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Let $K$ be some field and $A, B \in M_n(K)$. Prove that: If $B$ is nilpotent and $AB=BA$ then $\det(A+B)=\det(A)$.

I believe there is a nice solution here. However, it seems that this problem could be solved using Jordan Canonical Form of the matrices. I am wondering how. Could anyone give me a hint?

Rodrigo de Azevedo
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mathdoge
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    Didn't darij grinberg's hint under the linked question solve the problem? – user1551 Jun 07 '19 at 15:15
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    I don't think you want JNF, but simultaneously triangularization. Note that it is not true that $A,B$ can be simultaneously put into JNF. – user10354138 Jun 07 '19 at 15:25
  • I think the linked question has many good references. It is perhaps a bit misleading that the answers speaks about avoiding the "machinery of JNF". I also think, simultansous triangularization is what you want. – Dietrich Burde Jun 07 '19 at 15:27
  • Thank you! I only know simultaneous triangularization on $M_n(\mathbb{C})$, does it work still on any general field? – mathdoge Jun 07 '19 at 15:50
  • Oh I found that any algebraically closed field works! Thanks! – mathdoge Jun 07 '19 at 15:56

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