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Let $A\in M_{n}(\mathbb{F})$ be a triangularizable matrix with complex spectrum where $ \mathbb{F} $ is real numbers set. What is a form of minimal polynomial of $A$ that is irreducible?

Now, if $\mathbb{F} $ is complex numbers set. What is a form of minimal polynomial of $A$ that is irreducible?

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  • $\begingroup$ I don't understand the questions "what is a form of minimal polynomial of A that is irreducible"? What does it mean for the form of a polynomial to be irreducible? $\endgroup$ – Ben Grossmann Aug 21 '17 at 1:57
  • $\begingroup$ Based on a previous Question (and clarification) by the same user, I think the problem is asking what form an irreducible polynomial (that happens to be the minimal polynomial of $A$) would take. I'm guessing that despite the mention of "complex spectrum", the first paragraph is asking about irreducibility over the reals, and the second is asking about irreducibility over the field of complex numbers. $\endgroup$ – hardmath Aug 21 '17 at 2:04
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Case 1; $F=\mathbb{R}$. Then $A$ is a scalar matrix $A=aI_n$ (where $a\in\mathbb{R}$) or the minimal polynomial of $A$ is in the form $x^2+ax+b$, where $a^2-4b<0$, with roots $u\pm iv$ where $u\in\mathbb{R},v\in\mathbb{R}^*$. In the last case, $n=2p$ is even and $A$ is similar (over $\mathbb{R}$) to the matrix $diag(Z_1,\cdots,Z_p)$ where $Z_j=\begin{pmatrix}u&-v\\v&u\end{pmatrix}$.

Case 2. $F=\mathbb{C}$. Then $A$ is a scalar matrix $A=aI_n$ (where $a\in\mathbb{C}$).

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