I have some questions concerning a problem in linear algebra / abstract algebra.
A $ 6 \times 6$ matrix $A$ over $\mathbb{R}$ has the minimal polynomial $m(x) = (x^2+1)(x-2)(x-1)$. Using this fact, I want to determine all the possibilities for non-unit monic invariant factors for the matrix $A-xI_6$, and in each case to find the corresponding rational canonical form.
The characteristic polynomial of $A$, $\chi(x)=\det(A – xI_6)$, is the product of the invariant factors, and the minimal polynomial divides the characteristic polynomial, which is of degree 6. In addition, the characteristic polynomial is monic since 6 is an even number. Thus, since the minimal polynomial is of degree 4, there must be some polynomial of second degree $x^2+bx+c$ such that
$m(x)(x^2+bx+c) = \chi(x) = f_1(x) \cdots f_6(x)$
where $f_i(x)$ are the invariant factors of $A-xI_6$ with $f_1(x)| \cdots |f_6(x)$.
But now I'm stuck. I guess one possibility would be $f_1(x)=\cdots=f_5(x)=1$ and $f_6(x) = m(x)(x^2+bx+c)$. But then I get infinitely many possibilities for different $a$ and $b$, and I think the solution is simpler than that. So, is there something I am forgetting here?
Thanks in advance for all answers.
Best Answer
Every irreducible factor of the characteristic polynomial must divide the minimal polynomial.