# Equivalence Relation on Square Matrices induced by Positive Integer Powers

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## Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $S$ be the set of all square matrices of order $n$.

Let $\alpha$ denote the relation defined on $S$ by:

- $\forall \mathbf A, \mathbf B \in S: \mathbf A \mathrel \alpha \mathbf B \iff \exists r, s \in \N: \mathbf A^r = \mathbf B^s$

Then $\alpha$ is an equivalence relation.

## Proof

Checking in turn each of the criteria for equivalence:

### Reflexivity

We have that for all $\mathbf A \in S$:

- $\mathbf A^r = \mathbf A^r$

for all $r \in \N$.

It follows by definition of $\alpha$ that:

- $\mathbf A \mathrel \alpha \mathbf A$

Thus $\alpha$ is seen to be reflexive.

$\Box$

### Symmetry

\(\displaystyle \mathbf A\) | \(\alpha\) | \(\displaystyle \mathbf B\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \mathbf A^r\) | \(=\) | \(\displaystyle \mathbf B^s\) | for some $r, s \in \N$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \mathbf B^s\) | \(=\) | \(\displaystyle \mathbf A^r\) | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \mathbf B\) | \(\alpha\) | \(\displaystyle \mathbf A\) |

Thus $\alpha$ is seen to be symmetric.

$\Box$

### Transitivity

Let:

- $\mathbf A \mathrel \alpha \mathbf B$ and $\mathbf B \mathrel \alpha \mathbf C$

for square matrices of order $n$ $\mathbf A, \mathbf B, \mathbf C$.

Then by definition:

\(\displaystyle \mathbf A^r\) | \(=\) | \(\displaystyle \mathbf B^s\) | for some $r, s \in \N$ | ||||||||||

\(\displaystyle \mathbf B^u\) | \(=\) | \(\displaystyle \mathbf C^v\) | for some $u, v \in \N$ | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \mathbf A^{r u}\) | \(=\) | \(\displaystyle \mathbf B^{s u}\) | raising both sides to $u$th power | |||||||||

\(\displaystyle \mathbf B^{s u}\) | \(=\) | \(\displaystyle \mathbf C^{s v}\) | raising both sides to $s$th power | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \mathbf A^{r u}\) | \(=\) | \(\displaystyle \mathbf C^{s v}\) |

Thus $\alpha$ is seen to be transitive.

$\Box$

$\alpha$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $6$