I am having trouble understanding this proof that every operator has an upper-triangular matrix.
$\lambda=$ is an eigenvalue of $T$, for $T \in L(V)$ where $V$ is a vector space on $F^n$, they say :
suppose $U = \mathcal{R}(T-\lambda I)$, then $\dim(U)<\dim(V)$ because $U$ is not surjective.
$Tu=(T-\lambda I)u + \lambda u$ shows $T$ is invariant under $U$, because both terms are in U.
So above they are showing that since $T$ is an operator on $U$, $T$ has an upper triangular matrix with respect to some basis of $U$, $u_1,...u_m$ because in this is a claim they are trying to prove by induction.
Because $T$ has an upper triangular matrix, this means $Tu_j=(T|_u)(u_j)\in \operatorname{span}(u_1...u_j)$
I think I get all of this so far. Here is what I don't understand:
Extend $u_1...u_m$ to a basis of $V$, $u_1,...u_m,v_1,...,v_n$. For each $k$, $Tv_k=(T-\lambda I)v_k + \lambda v_k$
$(T-\lambda I)v_k \in U = \operatorname{span}(u_1,...,u_m) => Tv_k \in \operatorname{span}(u1,...,u_m,v_1,...,v_k)$
I guess $(T-\lambda I)v_k \in U $ because it equals zero, and zero is in $U?$ So why does it mean $Tv_k \in \operatorname{span}(u1,...,u_m,v_1,...,v_k)?$ $\lambda v_k$ is an eigenvector/eigenvalue, but is it true that there is only 1 independent eigenvector per eigenvalue? This one eigenvalue works on $v_1...v_k?$
By definition, $\mathrm{range}(A)=\{A.v ; v\in V\}$ for any $A \in \mathcal L(E)$. So $U=\mathrm{range}(T-\lambda I)=\{\left(T-\lambda I\right)v ; v\in V\}$. $v_k$ being a vector of $V$ like any other vectors of this vector space, you have $\left(T-\lambda I\right)v_k \in \mathrm{range}(T-\lambda I)=U$.
The rest follows.
The $v_k$ are not supposed in that proof to be related to the eigenvalue $\lambda$.