Matrix Equivalence is Equivalence Relation

From ProofWiki
Jump to navigation Jump to search

Theorem

Matrix equivalence is an equivalence relation.


Proof

Checking in turn each of the criteria for equivalence:


Reflexive

Let $\mathbf {I_m}$ and $\mathbf {I_n}$ denote the identity matrices of order $m$ and $n$ respectively.

Let $\mathbf A$ be an arbitrary $m \times n$ matrix.

Then:

$\mathbf A = \mathbf {I_m}^{-1} \mathbf A \mathbf {I_n}$

trivially.

Thus reflexivity holds.

$\Box$


Symmetric

Let $\mathbf A$ and $\mathbf B$ be arbitrary $m \times n$ matrices such that $\mathbf A \equiv \mathbf B$.

Then by definition there exist invertible matrices $\mathbf P$ and $\mathbf Q$ such that:

$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
\(\ds \mathbf Q \mathbf B \mathbf P^{-1}\) \(=\) \(\ds \mathbf Q \mathbf Q^{-1} \mathbf A \mathbf P \mathbf P^{-1}\)
\(\ds \) \(=\) \(\ds \mathbf{I_m} \mathbf A \mathbf{I_n}\)
\(\ds \) \(=\) \(\ds \mathbf A\)

That is:

$\mathbf B \equiv \mathbf A$

Thus symmetry holds.

$\Box$


Transitive

Let $\mathbf A$, $\mathbf B$ and $\mathbf B$ be arbitrary $m \times n$ matrices such that:

$\mathbf A \equiv \mathbf B$
$\mathbf B \equiv \mathbf C$

Then by definition there exist invertible matrices $\mathbf P_1$, $\mathbf P_2$, $\mathbf Q_1$ and $\mathbf Q_2$ such that:

$\mathbf B = \mathbf Q_1^{-1} \mathbf A \mathbf P_1$
$\mathbf C = \mathbf Q_2^{-1} \mathbf B \mathbf P_2$

Then:

$\mathbf C = \mathbf Q_2^{-1} \mathbf Q_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2$

From Inverse of Matrix Product:

$\mathbf Q_2^{-1} \mathbf Q_1^{-1} = \paren {\mathbf Q_1 \mathbf Q_2}^{-1}$

By Product of Matrices is Invertible iff Matrices are Invertible, both $\mathbf Q_1 \mathbf Q_2$ and $\mathbf P_1 \mathbf P_2$ are invertible.

Hence:

$\mathbf A \equiv \mathbf C$

Thus transitivity holds.

$\Box$


Hence the result by definition of equivalence relation.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Matrix_Equivalence_is_Equivalence_Relation&oldid=565711"