Consider matrix $A$:
$\begin{bmatrix} 2 & -2\\ 3 & -3\\ \end{bmatrix}$
After little computing, we find the eigenvectors (and their corresponding eigenvalues) to be equal to
$E_{\lambda=0}= \begin{bmatrix} 1\\ 1\\ \end{bmatrix},\ E_{\lambda=-1}= \begin{bmatrix} 2\\ 3\\ \end{bmatrix}$
What, if any, is the significance of $\lambda$'s -1 eigenvector? I ask this because it is the same as the first column in the original matrix $A$ and I am curious to see if their is a specific set of parameters for this to occur and/or if this is significant in any way.
If you were to interpret the matrix as representing a "dynamical system" where the state after time $t$ had elapsed was given by $A^{t}x_0$ ($x_0$ the initial position), the eigenvector is where all states would eventually tend to. The reason your eigenvector is a column of the matrix is because all columns of the matrix are parallel - in general if we let $$A=\begin{bmatrix}c_0\textbf{x}&c_1\textbf{x}&\dots &c_n\textbf{x}\end{bmatrix}$$ ($\textbf{x}$ a vector) then for any input $y$ we will get $$A\textbf{y}=c\textbf{x}$$ for some constant $c$. Clearly then, $\textbf{x}$ will be an eigenvector of $A$ (and perhaps one of its columns!) because it will get sent to something parallel to itself.
Note however that there is nothing specifically significant per se about the eigenvector actually being a column. You could just as well have said that the eigenvector was $c\times\text{eigenvector}$ for any non-zero scalar and then it would not have been an actual column in the matrix.