Number of Matrix Equivalence Classes

Theorem

Let $K$ be a field.

Let $\map {\MM_K} {m, n}$ be the $m \times n$ matrix space over $K$.

Let $\mathbf A$ be an $m \times n$ matrix of rank $r$ over $K$.

Then:

$\mathbf A \equiv \begin{cases}

\sqbrk {0_K}_{m n} & : r = 0 \\ & \\ \begin{bmatrix}

 \mathbf I_r & \bszero \\
 \bszero & \bszero 

\end{bmatrix} & : 0 < r < \min \set {n, m} \\ & \\ \begin{bmatrix}

 \mathbf I_r & \bszero 

\end{bmatrix} & : r = m < n \\ & \\ \begin{bmatrix}

 \mathbf I_r \\
 \bszero 

\end{bmatrix} & : r = n < m \\ & \\ \mathbf I_r & : r = m = n \end{cases}$

Thus there are exactly $\min \set {m, n} + 1$ equivalence classes for the relation of equivalence on $\map {\MM_K} {m, n}$, one of which contains only the zero matrix.

Proof

Follows from Equivalent Matrices have Equal Rank.

$\blacksquare$

Sources

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