Coefficients of a minimal polynomial are homogeneous polynomials. If we have an endomorphism $\varphi$ in a $n$-dimensional vector space, and the expression of his characteristic polynomial $P_{\varphi}(X)$ is as follows :
\begin{equation}
P_{\varphi}(X)= X^n + \alpha_1X^{n-1} + \dots + \alpha_{n-1}X + \alpha_n
\end{equation}
then, we know the $\alpha_i$'s are all homogeneous polynomials of degree $i$ of the coefficients of the matrix representation of $\varphi$.
We also know that the minimal polynomial $\mu_\varphi$ of $\varphi$ divides $P_\varphi(X)$.

My question is : does the coefficients of $\mu_\varphi$ also follows the same rules as the characteristic polynomials? i.e. if we have
\begin{equation}
\mu_\varphi(X)= X^r + \beta_1X^{r-1} + \dots + \beta_{r-1}X + \beta_{r}
\end{equation}

Are the $\beta_i$'s homogeneous polynomials of degree $i$ of the coefficients of the matrix representation of $\varphi$ ? If yes, how do we find that ?
No: indeed the $\beta_i$ are not even continuous functions of those coefficients, essentially because $r$ itself is not a continuous function of those coefficients. Consider for example the family of matrices
$$\bigg\{\begin{pmatrix}1&amp;t\\0&amp;1\end{pmatrix}\colon t\in\Bbb R\bigg\}.$$
Here the minimal polynomial is $X^2-2X+1$ for all $t\in\Bbb R\setminus\{0\}$, but $X-1$ for $t=0$.
Let $t$ be trancendental/indeterminate. 

Then 
$$
\mu_{t\phi}(X)\ |\ \mu_{\phi}(tX)\ |\ \mu_{t\phi}(X)
$$
and so these minimal polynomials are equal.

Now this gives us the characteristic property of homogeneity:
$$
\beta_j(t a_{11}, \dots, t a_{nn})=t^j\beta_j( a_{11}, \dots,  a_{nn})
$$
where the $a_{ij}$ are the matrix entries.