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Let $A\in M_{n}(C)$ such that $A^n=0$, but $A^{n-1}\neq0$.

If $B\in M_n(C)$ such that $AB=BA$, prove that $B=a_0+a_1A+a_2A^2+...+a_{n-1}A^{n-1}$ for some $a_0,a_1,a_2,...,a_{n-1}\in C$.

Note: $M_n(C)$ means a n*n matrix of complex numbers.

So, I know A is a Nilpotent matrix. I searched everyting I can for Nilpotent matrix in order to answer this question, but none of them had given me any clues. Any hints for me, please?

Mark
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  • $A$ can only have one possible Jordan block structure; from there you can replace $A$ with its Jordan normal form then try to compute directly what the condition $AB = BA$ implies about the coefficients of $B$. More abstract arguments are also possible. – Qiaochu Yuan Aug 06 '22 at 09:54
  • @QiaochuYuan I'm just doing this problem out of interest, and I was wondering: what allows us to replace $A$ with its JNF in the condition $AB=BA$? – Elliot Herrington Aug 06 '22 at 11:05
  • It's just a change of coordinates. The condition $AB = BA$ is invariant under conjugation (of both $A$ and $B$ simultaneously). – Qiaochu Yuan Aug 06 '22 at 11:44
  • Ohh of course, since they commute. Thanks! – Elliot Herrington Aug 06 '22 at 11:46
  • If a matrix $B$ commutes with a matrix $A$, then the matrix $B$ is a polynomial in $A$ if and only if $A$ has equal characteristic and minimum polynomial. For reference see for example this post. – Dietrich Burde Aug 06 '22 at 11:50

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The minimal polynomial for $A$ is $m(\lambda)=\lambda^n$. Therefore, there is a unit vector $x$ such that $A^{n-1}x\ne 0$, but $A^{n}x=0$. It follows that $\{ x,Ax,A^2x,\cdots,A^{n-1}x\}$ is a basis for $\mathbb{C}^n$. Now suppose $B\in M_n(\mathbb{C})$ commutes with $A$. Then there is some polynomial $p$ such that $$ Bx=p(A)x $$ Therefore $BA^kx=p(A)A^kx$ for $k=0,1,2,\cdots,n-1$, which forces $B=p(A)$ because $\{x,Ax,A^x,\cdots,A^{n-1}x\}$ is a basis.

Having commutativity and a cyclic basis is a powerful combination in linear algebra. And the Jordan form is a way of decomposing into cyclic subspaces.

Disintegrating By Parts
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