# Determinant of Unit Matrix

## Theorem

Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

The determinant of the unit matrix of order $n$ over $R$ is equal to $1_R$.

## Proof

Let $\mathbf I_n$ denote the unit matrix of order $n$ over $R$.

The proof proceeds by induction.

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:

$\map \det {\mathbf I_n} = 1_R$

By definition of Determinant of Order $1$:

$\begin {vmatrix} a_{1 1} \end {vmatrix} = a_{1 1}$

In this case $a_{1 1} = 1_R$.

Thus $\map P 1$ is seen to hold.

### Basis for the Induction

 $\ds \map \det {\mathbf I_2}$ $=$ $\ds \begin {vmatrix} 1_R & 0_R \\ 0_R & 1_R \end {vmatrix}$ $\ds$ $=$ $\ds 1_R \cdot 1_R - 0_R \cdot 0_R$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds 1_R$

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$\map \det {\mathbf I_k} = 1_R$

from which it is to be shown that:

$\map \det {\mathbf I_{k + 1} } = 1_R$

### Induction Step

This is the induction step:

 $\ds \mathbf I_{k + 1}$ $=$ $\ds \begin {bmatrix} 1_R & 0_R \\ 0_R & \mathbf I_n \end {bmatrix}$ Definition of Unit Matrix $\ds \leadsto \ \$ $\ds \map \det {\mathbf I_{k + 1} }$ $=$ $\ds \begin {vmatrix} 1_R & 0_R \\ 0_R & \mathbf I_n \end {vmatrix}$ Definition of Determinant of Matrix $\ds$ $=$ $\ds \map \det {\mathbf I_k}$ Determinant with Unit Element in Otherwise Zero Row $\ds$ $=$ $\ds 1_R$ Induction Hypothesis

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{\ge 0}: \map \det {\mathbf I_n} = 1_R$

$\blacksquare$