a_n$. Beginning with $a_r$ and $a_{r+1}$ successively interchange $a_r$ with the row that follows it until the rows are in the sequence

In all, $s-r$ interchanges of adjacent rows are needed to produce this sequence. Then successively interchange $a_s$ with the row that precedes it until the rows are in the order

This process requires an additional $s-r-1$ interchanges of adjacent rows and produces the matrix B. It follows from the preceding paragraph that

(b) Suppose that rows r and s of $A\in {\text{M}}_{n\times n}(F)$ are identical, where $r<s$ If $s=r+1,$ then $\delta (A)=0$ because $\delta$ is alternating and two adjacent rows of A are identical. If $s>r+1,$ let B be the matrix obtained from A by interchanging rows $r+1$ and s. Then $\delta (B)=0$ because two adjacent rows of B are identical. But $\delta (B)=-\delta (A)$ by (a). Hence $\delta (A)=0$.

Let $\delta :{\text{M}}_{n\times n}(F)\to F$ be an alternating n-linear function. If B is a matrix obtained from $A\in {\text{M}}_{n\times n}(F)$ by adding a multiple of some row of A to another row, then $\delta (B)=\delta (A)$.

Let B be obtained from $A\in {\text{M}}_{n\times n}(F)$ by adding k times row i of A to row j, where $j\ne i,$ and let C be obtained from A by replacing row j of A by row i of A. Then the rows of A, B, and C are identical except for row j. Moreover, row j of B is the sum of row j of A and k times row j of C. Since δ is an n-linear function and C has two identical rows, it follows that

The next result now follows as in the proof of the corollary to Theorem 4.6 (p. 216). (See Exercise 11.)

Let $\delta :{\text{M}}_{n\times n}(F)\to F$ be an alternating n-linear function. If $M\in {\text{M}}_{n\times n}(F)$ has rank less than n, then $\delta (M)=0$.

Exercise.

Let $\delta :{\text{M}}_{n\times n}(F)\to F$ be an alternating n-linear function, and let $E_1,E_2$ and $E_3$ in ${\text{M}}_{n\times n}(F)$ be elementary matrices of types 1, 2, and 3, respectively. Suppose that $E_2$ is obtained by multiplying some row of I by the nonzero scalar k. Then $\delta (E_1)=-\delta (I),\delta (E_2)=k·\delta (I),$ and $\delta (E_3)=\delta (I)$.

Exercise.

We wish to show that under certain circumstances, the only alternating n-linear function $\delta :{\text{M}}_{n\times n}(F)\to F$ is the determinant, that is, $\delta (A)=\text{det}(A)$ for all $A\in {\text{M}}_{n\times n}(F)$. Because any scalar multiple of an alternating n-linear function is also an alternating n-linear function, we need a condition that distinguishes the determinant among its scalar multiples. Hence the third condition that is used in the characterization of the determinant is that the determinant of the $n\times n$ identity matrix is 1. Before we can establish the desired characterization of the determinant, we must first prove a result similar to Theorem 4.7 (p. 223). The proof of this result is also similar to that of Theorem 4.7, and so it is omitted. (See Exercise 12.)

Exercise.

Let $\delta :{\text{M}}_{n\times n}(F)\to F$ be an alternating n-linear function such that $\delta (I)=1,$ and let $A\in {\text{M}}_{n\times n}(F)$. If A has rank less than n, then by Corollary 2 to Theorem 4.10, $\delta (A)=0$. Since the corollary to Theorem 4.6 (p. 217) gives $\text{det}(A)=0,$ we have $\delta (A)=\text{det}(A)$ in this case. If, on the other hand, A has rank n, then A is invertible and hence is the product of elementary matrices (Corollary 3 to Theorem 3.6 p. 158), say $A=E_m\cdots E_2{E}_1$. Since $\delta (I)=1,$ it follows from Corollary 3 to Theorem 4.10 and the facts on page 249 that $\delta (E)=\text{det}(E)$ for every elementary matrix E. Hence by Theorems 4.11 and 4.7 (p. 223), we have

Theorem 4.12 provides the desired characterization of the determinant: It is the unique function $\delta :{\text{M}}_{n\times n}(F)\to F$ that is n-linear, is alternating, and has the property that $\delta (I)=1$.