Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A question says, write down the possible minimal polynomials which have characteristic polynomial $(1-x)(1-x^3)$, and for each possibility find a specific example of a matrix having this minimal polynomial.

Previously we're told that a matrix with minimal polynomial $x^2 + x + 1$ is $\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$.

So factorising $(1-x)(1-x^3)$, gives, $-(x-1)^2(x^2 + x + 1)$.

I understand that a two possible minimal polynomial are $(x-1)^2(x^2 + x + 1)$ and $(x-1)(x^2 + x + 1)$ with matrices $$\begin{pmatrix}1&1&0&0\\0&1&0&0\\0&0&0&-1\\0&0&1&-1\end{pmatrix}$$ and $$\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&-1\\0&0&1&-1\end{pmatrix}$$ respectively, but why can't we have a minimal polynomial $(x^2 + x + 1)$ and find a specific matrix? (I have the answers to this question by the way and it doesn't give that as a minimal polynomial).

share|improve this question
    
The eigenvalue $1$ is not a root of $x^2 + x + 1$, so... –  J. M. Apr 25 '12 at 15:17
    
Ah of course, all eigenvalues must be a root of the minimal polynomial so it can't be! –  user26069 Apr 25 '12 at 15:30

1 Answer 1

up vote 2 down vote accepted

Not only does the minimal polynomial of $A$ always divide the characteristic polynomial of $A$, it is also the case that every irreducible factor of the characteristic polynomial of $A$ must divide the minimal polynomial. This includes the linear terms (which correspond to eigenvalues).

In your case, the irreducible factors of the characteristic polynomial are $x-1$ and $x^2+x+1$, so the minimal polynomial $\mu(x)$ must satisfy $$(x-1)(x^2+x+1) \text{ divides }\mu(x)\text{ divides }(x-1)^2(x^2+x+1)$$ which gives you only the two possibilities you've given.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.