Consider eigendecomposition $A=P\Lambda P^{-1}$ where $P$ consists of the eigenvectors of $A$. $$ \Lambda = \begin{bmatrix} \lambda_1 & & \\ & \lambda_2 & \\ ... \\ & & \lambda_n \end{bmatrix} $$ Here let's rank $|\lambda_1| > |\lambda_2| > ... > |\lambda_n| $
Then $$ A^k = P \begin{bmatrix} \lambda_1^k & & \\ & \lambda_2^k & \\ ... \\ & & \lambda_n^k \end{bmatrix} P^{-1} $$ where $v_i$ is the $i$th eigenvector.
Now consider the first column of $A^n$. What's the biggest element in terms of absolute value? It seems like, it should definitely be dominated by $\lambda_1$? What can we say about the biggest element in the first column of $A^k$ (suppose $k$ is very big)?
Edit: based on Christopher A. Wong's answer, we can write the power as $$ A^k e_1 = \lambda_1^k a_1 p_1 + \lambda_2^k a_2 p_2 + \ldots + \lambda_n^k a_n p_n. $$
So we can indeed see that the largest element is dominated by $\lambda_1$, when $a_1$ is not zero. But what can we say about the position of this largest element? Based on this expression, it seems like the position of the max abs value of the first column should be fixed? But doesn't look like that's the case based on my experiment.
If the biggest eigenvalue in terms of magnitude is complex, does that change anything?
It might be helpful to expand out the first column of $A^k$ explicitly. The first column is given by $Ae_1$. If $p_1, \ldots, p_n$ are a complete set of eigenvectors for $A$, then for some coefficients $a_1, \ldots, a_n$, $$ e_1 = a_1 p_1 + a_2 p_2 + \ldots + a_n p_n. $$ Then $$ Ae_1 = \lambda_1 a_1 p_1 + \lambda_2 a_2 p_2 + \ldots + \lambda_n a_n p_n, $$ and similarly $$ A^k e_1 = \lambda_1^k a_1 p_1 + \lambda_2^k a_2 p_2 + \ldots + \lambda_n^k a_n p_n. $$ Assuming the power $k$ is very large, then the term corresponding to the largest eigenvalue is dominant, except that you also have to consider the values of the coefficients $a_1, \ldots, a_n$. If $a_1 = 0$, then the dominant term will not be the first one. So instead the dominant term will be the term corresponding to the largest eigenvalue such that the corresponding scalar coefficient is also nonzero. Furthermore, the dominant term does not depend on how you order the eigenvalues.