Theorem about operators on complex vector spaces always having an upper triangular matrix

Question

I'm reading the following proof of a theorem in Linear Algebra done right

why is it obvious that $(T-\lambda I)u \in U$ from the definition of $U$? Surely if $u \in U$ it means there exists $v \in V$ such that $(T-\lambda I) v = u$. I don't see why it's then obvious that $(T - \lambda)u \in U$.

I'd really appreciate some clarity thanks!.

Answer 1

If $u \in U$, then $u \in V$ because $U \subset V$. Now if you apply $(T-\lambda I)$ to an element of $V$, the result is an element of the range, which is by definition $U.$

Answer 2

It should be clear that $(T - \lambda I)u \in range(T- \lambda I)$.

By definition $U=range(T- \lambda I) !$