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I'd like to show that for a square matrix A with the property that there exists a positive integer n such that $\dfrac{1}{n}A^n = \dfrac{1}{n+1}A^{n+1} = \dfrac{1}{n+2}A^{n+2}$, any eigenvalue of A must be 0.

I proceed as follows:

Let $\lambda$ be an eigenvalue of A, and $v$ be an eigenvector of A, such that $v \neq 0$.

Recall: $A^nv=\lambda^nv$, then since \begin{align} \dfrac{1}{n}A^n &=\dfrac{1}{n+1}A^{n+1} \\ \Rightarrow \dfrac{n+1}{n}A^nv-A^{n+1}v &=0 \\ \Rightarrow \lambda^n\left(\dfrac{n+1}{n}-\lambda\right)v &= 0 \\ \end{align} $\Rightarrow \lambda = 0$ or $\lambda = \dfrac{n+1}{n}$

Similarly: \begin{align} \dfrac{1}{n+1}A^{n+1} &=\dfrac{1}{n+2}A^{n+2} \\ \Rightarrow \dfrac{n+2}{n+1}A^{n+1}v-A^{n+2}v &=0 \\ \Rightarrow \lambda^{n+1}\left(\dfrac{n+2}{n+1}-\lambda\right)v &= 0 \\ \end{align} $\Rightarrow \lambda = 0$ or $\lambda = \dfrac{n+2}{n+1}$

My conclusion is since it's mathematically impossible that $\lambda = \dfrac{n+1}{n} = \dfrac{n+2}{n+1}$, the only eigenvalue of A must be 0.

Is my demonstration incomplete or incorrect?

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    Looks fine to me. – 2017-02-26
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    @ancientmathematician Could you be a bit more explicit? – 2017-02-26
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    @Zizou23, see comment below the answer. – 2017-02-26

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Essentially correct. You should stick to mathematical english or to $\Rightarrow$ language, though and not mix them up as you do in the first block of the proof.

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    Where does he mix them up? – 2017-02-26
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    @mmovice, s/he has a sentence which reads " ...then since [blah] implies [blah'] implies [blah'']". Either write this case like the second case, or replace the $\Rightarrow$s by english -- and not "implies". – 2017-02-26