Call the linear transformation $T$, and the vector space on which it acts, $V$. Assume $V$ is finite-dimensional.
In the special case where the $V$ has a basis $B$ consisting of eigenvectors of $T$, and there is a dominant eigenvalue $\lambda$, the following is true: if $v$ is any element of $V$, and if the expression for $v$ with respect to the basis $B$ has a nonzero component in the eigenspace $W$ of $\lambda$, then $\lim_{n\to\infty}\lambda^{-n}T^nv$ exists and is nonzero and is in $W$.
The easiest way to see this is to write $v$ as a linear combination of vectors in $B$ and then compute $\lim_{n\to\infty}\lambda^{-n}T^nv$ (this will also give you some idea what to do when the conditions are not met). That is, let $B=\{{x_1,x_2,\dots,x_r\}}$ and let $v=a_1x_1+a_2x_2+\cdots+a_rx_r$ where $Tx_i=\lambda_ix_i$. Then $T^nv=a_1\lambda_1^nx_1+a_2\lambda_2^nx_2+\cdots+a_r\lambda_r^nx_r$ Now suppose $\lambda_1$ is dominant, that is, $|\lambda_1|\gt|\lambda_i|$ for $i=2,3,\dots,r$. Then $\lambda_1^{-n}T^nv=a_1x_1+a_2(\lambda_2/\lambda_1)^nx_2+\cdots+a_r(\lambda_r/\lambda_1)^nx_r$ and when you take the limit as $n\to\infty$ you get $a_1x_1$.
Things get more complicated if there isn't a single dominant eigenvalue but you can still use this approach and get pretty far with it.