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Let $A=\begin{pmatrix} 1 &2 & 1 &0\\ -2 & 1 & 0 &1 \\ 0 &0 &1 &2 \\ 0 &0 & -2 & 1 \end{pmatrix}$. Then the characteristic polynomial of $A$ is $\chi (x)=(x^2-2x+5)^2$.

I want to find the minimal polynomial of $A$. How can I find this? Are there some formulaic algorithm?(That is, if some algorithm for this question exists, this algorithm can be applied to other matrices?)

user112018
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  • I see the $4 \times 4$ real Jordan form for complex e-value $\lambda = -1 \pm 2i$. This cannot be diagonalized even by complex similarity transformation as there are not enough e-vectors here. This is not an answer to your question. I'm sure someone will come along shortly for that. – James S. Cook Jan 06 '14 at 15:30
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    @user112018: Do you have access to a copy of Dummit and Foote? – Amzoti Jan 06 '14 at 15:32

1 Answers1

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The minimal polynomial must be a divisor of the characteristic polynomial.

You've already found a factorization of the characteristic polynomial into quadratics, and it's clear that $A$ doesn't have a minimal polynomial of degree $1$, so the only thing that remains is to check whether or not $x^2-2x+5$ is actually the minimal polynomial or not.

rschwieb
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  • We don't know what the base field is, it could be that $x^2-2x+5$ is reducible in $\mathbb K[X]$! – zarathustra Jan 06 '14 at 16:07
  • @Antoine Sure, but if its minimal polynomial was linear, then $A$ would have to be a diagonal matrix, but it is not... And besides that, the minimal and characteristic polynomials don't depend on the base field, and it's most likely the user has $\Bbb R$ in mind... – rschwieb Jan 06 '14 at 16:14
  • I was just commenting on "you've already found a prime factorization", of course not questioning the last statement! – zarathustra Jan 06 '14 at 16:20
  • @Antoine Yes, again, I was just betting they had $\Bbb R$ in mind. Students who don't mention are often blissfully unaware of matrices over other fields. I'll try to make modifications to avoid sounding unclear about this, though. – rschwieb Jan 06 '14 at 16:27
  • Thanks for answer. Your answer is to check all the possibilities - in this problem, fortunately, there are only two cases. If some matrix has the minimal polynomial $(x-1)^2(x-2)^2(x-3)^2(x-4)^2$, then I have to check 16 candidates, right? – user112018 Jan 06 '14 at 23:46
  • @user112018 you are very likely to hit upon it before you try all 16. You would probably try the four of highest degree first, and see where that leads. One might be able to make more observations if one finds the Jordan form, too. – rschwieb Jan 07 '14 at 03:41
  • @rschwieb What do you mean by saying ''A doesn't have a minimal polynomial of degree 1'' ? – Leyla Alkan Feb 18 '18 at 21:36
  • @LeylaAlkan Think about what it means to be a matrix whose minimal polynomial is of the form $X-\alpha$. You should be able to see that such matrices are easily identified by sight! – rschwieb Feb 18 '18 at 21:39