The characteristic polynomial of a matrix \(\displaystyle M\in \mathbb{F}^{n\times n}\) must be of degree \(\displaystyle n\), as far as I know.

Consider the following matrix \(\displaystyle M\in (\mathbb{Z}_2)^{2\times 2}\):

\(\displaystyle M=\begin{pmatrix}1&1&0\\1&0&0\\0&0&1\end{pmatrix}\)

If we try to calculate its characteristic polynomial, we get:

\(\displaystyle P_M(\lambda)=|M-\lambda I|=\begin{vmatrix}1-\lambda& 1 & 0\\ 1 & -\lambda & 0\\ 0 & 0 & 1-\lambda\end{vmatrix}=(1-\lambda)(\lambda^2-\lambda-1)\)

But now since \(\displaystyle \forall x\in\mathbb{Z}_2: x^2=x\), what we actually have here is: \(\displaystyle P_M(\lambda)=(\lambda-1)\)

How can the characteristic polynomial have degree less than 3?

Consider the following matrix \(\displaystyle M\in (\mathbb{Z}_2)^{2\times 2}\):

\(\displaystyle M=\begin{pmatrix}1&1&0\\1&0&0\\0&0&1\end{pmatrix}\)

If we try to calculate its characteristic polynomial, we get:

\(\displaystyle P_M(\lambda)=|M-\lambda I|=\begin{vmatrix}1-\lambda& 1 & 0\\ 1 & -\lambda & 0\\ 0 & 0 & 1-\lambda\end{vmatrix}=(1-\lambda)(\lambda^2-\lambda-1)\)

But now since \(\displaystyle \forall x\in\mathbb{Z}_2: x^2=x\), what we actually have here is: \(\displaystyle P_M(\lambda)=(\lambda-1)\)

How can the characteristic polynomial have degree less than 3?

Last edited: