Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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? – Daniel Fischer Sep 5 '13 at 11:34
up vote 4 down vote accepted

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).

share|cite|improve this answer
I am sorry, this is a question about complex vector space. – blue Sep 5 '13 at 11:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.