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Why doesn't the minimal polynomial of a matrix change if we extend the field? I appreciate any help or proof.

CnR
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  • Well, it can change: over some field it can be reducible to a product of linear factors, over other one it may not...what do you man by "change" – DonAntonio Nov 23 '13 at 21:39
  • Well, yes. The form of the minimal polynomial can indeed change, but the minimal polynomial cannot. – Doc Nov 23 '13 at 21:46
  • Well, you can write $;(x-i)(x+i);$ in $;\Bbb C[x];$ , yet you can not write that for $;x^2+1\in\Bbb R[x];$ ...imo it is more than just the polynomial's "form". – DonAntonio Nov 23 '13 at 21:51
  • @DonAntonio: They are still the same polynomial. That's like saying $-1$ changes of value in $\mathbb{C}$ because it then becomes $i ^2$. – Najib Idrissi Nov 23 '13 at 21:52
  • No, it's not the same. I'd say the correct analogy it's like saying that $;\sqrt{-1}=i;$ everywhere. – DonAntonio Nov 23 '13 at 21:54
  • @DonAntonio: I'm sorry, but I don't understand what you mean. Are you claiming that $x^2 + 1 \neq (x-i)(x+i)$? – Najib Idrissi Nov 23 '13 at 21:55
  • Of course I am, @nik: in $;\Bbb R[x];$ they are not the same. Are you claiming they are the same in that polynomial ring?? – DonAntonio Nov 23 '13 at 21:56
  • @DonAntonio: yes, I am claiming they are, IF you consider $\mathbb{R}[x]$ as a subset of $\mathbb{C}[x]$ (which IMO is legitimate). – Najib Idrissi Nov 23 '13 at 21:57
  • Of course you consider it so, but in $;\Bbb R;$ there does not exist any element $;i=\sqrt{-1};$ and thus the polynomials $;x\pm i;$ doesn't exist there. – DonAntonio Nov 23 '13 at 21:59
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    @DonAntonio: Yes, but the polynomial $(x+i)(x-i)$ does exist there. And it is equal to $x^2+1$. This discussion is somewhat sterile, I don't really see your point. – Najib Idrissi Nov 23 '13 at 22:00
  • The reduction to $;(x-i)(x+i);$ cannot be done in the ring $;\Bbb R[x];$ as none of the two factors belong to that ring, and you're right about this discussion. Final point now. – DonAntonio Nov 23 '13 at 22:03
  • @DonAntonio, see earlier question at the link in my "answer" – Will Jagy Nov 23 '13 at 23:51
  • Ah, great comments and answers there, @WillJagy. Thanks. – DonAntonio Nov 24 '13 at 00:18

2 Answers2

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If you will find such polynomial $f$ that $f(A)=0$, you have to solve a system of linear equations with elements of matrices $A,A^2,\ldots$ as coefficients. Hence the coefficients of $f$ must belong to the same field as elements of $A$.

Boris Novikov
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see Can a matrix in $\mathbb{R}$ have a minimal polynomial that has coefficients in $\mathbb{C}$?

The proof is easy if going from the rationals or reals by just adjoining $i = \sqrt {-1}.$ It gets a bit more intricate in arbitrary finite extensions, as we are not always guaranteed an automorphism of the extension field that fixes the base field and only the base field. Still true. I would actually like to see more detail on those three answers myself.

Community
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Will Jagy
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  • Let $m(x)\in F[x]$ be the minimal polynomial of $A$, and let $K$ be any field extension of $F$. Let $f_i(x)$ be the irreducible factors of $m(x)$ in $F[x]$. Then, despite the fact that $f_i(x)$ may factor in $K[x]$, the collective roots of its factors will be roots of $m(x)$. This is because the Galois group $Gal(K/F)$ of $K$ over $F$ leaves $f_i(x)$ invariant (it fixes the coefficients of $f_i(x)$ since they are elements of $F$). Since $f_i(x)$ is irreducible, $Gal(K/F)$ will transitively permute its roots. Thus each $f_i(x)$ will appear in $m(x)$ whether $f_i(x)$ factors or not. – Doc Nov 24 '13 at 06:39