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 $A$?


1 Answer 1


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 $N=3$.


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