Number of Matrix Equivalence Classes
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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