# Determinant of Upper Triangular Matrix

## Theorem

Let $\mathbf T_n$ be an upper triangular matrix of order $n$.

Let $\map \det {\mathbf T_n}$ be the determinant of $\mathbf T_n$.

Then $\map \det {\mathbf T_n}$ is equal to the product of all the diagonal elements of $\mathbf T_n$.

That is:

$\displaystyle \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$

## Proof

Let $\mathbf T_n$ be an upper triangular matrix of order $n$.

We proceed by induction on $n$, the number of rows of $\mathbf T_n$.

### Basis for the Induction

For $n = 1$, the determinant is $a_{11}$, which is clearly also the diagonal element.

This forms the basis for the induction.

### Induction Hypothesis

Fix $n \in \N$.

Then, let:

$\mathbf T_n = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ 0 & a_{2 2} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{n n} \\ \end {bmatrix}$

be an upper triangular matrix.

Assume that:

$\displaystyle \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$

This forms our induction hypothesis.

### Induction Step

Let $\mathbf T_{n + 1}$ be an upper triangular matrix of order $n + 1$.

Then, by the Expansion Theorem for Determinants (expanding across the $n + 1$th row):

$\displaystyle D = \map \det {\mathbf T_{n + 1} } = \sum_{k \mathop = 1}^{n + 1} a_{n + 1, k} T_{n + 1, k}$

Because $\mathbf T_{n + 1}$ is upper triangular, $a_{n + 1, k} = 0$ when $k < n + 1$.

Therefore:

$\map \det {\mathbf T_{n + 1} } = a_{n + 1 \, n + 1} T_{n + 1, n + 1}$

By the definition of the cofactor:

$T_{n + 1, n + 1} = \paren {-1}^{n + 1 + n + 1} D_{n + 1, n + 1} = D_{n n}$

where $D_{n n}$ is the order $n$ determinant obtained from $D$ by deleting row $n + 1$ and column $n + 1$.

But $D_{n n}$ is just the determinant of an upper triangular matrix $\mathbf T_n$.

Therefore:

$\map \det {\mathbf T_{n + 1} } = a_{n + 1, n + 1} \map \det {\mathbf T_n}$

and the result follows by induction.

$\blacksquare$