- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
3.5 $\quad$ The uniqueness theorem
$\quad$ In Example 3 of the previous section, we showed that as a consequence of Axioms 1, 2, and 3 that if $U$ is an upper-triangular matrix, then $\det U = u_{11}u_{22}\cdots u_{nn}$ and moreover, if $U$ is the matrix reached by applying the Gauss-Jordan elimination process to matrix $A,$ then we have \begin{align*} \det A &= (-1)^p (c_1c_2\cdots c_q)^{-1} \det U \end{align*} But this implies that for every $n \times n$ matrix $A,$ there is a scalar $c$ (depending on $A$) such that \begin{align*} d(A_1, \dots, A_n) &= c\,d(I_1, \dots, I_n) \end{align*} And because this formula is a consequence of Axioms 1, 2, and 3 alone, we can prove that there cannot be more than one determinant function.
$\quad$ Theorem 3.2. $\quad$ Uniqueness Theorem for Determinants. $\quad$ Let $d$ be a function satisfying all four axioms for a determinant function of order $n$ and let $f$ be a another function satisfying Axioms 1, 2, and 3. Then, for every choice of vectors $A_1, \dots, A_n$ in $n$-space, we have \begin{align*} f(A_1, \dots A_n) &= d(A_1, \dots, A_n)f(I_1, \dots, I_n). \end{align*} In particular, if $f$ also satisfies Axiom 4 we have $f(A_1, \dots, A_n) = d(A_1, \dots, A_n).$
$\quad$ Proof. $\quad$ Let $g(A_1, \dots, A_n) = f(A_1, \dots A_n) - d(A_1, \dots, A_n)f(I_1, \dots, I_n).$ We will prove that $g(A_1, \dots, A_n) = 0$ for every choice of $A_1, \dots, A_n.$ Because $f$ and $d$ satisfy Axioms 1, 2, and 3 of the determinant function, $g$ does as well. Thus, there exists a scalar $c,$ dependent on $A,$ such that \begin{align*} g(A_1, \dots, A_n) &= cg(I_1, \dots I_n) = cf(I_1, \dots, I_n) - c\,d(I_1, \dots, I_n)f(I_1, \dots, I_n) \end{align*} But because $d$ satisfies Axiom 4 of the determinant function, $d(I_1, \dots, I_n) = 1,$ which gives us \begin{align*} g(A_1, \dots, A_n) &= cg(I_1, \dots I_n) = cf(I_1, \dots, I_n) - cf(I_1, \dots, I_n) = 0. \end{align*} And because this is true for any choice of vectors $A_1, \dots A_n$ in $n$-space, it follows that \begin{align*} f(A_1, \dots A_n) &= d(A_1, \dots, A_n)f(I_1, \dots, I_n). \quad \blacksquare \end{align*}