Matrix Equivalence is Equivalence Relation

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Theorem

Matrix equivalence is an equivalence relation.


Proof

Checking in turn each of the criteria for equivalence:


Reflexive

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

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

From Unit Matrix is its own Inverse, both $\mathbf {I_m}$ and $\mathbf {I_n}$ are invertible matrices.

Then we have that Unit Matrix is Identity for Matrix Multiplication.

Hence there exist invertible matrices $\mathbf {I_m}, \mathbf {I_n}$ such that:

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

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:

$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
\(\ds \mathbf Q \mathbf B \mathbf P^{-1}\) \(=\) \(\ds \mathbf Q \paren {\mathbf Q^{-1} \mathbf A \mathbf P} \mathbf P^{-1}\) by hypothesis
\(\ds \) \(=\) \(\ds \paren {\mathbf Q \mathbf Q^{-1} } \mathbf A \paren {\mathbf P \mathbf P^{-1} }\) Matrix Multiplication is Associative
\(\ds \) \(=\) \(\ds \mathbf {I_m} \mathbf A \mathbf {I_n}\) Definition of Invertible Matrix
\(\ds \) \(=\) \(\ds \mathbf A\) Definition of Identity Matrix

That is:

$\mathbf B \equiv \mathbf A$

Thus symmetry holds.

$\Box$


Transitive

Let $\mathbf A$, $\mathbf B$ and $\mathbf C$ 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:

\(\ds \mathbf C\) \(=\) \(\ds \mathbf Q_2^{-1} \paren {\mathbf Q_1^{-1} \mathbf A \mathbf P_1} \mathbf P_2\) by hypothesis
\(\ds \) \(=\) \(\ds \paren {\mathbf Q_2^{-1} \mathbf Q_1^{-1} } \mathbf A \paren {\mathbf P_1 \mathbf P_2}\) Matrix Multiplication is Associative
\(\ds \) \(=\) \(\ds \paren {\mathbf Q_1 \mathbf Q_2}^{-1} \mathbf A \paren {\mathbf P_1 \mathbf P_2}\) Inverse of Matrix Product

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

\(\ds \paren {\mathbf Q_1 \mathbf Q_2} \mathbf C \paren {\mathbf P_1 \mathbf P_2}^{-1}\) \(=\) \(\ds \paren {\mathbf Q_1 \mathbf Q_2} \paren {\paren {\mathbf Q_1 \mathbf Q_2}^{-1} \mathbf A \paren {\mathbf P_1 \mathbf P_2} } \paren {\mathbf P_1 \mathbf P_2}^{-1}\) by hypothesis
\(\ds \) \(=\) \(\ds \paren {\paren {\mathbf Q_1 \mathbf Q_2} \paren {\mathbf Q_1 \mathbf Q_2}^{-1} } \mathbf A \paren {\paren {\mathbf P_1 \mathbf P_2} \paren {\mathbf P_1 \mathbf P_2}^{-1} }\) Matrix Multiplication is Associative
\(\ds \) \(=\) \(\ds \mathbf A\) Definition of Invertible Matrix

Hence:

$\mathbf A \equiv \mathbf C$

Thus transitivity holds.

$\Box$


Hence the result by definition of equivalence relation.

$\blacksquare$


Sources

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