If $T \in \mathcal{L}(V)$ is a linear operator on $V$, what are the conditions for which $T^k(u) \rightarrow v$ as $k \rightarrow \infty$? I'm looking for a decently general formal statement of a linear operator that does something roughly as follows:
Goal: We can use $T$ to model transition probabilities, and if $T^k(u)$ settles down (and we know the direction in which it settles down), that's a useful property to have.
Intuition: Assume we're dealing with real numbers only. I'm looking for the exact conditions on a linear operator $T$ such that there exists a $v \in V$ such that if I repeatedly apply $T$ to $u \in V$, the result $T^k(u)$ gets closer and closer to the direction of the vector $v$ as $k$ increases.
Now, if $V$ has a basis of eigenvectors of $T$, then $T$'s matrix can be diagonalized, then if it has $dim V$ distinct eigenvalues, then its clear that there exists an eigenvector $v$ whose absolute value of its corresponding eigenvalue is maximal, and thus such $T$ and eigenvector $v$ satisfies the property above.
Now if $T$ does not have a basis of eigenvectors, what are the minimum conditions such that I can find the $v$ where I get that nice property? Is it sufficient that $T$ has at least one eigenvalue? In particular, how can I relax the relatively strict condition of requiring $T$ to provide a basis of eigenvectors, such that I can still identify a $v$ with the nice property above?
 A: Fix a norm $\| \cdot \|$ on $V$ and pick a nonzero $u \in V$ and consider the sequence $u_n := T^n u$. We have
$$
\|u_n\| \leq \|T\|^n \|u\|,
$$
where $\|T\|$ is the induced operator norm (which is submultiplicative). If $\|T\| < 1$, we conclude that $u_n \to 0$ for any nonzero vector $u$.
Assume then that $\|T\| \geq 1$ and that the sequence $u_n := T^nu$ converges to $v \in V$. Given an $\epsilon > 0$ we can then find a $N$ such that $\|v - u_n\| < \epsilon$ for all $n \geq N$. We then get
$$
\|Tv - u_{n+1}\|
= \|T(v - u_n)\| \leq \|T\| \, \|v - u_n\| = \|T\| \epsilon
$$
for all $n \geq N$. Then $(u_n)$ also converges to $Tv$. We conclude that $Tv = v$ by uniqueness of limits. If $v = 0$ this is fine; if not we see that $v$ must be an eigenvector of $T$ with eigenvalue $1$.
For our sequence to have a nonzero limit, one of the eigenvalues of our operator must thus be $1$. This is not sufficient: consider an operator in three-dimensional space that fixes a line and rotates vectors perpendicularly to the given line. Explicitly, something like
$$
T = \begin{pmatrix}
1 & 0 & 0
\\
0 & \cos \theta & -\sin\theta
\\
0 & \sin\theta & \cos\theta
\end{pmatrix}.
$$
If we pick a vector $u$ that's orthogonal to the line (like $u = (0,1,0)$ here), our sequence will not converge to anything. It is also not enough for all eigenvalues to be real, as
$$
T = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}
$$
and the vector $u = (0,1)$ shows.
You mentioned stochastic matrices as a motivation. There are some general conditions for those under which these sequences converge to a nonzero limit for any starting probability vector. They're discussed on Wikipedia.
