- 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
3.6. $\quad$ Exercises
5. $\quad$ A lower triangular matrix $A = (a_{ij})$ is a square matrix with all entries above the main diagonal equal to $0;$ that is, $a_{ij} = 0$ whenever $i \lt j.$ Prove that the determinant of such a matrix is equal to the product of its diagonal entries: det $A = a_{11}a_{22} \cdots a_{nn}.$
Proof. $\quad$ This proof will use a similar argument to that used in Section 4 to prove that the determinant of an upper-triangular matrix is the product of its diagonal elements.
$\quad$ First, if $a_{11} = 0,$ then $A$ has a row of zeros, and from Theorem 3.1 (a), this means that $\det A = 0.$ More generally, if the $k^{th}$ diagonal element is zero, then we can use rows $1, \dots, k-1$ to make row $k$ zero. Hence, if any diagonal element is zero, then $\det A = 0.$
$\quad$ Assume then, that the diagonal elements of $A$ are nonzero. Then, we can represent the first row of $A,$ $A_1,$ as the sum of the two row vectors \begin{align*} V_1 &= [a_{11}, 0, \dots, 0], \qquad V'_1 = [0, 0, \dots, 0] \end{align*} As such, the determinant of $A$ can be written as the sum \begin{align*} \det A &= d(V_1, A_2, \dots, A_n) + d(V'_1, A_2, \dots, A_n) \end{align*} But because $d(V'_1, A_2, \dots, A_n)$ has a diagonal element equal to zero, the rightmost term vanishes and we have $\det A = d(V_1, A_2, \dots, A_n).$ And since $V_1 = a_{11}I_1,$ where $I_1$ is the first row of the $n \times n$ identity matrix, we have \begin{align*} \det A &= a_{11}d(I_1, A_2, \dots, A_n) \end{align*} Similarly, we can write $A_2$ as the sum of the following vectors in $n$-space \begin{align*} V_2 = [0, a_{22}, \dots, 0], \qquad V'_2 = [a_{21}, 0, \dots, 0] \end{align*} Then, we can write $\det A$ as \begin{align*} \det A &= a_{11}d(I_1, V_2, \dots, A_n) + a_{11}d(I_1, V'_2, \dots, A_n) \end{align*} But because $d(I_1, V'_2, \dots, A_n)$ has a zero diagonal element, the rightmost term vanishes once more and we get \begin{align*} \det A &= a_{11}d(I_1, V_2, \dots, A_n) \\ &= a_{11}a_{22}d(I_1, I_2, \dots, A_n) \end{align*} Repeating this process for the remaining rows of $A,$ noting that $\det I = 1,$ we find that $\det A$ is given by: \begin{align*} \det A &= a_{11} \cdots a_{nn} d(I_1, \dots, I_n) \\ &= a_{11} \cdots a_{nn} \det I \\ &= a_{11} \cdots a_{nn}. \quad \blacksquare \end{align*}