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Algebra Difficulty 5.2 AIME, harder Prove it Romania

Consider the matrices A∈Mm,n(C)A \in \mathcal{M}_{m,n}(\mathbb{C}), B∈Mn,m(C)B \in \mathcal{M}_{n,m}(\mathbb{C}) with n≤mn \le m. Given that rank(AB)=n\text{rank}(AB) = n and (AB)2=AB(AB)^2 = AB, find BABA.

Solution

Left-multiply with BB and right-multiply by AA the equality (AB)2=AB(AB)^2 = AB to obtain (BA)3=(BA)2(BA)^3 = (BA)^2. Recall that the rank of a matrix product does not exceed the rank of its factors to derive from ABAB=ABABAB = AB that rankBA≥n\text{rank}BA \ge n, hence rankBA=n\text{rank}BA = n.

The square matrix BABA of order nn is thus invertible, hence (BA)3=(BA)2(BA)^3 = (BA)^2, which implies BA=InBA = I_n.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.