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Let $V$ be a complex vector space and $T:V \to V$ linear and $m_T$ the minimal polynomial of $T$. If $c$ is the charactersitic polynomial then $c(x) = \prod_i (x-\lambda_i)$ where the $\lambda_i$ are eigen values of $T$ not necessarily different. Then $m_T$ divides $c$. Doesn't it imply that $m_T$ is a product of linear factors? Please can somebody show an example where $m_T$ is not a product of linear factors?

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    $\begingroup$ Is it possible that the word "distinct" is missing somewhere? Maybe you want an example where $m_T$ is not the product of distinct linear factors? $\endgroup$ – Daniel Fischer Sep 5 '13 at 11:34
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Edit: Adapted to complex vector spaces.

In the complex numbers, every polynomial is a product of linear factors. This result is known as the Fundamental Theorem of Algebra. Therefore, in particular, every minimal polynomial over complex numbers will be a product of linear factors.

In general vector spaces, this is not always true: Consider the matrix $\left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right)$ over the real numbers. Its characteristic polynomial is $x^2+1$ (easy calculation), and in the field $\mathbb R$, there is no way of decomposing that polynomial into linear factors. Furthermore, it turns out that the minimal polynomial is also $x^2+1$ (Why? Note that it can't have lower degree, because a polynomial of the form $a x + b$ with $a \neq 0$ will always have a non-zero anti-diagonal).

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  • $\begingroup$ I am sorry, this is a question about complex vector space. $\endgroup$ – blue Sep 5 '13 at 11:02

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