Similar Matrices are Equivalent

Theorem
If two square matrices over a ring with unity $R$ are similar, then they are equivalent.

However, if two square matrices of order $n > 1$ are equivalent, they are not necessarily similar.

It follows directly that every equivalence class for the relation of similarity on $\mathcal M_R \left({n}\right)$ is contained in an equivalence class for the relation of matrix equivalence.

Here, $\mathcal M_R \left({n}\right)$ denotes the $n \times n$ matrix space over $R$.

Proof
If $\mathbf A \sim \mathbf B$ then $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.

Let $\mathbf Q = \mathbf P$ and the first result follows.

If $\mathbf A \equiv \mathbf B$ then $\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$.

It is not necessarily the case that $\mathbf Q = \mathbf P$.

As a counterexample, let $\mathbf A = \mathbf I_n$ be the identity matrix of order $n > 1$.

Let $\mathbf B$ be any invertible matrix of order $n$ that is different from the identity matrix.

Then $\mathbf A \equiv \mathbf B$, as:


 * $\mathbf I_n^{-1} \mathbf A \mathbf B = \mathbf I_n^{-1} \mathbf I \mathbf B = \mathbf B$.

If $\mathbf P$ is an invertible matrix of order $n$, then:


 * $\mathbf P^{-1} \mathbf A \mathbf P = \mathbf P^{-1} \mathbf P = \mathbf I_n \ne \mathbf B$

Hence, $\mathbf A$ is not similar to $\mathbf B$.