📂Matrix Algebra

Determinant of a Triangular Matrix

English Translation

Theorem

The determinant of a triangular matrix is expressed as the product of its diagonal elements.

Proof ¹

Without loss of generality, assume that $A$ is an upper triangular matrix.

$$A := \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{bmatrix}$$

Laplace’s Expansion: Let a square matrix $A_{n \times n} = (a_{ij})$ be given. Considering the determinant $M_{ij}$ of the matrix obtained by removing the $i$-th row and the $j$-th row from the square matrix $A_{n \times n} = (a_{ij})$, where $C_{ij} := (-1)^{i + j} M_{ij}$ is referred to as the cofactor for the selected $j$-th column. The following holds true: $$\det A = \sum_{i=1}^{n} a_{ij} C_{ij}$$

Considering the uppermost minors $M_{1j}$, since they are at least $j \ne 1$, the leftmost column must be a zero vector, leading to $M_{1j} = 0$. By recursive application of Laplace’s expansion, we obtain the following: $$\begin{align*} \det A =& \det \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{bmatrix} \\ =& a_{11} \det \begin{bmatrix} a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{bmatrix} \\ =& a_{11} a_{22} \det \begin{bmatrix} a_{33} & \cdots & a_{3n} \\ \vdots & \ddots & \vdots \\ 0 & \cdots & a_{nn} \end{bmatrix} \\ =& \prod_{i=1}^{n} a_{ii} \end{align*}$$

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