Characteristic polynomial of a matrix polynomial Suppose $A\in M_n(F)$ is an $n\times n$ matrix such that $F$ is a finite field. Also suppose that the characteristic polynomial of $A$ is irreducible and is equal to its minimal polynomial. If $B\in P[A]$, then is there any relation between characteristic polynomials of $A$ and $B$? 
 A: Alamos is right. Yet, using $\chi_A$, $\chi_B$ can be calculated without knowing the $(\lambda_i)$. For instance, let $\chi_A(x)=x^4+3x^3-6x^2+2x-1$ and $B=P(A)=A^3-7A^2+3A-I$ (note that we may assume that $\deg(P)\leq 3$). Then $I,B,B^2,B^3,B^4$ are known polynomials in $A$ of degree $\leq 3$. Consequently they are linearly dependent and we deduce explicitly $\chi_B$. We obtain $\chi_B(y)=y^4+247y^3+1189y^2-1023y+487$. If you have Maple, then, using Grobner basis theory, you can automate the calculation.
EDIT. I give some details to my downvoter. We assume that $K$ is a field, $A\in M_n(K)$ and $B=P(A)$ where $P\in K[x]$. 
STEP 1. We write $\chi_A(x)=f_1\cdots f_k$ where the $(f_k)$ are irreducible  non-necessarily distinct-. We apply the above method to each $f_i(x)=\Pi (x-\lambda_j)$ and we obtain $g_i(y)$, that is, in general $\Pi (y-P(\lambda_j))$. 
STEP 2. If $deg(g_i)=deg(f_i)$, then we are done; otherwise let $deg(f_i)=p$ and $(\lambda_j)$ be the -conjugate over $K$- zeroes of$f_i$. Necessarily $P(\lambda_1)=\cdots =P(\lambda_p)=\alpha \in K$ and $g_i(y)=y-\alpha$ (why ?). Finally, the polynomial that we associate to $f_i$ is $(y-\alpha) ^p$.
STEP 3. $\chi_B$, the characteristic poynomial of $B$, is the product of the polynomials associated to the $f_i$. 
Example: $K=\mathbb{R},\chi_A(x)=x^2+x+1,P(x)=x^3$. Then $f_1(x)=x^2+x+1, g_1(y)=y-1$ and $\chi_B(y)=(y-1)^2$.
