SUMMARY:
A complex eigenvector is calculated from a $2 \times 2$ square matrix A. Expressed as a sum of real and imaginary parts
$\lambda_{A} = \begin{bmatrix}a+c&0\\0&a-c\end{bmatrix} + \begin{bmatrix}ib&0\\0&ib\end{bmatrix}$
If the eigenvectors are real (i.e. $b=0$), I know the transformation has the effect of scaling. What is the transformation associated with the "imaginary matrix"? I'm under the impression a rotation is involved, but it is not clear to me as to why or how to interpret is visually.
DETAIL:
I have two $2 \times 2$ square matrices (A and B) for which I obtain their eigenvalues ($\lambda_{A_{1,2}}$ and $\lambda_{B_{1,2}}$) and eigenvectors ($\textbf{v}_{A_{1,2}}$ and $\textbf{v}_{B_{1,2}}$). It is my objective to understand qualitatively how the two transformations differ given knowledge of their eigenvalues and eigenvectors.
Both matrices have the same set of eigenvectors ($\textbf{v}_{A_{1,2}} = \textbf{v}_{B_{1,2}}$), but different corresponding eigenvalues ($\lambda_{A_{1,2}} \neq \lambda_{B_{1,2}}$). In my particular case, the eigenvalues are determined to be $\lambda_{A_{1,2}} \in\mathbb C$ and $\lambda_{B_{1,2}} \in \mathbb R$. Both the eigenvectors are determined to be $\textbf{v}_{A_{1,2}},\textbf{v}_{B_{1,2}} \in \mathbb R^2$.
If I consider $\operatorname{Re}(\lambda_{A_{1,2}})$, I can make an "apples to apples" comparison with $\lambda_{B_{1,2}}$, in which the real part corresponds to isotropic scaling. For the sake of discussion, lets assume $\operatorname{Re}(\lambda_{A_{1,2}}) = \lambda_{B_{1,2}}$, as it is my intention to better understand the effects of the imaginary component.
Since $\lambda_{A_{1,2}} \in\mathbb C$, there is a additional transformation corresponding to $\operatorname{Im}(\lambda_{A_{1,2}})$. I believe imaginary eigenvalues correspond to a rotation but I'm having a hard time envisioning this when the eigenvectors lie in $\mathbb R^2$. If it is a rotation, what is the axis of rotation?
I'm mostly interested in qualitative/visual solutions here so I can get a feel for what is going on. Also,since I'm talking about linear transforms here, is there any value in thinking of it in a different way? In other words, impose the complex nature of the eigenvalue onto the eigenvector then compare eigenvectors in a visual manor? I don't know, I'm just trying to get a grasp on the imaginary effects.
EDIT: I've provided a very specific case to go along with my question
Let
$A = B = \begin{bmatrix}z&-c\\-c&z\end{bmatrix}$
in which $c \in \mathbb R$. Regardless of how $z$ is defined, the eigenvectors (for both matrices) are given by
$\textbf{v}_{A,B} = \begin{bmatrix}-1&1\\1&1\end{bmatrix}$
with components $(x_1,x_2) | x_i \in \mathbb R$. It is my understanding that these vectors are therefore said to lie in $\mathbb R^2$. (I could be wrong with this lose understanding?)
The eigenvalues on the other hand, depend on where $z = a+ib | a,b \in \mathbb R$ lies in the complex plane. For the A-matrix, I say that $b\neq0$, hence $\lambda_{A_{1,2}} \in\mathbb C$ and for the B-matrix, I say that $b=0$, hence $\lambda_{B_{1,2}} \in\mathbb R$. The corresponding eigenvalues are defined by the diagonals in the following
$\lambda_{A} = \begin{bmatrix}a+c+ib&0\\0&a-c+ib\end{bmatrix} \quad \lambda_{B} = \begin{bmatrix}a+c&0\\0&a-c\end{bmatrix}$
I should note that the eigenvalues for $\lambda_{A}$ are NOT complex conjugate pairs ($ib > 0$ for both). I'm pretty sure I calculated those right. Maybe for a complex matrix transformation, this is OK?
I believe this abstract example helps clarify my question. When comparing the transformation matrices, A and B, I'm trying to understand how to visualize the transformation effects from $\operatorname{Im}(\lambda_{A_{1,2}}) = ib$. Where does a resultant vector end up following the transformation ($\mathbb R^2$ or $\mathbb C$)? Also, please see my questions in the original posting above.