Spectral decomposition of an operator

Question

Given the operator: $$A=\begin{pmatrix} i & 0 & - 4\\ 0 & - 3i & 0\\ 2 & 0 & - i \end{pmatrix}$$ Now $\det(zI-A)=(z+3i)(z^2+9)=(z+3i)^2(z-3i)$,

I know a theorem that says that these four prepositions are equivalent:

1. Eigenvectors of $A$ form a basis,

2. There are no spectral operators in spectral decomposition of $A$, $\mathbb{E} $,

3. $R(z)$ has only poles of order one.

4. $A$ satisfies an algebric equation with all zeroes of order one ("simple roots").

Now to find the spectral representation I calculate the operator $R(z)=\frac{1}{zI-A}$: $$R(z)=\begin{pmatrix} \frac{z+i} {(z+3i)(z-3i)} & 0 & \frac{-4}{(z+3i)(z-3i)}\\ 0 & \frac{1}{z+3i} & 0\\ \frac{2}{(z+3i)(z-3i)} & 0 & \frac{z-i} {(z+3i)(z-3i)} \end{pmatrix} $$

And find that there are no spectral operators $\mathbb{E} $ respecting 1 and 3 of the theorem, but the $\det$ has not only zeroes of order one. How is this possible? How should I interpret the point 4 of the theorem?

Thanks you all