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Is there a matrix with real entries such that its minimal polynomial with coefficients in the complex field is different than its minimal polynomial over the real numbers?

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No, the minimal polynomial is independent of the field over which it is calculated, given that the entries are in the field. The minimal polynomial can, of course, be reducible over $\mathbb{C}$, while it is irriducible over $\mathbb{R}$.
To prove this fact, let $m_A^\mathbb{C}(x)$ denote the minimal polynomial of $A$ over $\mathbb{C}$. By the definition of $m_A^\mathbb{C}(x)$, it is the polynomial with the lowest degree such that $m_A^\mathbb{C}(A)=0$ and the leading coefficient is 1. Now, if there is a complex coefficient in $m_A^\mathbb{C}(x)$, then look at $m_A^\mathbb{C}(x)-\overline{m_A^\mathbb{C}(x)}$ (complex conjugate). This will be a non-zero polynomial of smaller degree which will be zero at $A$. That is a contradiction.
Hence all the coefficients of $m_A^\mathbb{C}(x)$ are real, so $m_A^\mathbb{C}(x)=m_A^\mathbb{R}(x)$

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How do you prove this? –  user25640 Jun 14 '12 at 9:12
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Perhaps you should also remind the reader that as $A$ has real entries, it will also be a zero of $\overline{m_{A}^{\mathbb{C}}(x)}$. –  Jyrki Lahtonen Jun 14 '12 at 10:31

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