TITLE: How to get the upper triangular form for any linear map?
QUESTION [5 upvotes]: Let $T:V \to V$ be linear. $V$ is a complex vector space of dimension $k$. Then there exists a basis so that the matrix generated by $T$ under that basis is upper triangular. The proof is by induction on $k$. 
But how to generate such a basis. The first step is that because $v$ is a complex vector space, so there exists a non-zero vector $v$ such that $Tv=\lambda v$ for some $\lambda \in \Bbb F$. So, take $e_1=v$ because then $T(e_1)=\lambda e_1$. So, in the first column of the matrix the rows 2 to n will be zero. Now if I extend $e_1$ to any basis of $v$ that basis may not be a basis for which the matrix is upper triangular. Then how to construct it ?

REPLY [2 votes]: Hint: Apply the induction hypothesis to the quotient vector space $V'=V/\text{span}(e_1)$ and the induced linear map $\overline{T}:V'\to V'$. 
You should check (a) the dimension of $V'$ is strictly less than the dimension of $V$ (b) $\overline{T}:V'\to V'$ is a well-defined linear map and (c) an application of the induction hypothesis to $\overline{T}:V'\to V'$ results in an upper triangular matrix representation of $T:V\to V$ (you should also be precise about the particular basis with respect to which $T$ is upper triangular).
I hope this helps!