# Matrix problem similar to SEEMOUS 2019 3

Let $$A, B\in M_n(\mathbb{C} )$$ such that $$B^2=B$$. Prove that $$\operatorname{rank} (AB-BA) \le \operatorname{rank} (AB+BA)$$.
The problem above is the one that appeared in the SEEMOUS 2019,which is a contest for college students. In a contest for 11th graders in my country(here we study both linear algebra and real analysis in the 11th and the 12th grade), the same problem appeared, only that $$A, B\in M_n(\mathbb{R})$$. Now, I have seen solutions to the SEEMOUS problem which use pretty advanced techniques, which are not taught in high school here, but I want to find one that is appropriate for my level. I suppose there should be one since it appeared in that contest for 11th graders, yet I can't find one and there is also no official solution.
To sum up, I am looking for a solution to this which may only use basic rank inequalities, linear independence, diagonalization, block matrices and other things like these.
EDIT: From $$B^2=B$$ I deduced that $$B$$ is diagonalizable and its eigenvalues are $$0$$ or $$1$$(I doubt that eigenvalues help here).
If $$B$$ is invertible we have that $$B=I_n$$ and the conclusion follows.
In the other case, I think that we may use that $$B$$ is diagonalizable, but I don't know if this helps me express $$AB$$ and $$BA$$ somehow.

As $$B^2=B$$ so $$B$$ is diagonalizable. Now for $$B=0$$ or $$B=I$$, the result is trivial. So let us assume that $$B\neq0,I$$. Note that eigenvalues of $$B$$ can be only $$0$$ or $$1$$. Since $$B$$ is diagonalizable, we can write $$B=UDU^{-1}$$ where $$D$$ is the diagonal matrix with $$1'$$s at the first diagonal entries and $$0'$$s at the rest of them. Without loss of generality, assume that $$D=\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}$$ where $$I_r$$ is the $$r\times r$$ identity matrix. Since rank is invariant under similar matrices, we have that $$\mbox{ rank }(AB-BA)=\mbox{ rank }\left(U^{-1}(AB-BA)U\right)=\mbox{ rank }(CD-DC)$$ and $$\mbox{ rank }(AB+BA)=\mbox{ rank }\left(U^{-1}(AB+BA)U\right)=\mbox{ rank }(CD+DC)$$ where $$C=U^{-1}AU$$. So it is enough to prove that $$\mbox{ rank }(CD-DC)\leq\mbox{ rank }(CD+DC)$$ Now let $$C=\begin{pmatrix} P_{r\times r} & Q_{r\times n-r} \\ R_{n-r\times r} & S_{n-r\times n-r} \end{pmatrix}$$ where the partition have been done conformally to $$D$$. Now $$CD-DC=\begin{pmatrix} 0 & -Q \\ R & 0 \end{pmatrix}$$ and $$CD+DC=\begin{pmatrix} 2P & Q \\ R & 0 \end{pmatrix}$$ From this it, it follows that $$\mbox{ rank }(CD-DC)\leq\mbox{ rank }(CD+DC)$$
• Your proof isn't complete. You need to explain why the rank of $\pmatrix{0&-Q\\ R&0}$ is dominated by the rank of $\pmatrix{2P&Q\\ R&0}$. Since this was shown in Stelios Sachpazis' answer to your duplicate question, I don't see the purpose of your incomplete answer here. – user1551 May 4 at 17:13