An \(n \times n\) matrix \(A\) is called triangular if \(A_{ij} = 0\) whenever \(i > j\) or if \(A_{ij} = 0\) whenever \(i < j\). Prove that the determinant of a triangular matrix is the product \(A_{11}A_{22} \cdots A_{nn}\) of its diagonal entries.
Solution
Goal. Show \(\det A = A_{11} A_{22} \cdots A_{nn}\) for a triangular matrix.
1. It suffices to consider the upper-triangular case (the lower-triangular case is identical by transposition, since \(\det A = \det A^T\)).
\(\det A = \det A^T\), and the transpose of a lower-triangular matrix is upper-triangular.
2. Expand the determinant along the first column.
2.1. In an upper-triangular matrix, the first column has only one nonzero entry, \(A_{11}\).
\(A_{i1} = 0\) for \(i > 1\).
2.2. Hence \(\det A = A_{11} \cdot \det A_{11}\), where \(A_{11}\) is the \((n-1)\times(n-1)\) minor obtained by deleting the first row and column.
cofactor expansion along the first column.
2.3. \(A_{11}\) is again upper-triangular with diagonal \(A_{22}, \dots, A_{nn}\).
deleting the first row and column of an upper-triangular matrix gives an upper-triangular matrix.
3. Induct on \(n\).
3.1. Base case \(n = 1\): \(\det A = A_{11}\).
a \(1\times 1\) matrix \(A = [A_{11}]\) has determinant equal to its single entry, and the product of its diagonal entries is \(A_{11}\).
3.2. Inductive step: \(\det A = A_{11} \det A_{11} = A_{11} (A_{22} \cdots A_{nn})\).
by 2.2 and the induction hypothesis applied to \(A_{11}\).
4. Q.E.D.
3.2 gives \(\det A = A_{11} A_{22} \cdots A_{nn}\).