I'm confused. Characteristic polynomial of degree less than n... Impossible? well...

Mar 2015
32
0
Australia
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?
 
Last edited:
Sep 2012
1,061
434
Washington DC USA
Re: I'm confused. Characteristic polynomial of degree less than n... Impossible? well

You're confusing a polynomial with a function that's giving its evaluation. When you do, say, calculus, those are the same, but here you need to keep track of the distinction.
Polynomials are formal sums with coefficients in the ring. They aren't functions, but rather elements in a set.
Example: Let \(\displaystyle f(x) = x^2 + 1 \in \mathbb{Z}_2[x], g(x) = x + 1 \in \mathbb{Z}_{2}[x]\). Then \(\displaystyle f(a) = g(a) \ \forall \ a \in \mathbb{Z}_2\), but \(\displaystyle f \ne g\).
You never apply the ring relations to the \(\displaystyle x\) in a polynomial ring \(\displaystyle R[x]\). It's only when you evaluate that polynomial that use the ring to simplify.
So \(\displaystyle a^2 = a \ \forall \ a \in \mathbb{Z}_2[x]\), but \(\displaystyle x^2 \ne x\) when considering \(\displaystyle x^2, x \in \mathbb{Z}_{2}[x]\).
 
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Mar 2015
32
0
Australia
Re: I'm confused. Characteristic polynomial of degree less than n... Impossible? well

Great explanation. Thanks!