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

\(\ds \mathbf A\) \(\alpha\) \(\ds \mathbf B\)
\(\ds \leadsto \ \ \) \(\ds \mathbf A^r\) \(=\) \(\ds \mathbf B^s\) for some $r, s \in \N$
\(\ds \leadsto \ \ \) \(\ds \mathbf B^s\) \(=\) \(\ds \mathbf A^r\)
\(\ds \leadsto \ \ \) \(\ds \mathbf B\) \(\alpha\) \(\ds \mathbf A\)

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:

\(\ds \mathbf A^r\) \(=\) \(\ds \mathbf B^s\) for some $r, s \in \N$
\(\ds \mathbf B^u\) \(=\) \(\ds \mathbf C^v\) for some $u, v \in \N$
\(\ds \leadsto \ \ \) \(\ds \mathbf A^{r u}\) \(=\) \(\ds \mathbf B^{s u}\) raising both sides to $u$th power
\(\ds \mathbf B^{s u}\) \(=\) \(\ds \mathbf C^{s v}\) raising both sides to $s$th power
\(\ds \leadsto \ \ \) \(\ds \mathbf A^{r u}\) \(=\) \(\ds \mathbf C^{s v}\)

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

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