How to conclude that the minimal polynomial is the characteristic? I am given the following matrix
$$A=\begin{bmatrix}
0 & 0 & 4 & 1\\
0& 0 & 1 & 4\\
4 & 1 & 0 &0\\
1 & 4 & 0 & 0
\end{bmatrix}$$
And I have to find the minimal polynomial of the matrix. The characteristic polynomial is $$K(\lambda)=-(\lambda -5)(\lambda +5)(\lambda-3)(\lambda +3)$$
The minimal polynomial $m(\lambda)$ divides the characteristic polynomial. I know that the characteristic polynomial is the minimal, but how do i eliminate the possibilities of the linear, quadratic and qubic factors in the polynomial. When do i know the minimal is actually the characteristic polynomial?

In this case you can do easily without the characteristic polynomial. It is easy to see that even powers of $A$ will have their nonzero entries in the $2\times2$ blocks at the top left and bottom right, and the odd powers of $A$ have them (like $A$ itself) in the bottom left and top right $2\times2$ blocks. You are looking for a power$~A^k$ that equals a linear combination of lower powers, and by what I just remarked the exponents of the contributing lower powers will have the same parity as$~k$. Since $A$ is nonsingular a minimal polynomial with only odd degree terms is not possible (it would have $0$ as a root), so the smallest $k$ for which a relation exists will be even. Also you know that the minimal polynomial has degree at most$~4$, the size of the matrix. So all there is to it is compute $A^0=I$, $A^2$ and $A^4$ and find a linear combination.
Concretely $A^2$ has as top left block $B=(\begin{smallmatrix}17&8\\8&17\end{smallmatrix})$ (repeated at the bottom right), and $A^4$ has as as top left block $B^2=(\begin{smallmatrix}353&272\\272&353\end{smallmatrix})$. Now $B^2$ minus $272/8=34$ times $B$ is a multiple of the identity, or you can apply the Cayley-Hamilton theorem for $B$; either way you find $B^2-34B+225I_2=0$ whence $A^4-34A^2+225I_4=0$, and $X^4-34X^2+225$ is the minimal polynomial you were after.
