Determinant of Unit Matrix

From ProofWiki
Jump to navigation Jump to search

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:

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Determinant_of_Unit_Matrix&oldid=476589"