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Suppose $A$ is a $n\times n$ nilpotent matrix of index $m$. i.e. $A^m=0$ but $A^{m-1}\neq0$.

Construct $M=A+\lambda I_n$. It is known that $M^{-1}=\sum_{i=1}^{m}\lambda^{-i}(-A)^{i-1}$ I want to prove $M^{-1}= B+\lambda^{-1}I_{n}$, where B is a nilpotent matrix of index $m$, i.e. $B^m=0$ but $B^{m-1}\neq0$. By the way, I want to get the characteristic polynomial and minimal polynomial of $M^{-1}$

It is easy to show $B^m=0$. However, I do not have a clue how to prove $B^{m-1}\neq0$.

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Let $I=I_n$, let $M = A + \lambda I$, and let $B = M^{-1} - \lambda^{-1}I$.

You want to show $B^{m-1} \ne 0$.

\begin{align*} \text{Then}\;\;&MM^{-1}=I\\[4pt] \implies\;&(A+\lambda I)(B+\lambda^{-1}I)=I\\[4pt] \implies\;&AB+\lambda^{-1}A + \lambda B = 0\\[4pt] \implies\;&B(A + \lambda I) = -\lambda^{-1}A\\[4pt] \implies\;&BM = -\lambda^{-1}A\\[4pt] \implies\;&B=(-\lambda^{-1}A)M^{-1}\\[4pt] \implies\;&B^{m-1}=(-\lambda^{-1}A)^{m-1}M^{-(m-1)} &&\text{[since $M$ commutes with A]}\\[4pt] \implies\;&B^{m-1}\ne 0 &&\text{[since $A^{m-1} \ne 0$]}\\[4pt] \end{align*} Note that $M^{-1}$ has only one distinct eigenvalue, namely $\lambda^{-1}$, hence the characteristic polynomial for $M^{-1}$ is $(x-\lambda^{-1})^n$.

But $M^{-1}$ satisfies $(x-\lambda^{-1})^m=0$, since $B^m=0$, and $M^{-1}$ does not satisfy $(x-\lambda^{-1})^{m-1}=0$, since $B^{m-1} \ne 0$. It follows that the minimal polynomial for $M^{-1}$ is $(x-\lambda^{-1})^m$.

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Hint Writing out explicitly the first few terms of your summation expression gives $$M^{-1} = B + \lambda^{-1} I_n = \lambda^{-1} I_n - \lambda^{-2} A + p A^2$$ for some expression $p$ polynomial in $A$. So, we can write $B^{m - 1}$ as $$B^{m - 1} = (- \lambda^{-2} A + p A^2)^{m - 1} .$$

When expanding this expression, every term has contains a factor of $A^m$ (and hence by hypothesis is zero) except for $(-\lambda)^{(2 - m)} A^{m - 1}$.

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