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I have the following question:

Let $V$ be a vector space over a field $F$, and let $A$ be an endomorphism $V\rightarrow V$. Prove there is at most one pair of linear maps $D$ and $N$ such that $D+N=A$, $D$ is diagonalizable, $N$ is nilpotent, and every map that commutes with $D+N$ commutes with $D$ and $N$.

If there are two pairs, $D_1$, $N_1$ and $D_2$, $N_2$ that satisfy these properties, then $D_1-D_2 = N_2-N_1$. Since the sum of commuting diagonalizable and nilpotent matrices are diagonalizable and nilpotent respectively, the theorem is proved if I can show that $D_1, D_2$ and $N_1$, $N_2$ commute. But this is where I'm stuck.

Any help would be greatly appreciated. Thank you!

Dorothy Kimmel
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1 Answers1

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We have $A=D_1+N_1=D_2+N_2$ with the given property. And, of course $A$ commutes with $A$, so $A$ commutes with $D_1$ and $N_1$ (and also with $D_2$ and $N_2$). Writing again, $D_1$ commutes with $A=D_2+N_2$, so by the hypothesis again, $D_1$ commutes with both $D_2$ and $N_2$.

Berci
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  • Hi Berci, I am trying to follow this proof but I'm stuck on why $A (D_1 + N_1) = (D_1 + N_1)A \implies A D_1 = D_1 A$ etc. Could you please expand on the steps just a little bit? Thanks – muzzlator Mar 21 '13 at 07:25
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    This is part of the hypothesis on $D_i$ and $N_i$: 'If $M$ commutes with $A=D_i+N_i$ then $M$ commutes with both $D_i$ and $N_i$'. – Berci Mar 21 '13 at 12:44
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    Oh! I read that part as if it were an additional question to be proven. Thanks – muzzlator Mar 21 '13 at 12:46
  • @Berci I don't think your final sentence is justified. Just because $D_1$ commutes with $D_2 + N_2$ doesn't guarantee that $D_1$ commutes with $D_2$, unless we already know for sure that $D_1$ and $N_2$ commute. But this is not part of the hypothesis. – WillG Oct 04 '19 at 17:53