What's the mistake in the annihilating polynomial?

Question

For the matrix

$$A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$$

an annihilating polynomial is $x^3-1$, but its characteristic polynomial is $x^3+1$.

Is this a contradiction, since roots of characteristics polynomial should be roots of annihilating polynomial too?

Answer 1

You just made a mistake calculating the characteristic polynomial. The characteristic polynomial is $-x^3 + 1$, which is just your annihilating polynomial times $-1$.

Answer 2

As Elvorfirilmathredia pointed out, you have made a mistake calculating the characteristic polynomial. Here are the calculations you should have made so you can find where you went wrong:

To find the characteristic polynomial, we need to find $\det(A-\lambda I)$, which is

\begin{align}\det(A-\lambda I)&= \det\left(\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}-\lambda\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\right)\\ &= \det\left(\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}-\begin{pmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{pmatrix}\right) \\ &=\det\begin{pmatrix}-\lambda&1&0\\0&-\lambda&1\\1&0&-\lambda\end{pmatrix}\\ &=-\lambda\det\begin{pmatrix}-\lambda&1\\0&-\lambda\end{pmatrix}-1\det\begin{pmatrix}0&1\\1&-\lambda\end{pmatrix}+0\det\begin{pmatrix}0&-\lambda\\1&0\end{pmatrix}\\ &=-\lambda(\lambda^2-0)-1(0-1)+0(0+\lambda)\\ &=-\lambda^3+1+0\\ &=1-\lambda^3 \end{align}