# Cayley-Hamilton-Theorem - Possible characteristic polynomial

Let $$A: \mathbb{R}^3 \to \mathbb{R}^3$$ s.t.

$$A^3-2A^2+A= 0$$

The Cayley-Hamilton-Thm. states that if I put $$A$$ into its characteristic polynomial it'll equal $$0$$.

But am I allowed to conclude from the given equation $$A^3-2A^2+A= 0$$ that $$\lambda^3-2 \lambda^2+\lambda$$ is the characteristic polynomial of $$A$$?

No you're not. What if your matrix $$A$$ was a $$3\times 3$$ zero matrix (all elements are $$0$$). Then your equation would be valid but the characteristic polynomial is $$\lambda^3 = 0$$.
However you know that if $$\lambda$$ is an eigenvalue of $$A$$ then $$\lambda=0$$ or $$\lambda =1$$, which are the only two roots of $$P$$, where $$P(\lambda) = \lambda^3 - 2\lambda^2 + \lambda = \lambda (\lambda-1)^2\,.$$ So the characteristic polynomial can have at most $$0$$ and $$1$$ as roots. So all you know is that the characteristic polynomial $$Q$$ must be in the form $$Q(\lambda) = \lambda^n (\lambda-1)^m$$ with $$n\in\mathbb{N}$$, $$m\in\mathbb{N}$$ such that $$m+n=N$$, where $$N$$ is the number of rows (and so columns) of $$A$$. In your case $$N=3$$.