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I'm learning linear algebra and need help with the following problem:

Let $A = \begin{pmatrix}-2 & 4 & 3\\0 & 0 & 0\\-1 & 5 & 2\end{pmatrix} \in M_{3x3}(\mathbb{R})$. Show that $A^{2016}-2A^{3016}+A=0$.

I guess this is a direct application of the Cayley-Hamilton theorem which states that every matrix satisfies its own characteristic equation. The characteristic polynomial of $A$ is $p_{A}(\lambda) = \lambda - \lambda^{3}$ (I skipped the easy computation of the determinant to save me some time). Hence, by the Cayley-Hamilton theorem $$A - A^{3} = 0 \tag{*}$$

How should I make use of $(*)$ and continue from here to prove the identity? I thought I could write $(*)$ as $A = A^{3}$ and then appropriately multiply both sides of latter equality but I got stuck. I'm also interested to know if there are other methods to solve this problem.

EDIT: As A.G. demonstrated, the identity is not true. It would be true for odd powers, e.g. $A^{2017}-2A^{3017}+A=0$. This is an unfortunate typo from my teacher's notes. I apologize to the users who gave answers prior to this edit.

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    I don't find $A^{2016}-2A^{3016}+A=0$ - are you sure you have the right polynomial in $A$? – 2017-02-07
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    @AloneAndConfused You are right! This is an unfortunate typo from my teacher's notes. The identity is true for odd powers, e.g. $A^{2017}-2A^{3017}+A=0$. – 2017-02-07

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Yes, you are totally correct till now. And you have anticipated the right equations too. You can proceed as follows:

$$A=A^3 \implies A\cdot A=A^3\cdot A \implies A^2=A^4 $$ Similarly, $$A^2=A^4 \implies A^2\cdot A^2=A^4\cdot A^2 \implies A^4=A^6=A^2$$

So, we can mathematically induce $$A^2=A^4=A^6= \ldots = A^{2n} \,\,\,\,\, \text{for any n} \in \mathbb{Z^+}$$

Hence $A^{2016}=A^{3016}=\color{red}{A^2}$

Does it simplify anything?

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    Based on your conclusion that $A^{2016}=A^{3016}={A^2}$, the identity is not true. It is true for odd powers, e.g. $A^{2017}-2A^{3017}+A=0$, which I think is what my professor meant to write in the first place. Your method still works like a charm to prove the identity is true for odd powers. I do however apologize to you for having to write your answer based on a false statement. – 2017-02-07
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    @Elix No need to apologise .. Its okay . Hope my answer helped you. – 2017-02-07
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Another method to solve the problem.

The identity for $A$ is equivalent to the identity for the Jordan form, which is diagonal in our case (distinct eigenvalues $\pm 1$ and $0$) $$ J=\pmatrix{1 & 0 & 0\\0 & -1 & 0\\0 & 0 & 0}. $$ Now $$ J^{2016}-2J^{3016}+J=\pmatrix{1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0}-2\pmatrix{1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0}+\pmatrix{1 & 0 & 0\\0 & -1 & 0\\0 & 0 & 0}=\pmatrix{0 & 0 & 0\\0 & -2 & 0\\0 & 0 & 0}. $$ Well, the identity is not true. It would be true for odd powers, for example.

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    Thank you! This was driving me crazy. I did the computations in Mathematica and realise the identity did not hold. My teacher's notes have many typos. – 2017-02-07
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    @Elix Another way to get it true $A^{2016}-2A^{3016}+A^2=0$. – 2017-02-07
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$A-A^3= 0$ does indeed imply $A^3 = A$. This means that $$ A^{2016} = A^3A^{2013} = AA^{2013} = A^{2014} $$ Now keep going (you don't actually have to do this explicitly a thousand and six more times, just observe the pattern and note where it ends), and the equation should eventually be much simplified.