If A is nilpotent, and B is an invertible matrix of the same order as A and AB =BA, then B- A is invertible.

Question

proof
I am quite sure that my proof is wrong, because I did not use the fact that AB= BA in the proof, is there a way to prove that B-A is singular without using determinant?

JCAA · Accepted Answer

First notice that $C=AB^{-1}$ is nilpotent (here you use the fact that matrices commute). Then $(I-C)(1+C+C^2+...)=I$. So $(I -AB^{-1}) $ is invertible. Multiply it by invertible $B$ and get that $B-A$ is invertible.

Correct answer by JCAA on January 20, 2021

If A is nilpotent, and B is an invertible matrix of the same order as A and AB =BA, then B- A is invertible.

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