How would one prove following?
If $A^2 = A$ and $B^2 = B$, then show that $$\mbox{rank} (A − B) = \mbox{rank} (A − AB) + \mbox{rank} (B − AB)$$
2026-03-25 23:11:01.1774480261
Prove $\mbox{rank} (A − B) = \mbox{rank} (A − AB) + \mbox{rank}(B − AB)$, if $A^2 = A$ and $B^2 = B$
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Since $A - B = A(A - B) - (B-A)B$ we have by subadditivity that $$\mbox{rank}(A - B) \leq \mbox{rank} (A(A-B)) + \mbox{rank} ((B-A)B) = \mbox{rank} (A - AB) + \mbox{rank} (B - AB)$$
On the other hand, using Frobenius' rank inequality and noting that $A(A - B)B = 0$:
$$\mbox{rank} (A(A -B)) + \mbox{rank} ((A-B)B) \leq \mbox{rank} (A - B) + \mbox{rank} (A(A-B)B) = \mbox{rank} (A - B))$$