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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.

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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 $$\mbox{ rank }(CD-DC)\leq\mbox{ rank }(CD+DC)$$

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    $\begingroup$ 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. $\endgroup$ – user1551 May 4 at 17:13

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