- 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
3.7 $\quad$ The product formula for determinants
$\quad$ We will use the uniqueness theorem of the determinant function to prove that the determinant of a product of two square matrices is equal to the product of their determinants, $$ \det (AB) = (\det A) (\det B),$$ assuming that a determinant function exists.
$\quad$ We recall that the product $AB$ of two matrices $A = (a_{ij})$ and $B = (b_{ij})$ is the matrix $C = (c_{ij})$ whose $ij^{th}$ entry is given by the formula \begin{align*} c_{ij} &= \sum_{k=1}^n a_{ik}b_{kj}. \end{align*} The product is defined only if the number of columns in $A$ is equal to the number of rows in $B.$ If $A$ and $B$ are square matrices of the same size, this is always the case.
$\quad$ The proof of the product formula will make use the following lemma which holds between the rows of $AB$ and the rows of $A.$ As usual, we let $A_i$ denote the $i^{th}$ row of $A.$
$\quad$ Lemma 3.3. $\quad$ If $A$ is an $m \times n$ matrix and $B$ is an $n \times p$ matrix, then we have \begin{align*} (AB)_i &= A_iB. \end{align*} That is, the $i^{th}$ row of the product $AB$ is equal to the product of the $i^{th}$ row of $A$ with $B.$
$\quad$ Proof. $\quad$ By definition, the $i^{th}$ row of $AB$ is the vector $C_i = [c_{i1}, \dots, c_{ip}],$ where each value is given by \begin{align*} c_{ij} &= \sum_{k=1}^n a_{ik}b_{kj}, \end{align*} for $j = 1, \dots, p.$ In other words, $c_{ij}$ is the dot product of the $i^{th}$ row of $A,$ $A_i,$ and the $j^{th}$ column of $B,$ $B^j:$ \begin{align*} c_{ij} &= [a_{i1}, \dots, a_{1n}] \begin{bmatrix} b_{1j} \\ \vdots \\ b_{nj} \end{bmatrix} \\ &= A_i \cdot B^j \end{align*} But if this is the case, it follows that the $1 \times p$ matrix $C_i$ is formed by taking the product of the $1 \times n$ matrix $A_i$ with the $n \times p$ matrix $B:$ \begin{align*} C_i &= [A_i \cdot B^1, \dots, A_i \cdot B^p] = [a_{i1}, \dots, a_{1n}] \begin{bmatrix} b_{11} & \cdots & b_{1p} \\ \vdots & \ddots & \vdots \\ b_{n1} & \cdots & b_{np} \end{bmatrix} = A_i B. \quad \blacksquare \end{align*}
$\quad$ Theorem 3.4. $\quad$ Product Formula for Determinants. $\quad$ For any $n \times n$ matrices $A$ and $B,$ we have \begin{align*} \\ \det (AB) &= (\det A)(\det B). \end{align*}
$\quad$ Proof. $\quad$ As proven in Lemma 3.3, we know that $(AB)_i = A_i B.$ Thus, to prove that $\det (AB) = (\det A)(\det B),$ we will prove that \begin{align*} d(A_1 B, \dots, A_n B) &= d(A_1, \dots, A_n) d(B_1, \dots, B_n) \end{align*} Applying Lemma 3.3 to $B,$ we have $B_i = I_iB,$ where $I$ is the $n \times n$ identity matrix. Thus, the above equation becomes \begin{align*} d(A_1 B, \dots, A_n B) &= d(A_1, \dots, A_n) d(I_1B, \dots, I_nB) \end{align*} Keeping $B$ fixed, we define a new function $f$ such that \begin{align*} f(A_1, \dots, A_n) &= d(A_1B, \dots, A_nB) \end{align*} The equation we wish to prove states that $f(A_1, \dots, A_n) = d(A_1, \dots, A_n)f(I_1, \dots, I_n).$ But we can easily verify that $f$ satisfies Axioms 1, 2, and 3 of the determinant function. Hence, by the uniqueness theorem, the equation $f(A_1, \dots, A_n) = d(A_1, \dots, A_n)f(I_1, \dots, I_n)$ holds for all $n \times n$ matrices $A.$