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Two matrices of the same size are equivalent if and only if they have the same rank.

Created over 8 years ago, updated 10 days ago

Theorem: Two $m \times n$ matrices $A$ and $B$ are equivalent if and only if they have the same rank.

$$A \sim B \iff \text{rank}(A) = \text{rank}(B)$$

Proof:

  • ($\Rightarrow$) If $B = PAQ$ with $P, Q$ invertible, then rank is preserved because multiplying by invertible matrices does not change rank: $\text{rank}(B) = \text{rank}(PAQ) = \text{rank}(A)$.
  • ($\Leftarrow$) If $\text{rank}(A) = \text{rank}(B) = k$, then both are equivalent to the canonical form $D_k$, so by transitivity $A \sim B$.

Corollary: The equivalence classes of $m \times n$ matrices are in one-to-one correspondence with possible ranks $\{0, 1, 2, \ldots, \min(m,n)\}$. There are exactly $\min(m,n) + 1$ equivalence classes.