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
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:

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

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