Actually my question is that whether the minimal polynomial and the characteristic have the same root over the field of the vectorspace or do they have the same root over any extension field of F.
For example if the characteristic polynomial for a linear operator is $(x-2)(x^2+1)$ over $R$, is the only possibility of minimal polynomial the characterictic polynomial itself or are $(x-2)$ and $(x^2+1)$ also possible ?
If $A$ is an $n\times n$ matrix with entries in a field $k$, and $K$ a (commutative) field containing $k$, then a $K$-linear operator $\phi_{A,K}$ on $K^n$ is defined by left-multiplication by $A$, that is $\phi_{A,K}:v\in K^n\mapsto Av\in K^n$. A scalar $\lambda\in K$ is an eigenvalue of $\phi_{A,K}$ if and only if $\phi_{A,K}-\lambda I_{K^n}$ fails to be injective, in other words if $\dim\ker(\phi_{A,K}-\lambda I_{K^n})>0$; this happens whenever $\det(A-\lambda I_n)=0$ or equivalently when the characteristic polynomial $\chi_A$ of $A$, interpreted as an element of $K[X]$, admits $\lambda$ as root. (Nothing surprising here.)
Before going on to show that "$\lambda$ is an eigenvalue of $\phi_{A,K}$" is also equivalent to $\lambda$ being root of the minimal polynomial of $A$, it is important to establish that the minimal polynomial of $A$ over $K$ is the same as its minimal polynomial over $k$. This is the case because whether or not some power $A^d$ of $A$ can be expressed as a linear combination of lower powers $A^0=I_n,A,\ldots,A^{d-1}$ is a linear problem: it is asking whether a system of $n^2$ equations in $d$ unknown scalars (the coefficients of the linear combination sought) has any solution. Such a system has coefficients in$~k$ (since they come from the matrices $A^0,\ldots,A^{d-1}$), and if it has any solution over an extension field $K$ of $k$, it has a solution over$~k$; in particular for the lowest power $A^d$ for which a solution exists, and for which the solution is unique by the linear independence of $A^0,\ldots,A^{d-1}$, that solution exists over$~k$. That solution gives the minimal polynomial of $A$ over$~k$, which therefore is also the minimal polynomial of $A$ over any extension field$~K$.
So we now come to proving that eigenvalues of $A$ in $K$ are precisely the roots in $K$ of the minimal polynomial$~\mu_A$. If $\lambda$ is an eigenvalue of $A$ in $K$ then there exist eigenvectors for $\phi_{A,K}$: nonzero vectors $v\in K^n$ with $Av=\lambda v$. For such a vector and any polynomial $P\in K[X]$ one has $P[A]v=P[\lambda]v$, so in particular for $P=\mu_A$ it follows that $0=\mu_A[A]v=\mu_A[\lambda]v$, so $\mu_A[\lambda]=0$ (as $v$ is nonzero) and $\lambda$ is a root of $\mu_A$. Conversely if $\lambda\in K$ is a root of $\mu_A$, then $X-\lambda$ divides $\mu_A$, say $\mu_A=(X-\lambda)Q$. Now $Q[A]\neq0$ by minimality of $\mu_A$, which is equivalent to $Q[\phi_{A,K}]\neq0$, but $0=((X-\lambda)Q)[\phi_{A,K}]=(\phi_{A,K}-\lambda I)\circ Q[\phi_{A,K}]$, which means that $\ker(\phi_{A,K}-\lambda I)$ contains the image of $Q[\phi_{A,K}]$, which is of nonzero dimension. This establishes $\dim\ker(\phi_{A,K}-\lambda I_{K^n})>0$, and $\lambda$ is an eigenvalue of$~\phi_{A,K}$, and an eigenvalue of $A$ in$~K$.