Invariant Subspace on $\mathbf{C}^3$ 
Find all the invariant subspaces of $A$ viewed as a linear map on  $\mathbf{C}^3$
  when A is
  $$\begin{bmatrix}5 & 1&-1\\0 & 4&0\\1&1&3\end{bmatrix}.$$

I know how to find the invariant subspaces of $A$ when it is on $\mathbf{R}^3$. It is the kernel and the eigenspaces and $\mathbf{R}^3$. But how about $\mathbf{C}^3$?
The characteristic polynomial is $-\lambda^3+12\lambda^2-48\lambda+64=-(\lambda-4)^3$. The eigenvectors are $\begin{bmatrix}-1 & 1&0\end{bmatrix}^T$ 
 and$\begin{bmatrix}1 & 0&1\end{bmatrix}^T$.
So for $\mathbf{R}^3$, the invariant spaces are its kernel and its eigenspaces (both 1 dimension and 2 dimension), and $\mathbf{R}^3$. For $\mathbf{C}^3$, is it just adding $\mathbf{C}^3$, since I don't have complex roots for the determinant to add?

on $C^3$ you have next problem for $w = u + iv \in C^3$:
$Aw = Au + iAv = \lambda u + i\lambda v$, and actually we need to solve
$Re(Aw) = \lambda u ,$
$Im(Aw) =\lambda v$
but as we saw that $SpA = \{2, 4\}$, then our problem represents like 
$Au = \lambda u ,$
$Av =\lambda v$.
Where $v,u \in R^3$. So you have 2 Subspaces: $R^2\times iR^2$ for $\lambda = 2$ and $R\times iR$ for $\lambda = 4$
