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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.

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1 Answer 1

Every irreducible factor of the characteristic polynomial must divide the minimal polynomial.

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Thanks for the answers. There was a detail that I wasn't aware of, namely that the last invariant factor equals the minimal polynomial, i.e. $f_6(x) = m(x) = (x^2+1)(x-1)(x-2)$. As far as I can tell, then there will only be three possibilities: a) $f_4(x) = f_5(x)=x-1$, the others 1, b) $f_4(x) = f_5(x)=x-1$, the others 1, c) $f_5(x) = x^2+1$, the others 1. –  Svinepels Dec 14 '12 at 21:10

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