Matrix Similarity is Equivalence Relation/Proof 2

Proof
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.

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

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

So matrix similarity is symmetric.

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.

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