Matrix Similarity is an Equivalence

Theorem

Matrix similarity is an equivalence relation.

Proof

Follows directly from Matrix Equivalence is an Equivalence.

Alternatively, checking in turn each of the critera for equivalence:

Reflexive

$\mathbf A = \mathbf{I_n}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all order $n$ square matrices $\mathbf A$.

So matrix similarity is reflexive.

$\Box$

Symmetric

Let $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.

As $\mathbf P$ is invertible, we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \mathbf P \mathbf B \mathbf P^{-1}\)	\(=\)	\(\displaystyle \mathbf P \mathbf P^{-1} \mathbf A \mathbf P \mathbf P^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \mathbf{I_n} \mathbf A \mathbf{I_n}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \mathbf A\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So matrix similarity is symmetric.

$\Box$

Transitive

Let $\mathbf B = \mathbf P_1^{-1} \mathbf A \mathbf P_1$ and $\mathbf C = \mathbf P_2^{-1} \mathbf B \mathbf P_2$.

Then $\mathbf C = \mathbf P_2^{-1} \mathbf P_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2$.

The result follows from the definition of invertible matrix, that the product of two invertible matrices is itself invertible.

So matrix similarity is transitive.

$\Box$

So, by definition, matrix similarity is an equivalence relation.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 29$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 16$ (Worked Example)