Relationship between a linear-transformation and invariant subspace Let $V$ a vector over $F$  and $T: V\rightarrow V$ a linear transformation.
Prove that if $T$ is triangulizable, than it has a non-trivial invariant subspace.
Is this simply a case of taking a basis $B = \{v_1,...,v_n\}$ such that $[T]_B$ is triangular, than $Span(v_1)$ is an invariant sub-space for $T$?
If so - what is the logic behind this? Why is the triangular part plays a significant role here?
 A: Your logic is just fine. The triangulizable condition is just a simple sufficient condition for the existence of a nontrivial invariant subspace. 
Of course, as you seem to be noticing, one might now wish to ponder some followup questions. For example, is triangulatizability a necessary condition for existence of a nontrivial invariant subspace? If not, are there more interesting sufficient conditions? Is there an interesting necessary and sufficient condition? Pondering these questions indeed leads to some very significant and important mathematics.
A: As Lee Mosher already noted your reasoning is perfectly fine.
To see what significance the triangular part plays, recall that the $i$-th column of a matrix $[T]_B$ in a given ordered basis $B = (v_1, \dots, v_n)$ corresponds to the the image of the $i$-th basis vector under the linear transformation $T$.
If your matrix is for example upper-triangular, this yields that $v_1$ gets mapped to a scalar multiple of itself. Similarly $v_2$ gets mapped to a linear combination of $v_1$ and $v_2$ and so on until $v_n$, which is a linear combination of all basis vectors. 
In fact the existence of an ascending chain of invariant subspaces with dimensions $i = 1, \dots, n$ is a necessary and sufficient condition for a matrix to be triangulizable (see this question).
