# roots of minimal and characteristic polynomial

Why is it, that for the matrix

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

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

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 $\mu_A(\lambda)=0$.