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 $\left( \matrix{\lambda&0&...&0 \\ 0&\lambda&...&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&...&\lambda} \right)$.