Matrix Equivalence is Equivalence Relation
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices