Every square matrix is similar to a matrix in what's called Jordan Canonical Form. This has various properties, but most important here is that it is upper triangular, and the eigenvalues (of both the new and original matrix) are on the diagonal of the resulting matrix.
The way to think about this process is that we change bases, and in that new basis the matrix becomes diagonal. This will help the existing intuition given in the OP about iterating the matrix, because iterating a triangular matrix will simply exponentiate the diagonal entries (and do predictable but somewhat messy stuff to the part above the diagonal).
Now, if an eigenvalue is complex, all the above still holds. However if the original matrix had real entries, then that eigenvalue must be paired with its complex conjugate -- both $a+bi$ and $a-bi$ must be eigenvalues. Then, we can rearrange the standard JCF matrix, as described above, into what is called Real Jordan Canonical Form. Now, instead of the matrix being entirely upper triangular, there will be some $2\times 2$ blocks along the diagonal, a rearrangement of what was a complex eigenvalue-plus-its-conjugate pair in the original JCF. The entries of these $2\times 2$ blocks are exactly the real and imaginary parts of these two complex eigenvalues.
Each of these $2\times 2$ blocks performs a rotation in the two-dimensional space spanned by those two basis elements. Hence, a $6\times 6$ matrix with six complex eigenvalues might be doing three different rotations in three different two-dimensional directions at once.
In the original $2\times 2$ case, the reason that a complex eigenvalue leads to a rotation is that it must appear with its complex conjugate, assuming the original matrix has real entries. As a pair they give a rotation. If the original matrix did not have real entries, then the matrix need not represent a pure rotation, because the two eigenvalues need not be related.
Operator
$$
\mathbf{T} =
\left[
\begin{array}{rrr}
-1 & 2 & 0 \\
0 & 3 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
Eigenvalues
To find the eigenvalues, compute $p(\lambda)$, the characteristic polynomial.
$$
p (\lambda ) = \det \left( \mathbf{T} - \lambda \mathbf{I}_{3} \right) = \det
\left[
\begin{array}{ccr}
-\lambda -1 & 2 & 0 \\
0 & 3-\lambda & 0 \\
0 & 0 & -\lambda \\
\end{array}
\right]
=
-\lambda \left( 3 - \lambda \right) \left( -1 - \lambda \right)
$$
The roots $p(\lambda) = 0$ are the eigenvalues: $\lambda = \left\{ 3, -1, 0 \right\}$.
Eigenvectors
Solve
$$
\mathbf{T} v_{k} = \lambda_{k} v_{k} \qquad \Rightarrow \qquad \left( \mathbf{T} - \lambda_{k} \mathbf{I}_{3} \right) v_{k} = \mathbf{0}
$$
$\lambda = 3$
$$
\begin{align}
\left( \mathbf{T} - 3 \mathbf{I}_{3} \right) v_{1} &= \mathbf{0} \\
\left[
\begin{array}{rrr}
-4 & 2 & 0 \\
0 & 0 & 0 \\
0 & 0 & -3 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
v_{1_{1}} \\
v_{1_{2}} \\
v_{1_{3}}
\end{array}
\right]
&=
\left[
\begin{array}{c}
0 \\
0 \\
0
\end{array}
\right]
%
\qquad \Rightarrow \qquad
v_{1} =
\left[
\begin{array}{c}
1 \\
2 \\
0
\end{array}
\right]
%
\end{align}
$$
$\lambda = -1$
$$
\begin{align}
\left( \mathbf{T} + \mathbf{I}_{3} \right) v_{2} &= \mathbf{0} \\
\left[
\begin{array}{rrr}
0 & 2 & 0 \\
0 & 4 & 0 \\
0 & 0 & 1
\end{array}
\right]
%
\left[
\begin{array}{c}
v_{2_{1}} \\
v_{2_{2}} \\
v_{2_{3}}
\end{array}
\right]
&=
\left[
\begin{array}{c}
0 \\
0 \\
0
\end{array}
\right]
%
\qquad \Rightarrow \qquad
v_{2} =
\left[
\begin{array}{c}
1 \\
0 \\
0
\end{array}
\right]
%
\end{align}
$$
$\lambda = 0$
$$
\begin{align}
\left( \mathbf{T} + 0 \mathbf{I}_{3} \right) v_{3} &= \mathbf{0} \\
\left[
\begin{array}{rrr}
-1 & 2 & 0 \\
0 & 3 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
v_{3_{1}} \\
v_{3_{2}} \\
v_{3_{3}}
\end{array}
\right]
&=
\left[
\begin{array}{c}
0 \\
0 \\
0
\end{array}
\right]
%
\qquad \Rightarrow \qquad
v_{3} =
\left[
\begin{array}{c}
0 \\
0 \\
1
\end{array}
\right]
%
\end{align}
$$
Best Answer
Each eigenvalue appears once in $D$ for each eigenvector it has in your collection of eigenvectors. So in this case, $D$ has $2$ twice along the diagonal. A few requirements, though:
As long as these are fulfilled, the are no issues.