Equivalent Matrices may not be Similar

Theorem

If two square matrices of order $n > 1$ over a ring with unity $R$ are equivalent, they are not necessarily similar.

Proof

Proof by Counterexample:

Let $\mathbf A = \mathbf I_n$ be the unit matrix of order $n > 1$.

Let $\mathbf B$ be an arbitrary invertible matrix over $R$ of order $n$ that is different from the unit matrix.

Then:

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

showing that $\mathbf A$ and $\mathbf B$ are equivalent.

Let $\mathbf P$ be an invertible square 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 $\mathbf B$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices