- 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
3.3 $\quad$ A set of axioms for the determinant function
$\quad$ The properties of the scalar triple product described in the previous section can be generalized and used as axioms for determinants of order $n.$ If $A = (a_{ij})$ is an $n \times n$ matrix with real or complex entries, we denote its rows by $A_1, \dots, A_n.$ Thus, the $i^{th}$ row of $A$ is a vector in $n$-space given by \begin{align*} A_i &= (a_{i1}, \dots, a_{in}) \end{align*} We regard the determinant as a function of the $n$ rows $A_1, \dots, A_n$ and denote its values by $d(A_1, \dots, A_n)$ or by $\det A.$
$\quad$ Axiomatic definition of a determinant function. $\quad$ A real- or complex-valued function $d,$ defined for each ordered $n$-tuple of vectors $A_1, \dots, A_n$ in $n$-space is called a determinant function of order $n$ if it satisfies the following axioms for all choices of vectors $A_1, \dots, A_n$ and $C$ in $n$-space:
$\quad$Axiom 1. $\quad$ Homogeneity in each row. $\quad$ If the $k^{th}$ row $A$ is multiplied by a scalar $t,$ then the determinant is also multiplied by $t.$
\begin{align*}
d(\dots, tA_k, \dots) &= t\,d(\dots, A_k, \dots)
\end{align*}
$\quad$Axiom 2. $\quad$ Additivity in each row. $\quad$ For each $k$ we have
\begin{align*}
d(A_1, \dots, A_k + C, \dots, A_n) &= d(A_1, \dots, A_k, \dots, A_n) + d(A_1, \dots, C, \dots, A_n)
\end{align*}
$\quad$Axiom 3. $\quad$ The determinant vanishes if any two rows are equal. $\quad$
\begin{align*}
d(A_1, \dots, A_n) &= 0 \qquad \text{if $A_i = A_j$} \qquad \text{for some $i$ and $j$ with $i \neq j.$}
\end{align*}
$\quad$Axiom 4. $\quad$ The determinant of the identity matrix is equal to $1.$ $\quad$
\begin{align*}
d(I_1, \dots, I_n) &= 1 \qquad \text{where $I_k$ is the $k^{th}$ unit coordinate vector.}
\end{align*}
The first two axioms state that the determinant of a matrix is a linear function of its rows. This is often described by saying that the determinant is a multilinear function of its rows. By repeated application of linearity in the first row, we can write
\begin{align*}
d\left[\left(\sum_{k=1}^p t_kC_k\right), A_2, \dots, A_n \right] &= \sum_{k=1}^p t_k\,d(C_k, A_2, \dots, A_n)
\end{align*}
where $t_1, \dots, t_p$ are scalars and $C_1, \dots, C_p$ are any vectors in $n$-space.
$\quad$ Sometimes a weaker version of Axiom 3 is used:
$\quad$Axiom 3'. $\quad$ The determinant vanishes if two adjacent rows are equal. $\quad$
\begin{align*}
d(A_1, \dots, A_n) &= 0 \qquad \text{if $A_i = A_j$} \qquad \text{for some $i$ and $j$ with $i \neq j.$}
\end{align*}
$\quad$ Theorem 3.1 $\quad$
A determinant function satisfying Axioms 1, 2, and 3' has the following further properties
$\quad$ (a) $\quad$ The determinant vanishes if some row is $0:$
\begin{align*}
d(A_1, \dots, A_n) &= 0 \qquad \text{if $A_k = O$} \qquad \text{for some $k$}.
\end{align*}
$\quad$ (b) $\quad$ The determinant changes sign if two adjacent rows are interchanged:
\begin{align*}
d(\dots, A_k, A_{k+1}, \dots) &= -d(\dots, A_{k+1}, A_{k}, \dots).
\end{align*}
$\quad$ (c) $\quad$ The determinant changes sign if any two rows $A_i$ and $A_j$ with $i \neq j$ are interchanged.
$\quad$ (d) $\quad$ The determinant vanishes if any two rows are equal:
\begin{align*}
d(A_1, \dots, A_n) &= 0 \qquad \text{if $A_i = A_j$} \qquad \text{for some $i$ and $j$ with $i \neq j$}.
\end{align*}
$\quad$ (e) $\quad$ The determinant vanishes if its rows are dependent.
$\quad$ Proof. $\quad$ (a) is a direct consequence of Axioms 1 and 2, as for any row $k,$ we have \begin{align*} d(\dots, A_k - A_k, \dots) &= d(\dots, A_k, \dots) - d(\dots, A_k, \dots) = 0. \end{align*} We can also use the linearity of the determinant function in conjunction with Axiom 3' to derive (b): \begin{align*} d(\dots, A_k, A_{k+1}, \dots) &= d(\dots, A_k + A_{k+1}, A_k + A_{k+1}, \dots) \\ &- d[\dots, A_{k+1}, (A_k + A_{k+1}), \dots] \\ &= -d[\dots, A_{k+1}, A_{k+1}, \dots] - d(\dots, A_{k+1}, A_k, \dots) \\ &= -d(\dots, A_{k+1}, A_k, \dots) \end{align*} $\quad$ To derive (c), we note that we can interchange any two rows $A_i, A_j,$ in $A$ with $i \lt j$ by interchanging adjacent rows until $A_j$ is in row $i$ and $A_i$ is in row $j.$ To get $A_j$ into the $i^{th}$ row, it will take $j - i$ interchanges, leaving $A_i$ in row $i + 1.$ Then, to get $A_i$ into row $j,$ it will take $j - i - 1$ additional interchanges. Thus, interchanging two rows will take $2(j - i) - 1$ interchanges. Moreover, this means that the determinant will change sign $2(j - i) - 1$ times. But for any positive integers $i, j$ with $i \lt j,$ $2(j - i) - 1$ is an odd number. Thus, interchanging any two rows of $A$ changes the sign of its determinant.
$\quad$ If $A_i$ and $A_j$ are any two distinct rows of $A,$ we can apply property (c) to get \begin{align*} d(\dots, A_i, \dots, A_j, \dots) &= -d(\dots, A_j, \dots, A_i, \dots) \end{align*} But if $A_i = A_j,$ this means that \begin{align*} d(\dots, A_i, \dots, A_j, \dots) &= -d(\dots, A_i, \dots, A_j, \dots) \end{align*} Hence, $d(\dots, A_i, \dots, A_j, \dots) = 0,$ thus proving (d). As an extension of this result, if $A$ is such that some row $A_k$ is a linear combination of the other rows, then we can write $\det A$ as a linear combination of determinants having two equal rows, making each term of the sum zero, and thus, $\det A = 0,$ proving (e). This completes the proof.