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Why is it, that for the matrix

$A \in \text{Mat}(n\times n, \mathbb{C})$

the characteristic polynomial $\chi_A(t)$ and the minimal polynomial $\mu_A(t)$ have same roots?

Since $\chi_A(t) = \mu_A(t) \cdot p(t)$ it should be easy to follow, that $\chi_A(t)$ has roots where $\mu_A(t)$ has roots.

But why can't $\chi_A(t)$ have roots where $\mu_A(t)$ hasn't?

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Actually, in general, the characteristic polynomial and the minimal polynomial have the same irreducible factors. –  lhf Jan 22 '12 at 12:52

2 Answers 2

First, show the following easy claim: Let $f(x)\in\mathbb{C}[x]$ be any polynomial. If $\lambda$ is an eigenvalue of $A$ associated with the eigenvector $v$, then $f(\lambda)$ is an eigenvalue of $f(A)$ associated with the eigenvector $v$. (prove by calculating $f(A)v$, using distributivity of multiplication over addition).
Now let $\lambda$ be a root of $\chi_A(t)$. Thus $\lambda$ is an eigenvalue and so has an associated eigenvector $v\neq 0$. Using the claim, we get $0=\mu_A(A)\cdot v=\mu_A(\lambda)\cdot v$ (since $\mu_A(A)=0$). Since $v\neq 0$, we get $\mu_A(\lambda)=0$.

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I remember of something like this : if $\mu_A(t)=\prod_{\lambda \in Sp(A)}(t-\lambda)$, then $\chi_A(t)=\prod_{\lambda \in Sp(A)}(t-\lambda)^{\nu_\lambda}$. The two polynomials have the same roots, the eigenvalues of $A$, and the order of multiplicity of $\lambda$ is $1$ in the minimal polynomial and $\nu_\lambda$ in the characteristic polynomial.

This is proved by Dennis Gulko's answer.

Further, the order of multiplicity $\nu_\lambda$ is the the dimension of the eigenspace $\nu_\lambda = \dim E_\lambda=\dim \ker (\lambda I - A)=\dim \{x|Ax=\lambda x\}$.

This is shown by looking at the restriction of $A$ to $E_\lambda$, where it acts as the homothety $x \rightarrow \lambda x$ and its matrix is the square diagonal matrix $\left( \matrix{\lambda&0&...&0 \\ 0&\lambda&...&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&...&\lambda} \right)$.

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