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).

  • $\begingroup$ The eigenvalue $1$ is not a root of $x^2 + x + 1$, so... $\endgroup$ – J. M. isn't a mathematician Apr 25 '12 at 15:17
  • $\begingroup$ Ah of course, all eigenvalues must be a root of the minimal polynomial so it can't be! $\endgroup$ – user26069 Apr 25 '12 at 15:30

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy