Using the cofactor expansion, explain why the determinant of a triangular matrix is the product of the elements on its diagonal.
Is it the fact that there are $0$'s in the $L$ or $U$ part of the matrix and that somehow comes into play to where only the diagonal is accounted for? I'm not quite sure.
2 Answers
Let $$A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ & a_{22} & \cdots & a_{2n} \\ & & \ddots & \\ & & & a_{nn}\end{pmatrix}$$
be your upper triangular matrix. Expanding the left most column, the cofactor expansion formula tells you that the determinant of $A$ is
$$a_{11} \cdot \textrm{det} \begin{pmatrix} a_{22} & a_{22} & \cdots & a_{2n}\\ & a_{33} & \cdots & a_{3n} \\ & & \ddots & \\ & & & a_{nn}\end{pmatrix}$$ Now this smaller $(n-1)$ by $(n-1)$ matrix is also upper triangular, so you can compute it as $a_{22}$ times an $(n-2)$ by $(n-2)$ upper triangular determinant:
$$\textrm{det } A = a_{11} a_{22} \cdot \textrm{det} \begin{pmatrix} a_{33} & a_{34} & \cdots & a_{3n}\\ & a_{44} & \cdots & a_{4n} \\ & & \ddots & \\ & & & a_{nn}\end{pmatrix}$$
Iterating this argument, you're eventually going to get
$$\textrm{Det } A = a_{11} \cdots a_{n-2,n-2} \cdot \textrm{det} \begin{pmatrix} a_{n-1,n-1} & a_{n-1,n} \\ & a_{nn} \end{pmatrix} = a_{11} \cdots a_{nn}$$
Let ${\rm U}_n$ be an invertible $n \times n$ upper triangular matrix. Let
$${\rm U}_{n+1} := \begin{bmatrix} {\rm U}_n & {\rm c}_{n+1}\\ {\rm 0}_n^\top & u_{n+1}\end{bmatrix}$$
Using the Schur complement,
$$\det \left( {\rm U}_{n+1} \right) = \det \begin{bmatrix} {\rm U}_n & {\rm c}_{n+1}\\ {\rm 0}_n^\top & u_{n+1}\end{bmatrix} = \left( u_{n+1} - {\rm 0}_n^\top {\rm U}_n^{-1} {\rm c}_{n+1} \right) \det \left( {\rm U}_{n} \right) = u_{n+1} \det \left( {\rm U}_{n} \right)$$
Let ${\rm U}_{1} =: u_1$. Hence,
$$\begin{aligned} \det \left( {\rm U}_{1} \right) &= u_1\\ \det \left( {\rm U}_{2} \right) &= u_2 \, u_1\\ &\vdots\\ \det \left( {\rm U}_{n} \right) &= \color{blue}{u_n \, u_{n-1}\cdots u_2 \, u_1} \end{aligned}$$
The case where ${\rm U}_n$ is non-invertible is left as an exercise for the reader.
