Determinant of Triangular Matrix

Theorem
Let $\mathbf T_n$ be a triangular matrix (either upper or lower) of order $n$.

Let $\det \left({\mathbf T_n}\right)$ be the determinant of $\mathbf T_n$.

Then $\det \left({\mathbf T_n}\right)$ is equal to the product of all the diagonal elements of $\mathbf T_n$.

That is:
 * $\displaystyle \det \left({\mathbf T_n}\right) = \prod_{k \mathop = 1}^n a_{kk}$

Proof
In the light of Determinant of Transpose, we only need to show this for either the upper or lower triangular matrix, as one is the transpose of the other.

So, let's prove this for the upper triangular matrix.

Let $\mathbf T_n = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix}$ be an upper triangular matrix.

Let $A_{pq}$ be the cofactor of an element $a_{pq} \in \mathbf T_n$.

Then by the Expansion Theorem for Determinants:
 * $\displaystyle \det \left({\mathbf T_n}\right) = \sum_{k \mathop = 1}^n a_{nk} A_{nk}$

As $a_{nk} = 0$ when $k \ne n$, this reduces to:


 * $\det \left({\mathbf T_n}\right) = a_{nn} A_{nn}$

But from the definition of the cofactor, $A_{nn} = \left({-1}\right)^{n+n} D_{nn} = D_{nn}$, where $D_{nn}$ is the order $n-1$ determinant obtained from $D$ by deleting row $n$ and column $n$.

But $D_{nn}$ is simply the upper triangular matrix $\mathbf T_{n-1}$.

Thus, without being too formal about it, the result follows by induction.

The base case $\det \left({\mathbf T_1}\right)$ is trivial, while $\det \left({\mathbf T_2}\right)$ calculates directly:


 * $\det \left({\mathbf T_2}\right) = a_{11} a_{22} - a_{12} a_{21} = a_{11} a_{22}$