# Equivalence Relation on Square Matrices induced by Positive Integer Powers

## 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.

$\blacksquare$