Minimal polynomial problems Let $V = \mathbb R^4, \phi \in$ End(V)
$$D(\phi)= \begin{pmatrix}
1 & -3 & 1 &-2  \\
2 & 1 & 1 & 1  \\
-2 & 4 & -2 & 3 \\
-2 & -2 & -1 & -2 
\end{pmatrix}$$
The characteristic polynomial is $$h_\phi(x) = (x^2+1)(x+1)^2$$
After Cayley-Hamilton i know that the minimal polynomial divides the characteristic polynomial.
So let g(x) be the minimal polynomial. 
Then it follows that $$g(x) = (x^2+1)(x+1)^2 \vee (x^2+1)(x+1) $$
I know that $g(\phi) = 0$ but neither of the suggested polynomials for equal $0$ for $\phi$.
Have i done something wrong?

I have made a mistake when calculating.
The minimal polynomial is indeed $g(x)=(x^2+1) \cdot (x+1)^2$
I still have another question.
Let $p_i^{e^i}$ be a primary factor of the characteristic polynomial.
Determining $V_i = \operatorname{Ker}(p_i^{e^i}(\phi))$, I get
$$V_1 = \left<\begin{pmatrix}1 \\ 0 \\ -1 \\ 0\end{pmatrix}
           \begin{pmatrix}0 \\ 1 \\ 0 \\ -1\end{pmatrix} \right>,
  V_2 = \left<\begin{pmatrix}1 \\ 0 \\ -2 \\ 0\end{pmatrix}
              \begin{pmatrix}0 \\ 1 \\ 0 \\ -2\end{pmatrix}\right>
$$
If $T = V_1 \cup V_2$ then $T = 
  \left<\begin{pmatrix}1 \\ 0 \\ -1 \\ 0\end{pmatrix}
        \begin{pmatrix}0 \\ 1 \\ 0 \\ -1\end{pmatrix}
        \begin{pmatrix}1 \\ 0 \\ -2 \\ 0\end{pmatrix}
        \begin{pmatrix}0 \\ 1 \\ 0 \\ -2\end{pmatrix}\right>$ since they are linearly independent.
Is the Transformation matrix $D_T(\phi) = 
\begin{pmatrix}
1 &0 & 1 & 0 \\
0 & 1 & 0 &1 \\
-1 & 0 & -2 & 0 \\
0 & -1 & 0 & -2
\end{pmatrix}
  $
correct ?
