$A,B$ is square matrix. I have to prove that $B^2=B$ implies $rank(AB-BA)\leq rank(AB+BA)$
I know $(I-B)(AB+BA)=AB-BAB=(AB-BA)B$. But I'm not sure what to do next.
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I haven't figured out a full proof yet, but a general strategy that works for a lot of linear algebra problems like this is to a) prove it for diagonal matrices, b) prove that the property holds when you conjugate (which is of course true since we're talking about rank of something), and then c) prove that it's stable under taking limits, so that you can approximate an arbitrary matrix by a sequence of diagonalizable matrices and just take the limit to get the result in general. – A. Thomas Yerger May 08 '22 at 00:23
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1The answer given on this post is better – Ben Grossmann May 08 '22 at 00:45