Undergrad Question about an eigenvalue problem: range space

[bluesky314]

A theorem from Axler's Linear Algebra Done Right says that if 𝑇 is a linear operator on a complex finite dimensional vector space 𝑉, then there exists a basis 𝐵 for 𝑉 such that the matrix of 𝑇 with respect to the basis 𝐵 is upper triangular.
In the proof, he defines U=range(T-𝜆I) (as we have proved atleast one eigenvalue must exist)and says that this is invariant under T.
First I want understand what the range space is here. Suppose we are in 2D and I choose some basis consisting on an eigenvector(ev) and another free vector(v): ( ev, v) Now this v can be anything not necessarily an eigenvector. So there are many possibilities. Are all contained in U? U can have dimension 1 but all these possibilities are not linearly independent so what exactly is U? Is it a set of all vectors not in the span of ev? Does this not conflict with U having dimension 1?
He then proves this by saying if u belongs to U then Tu=(T-𝜆I)u + Tu and both (T-𝜆I)u and Tu belong to U. But if I had chosen a non-eigenvalue for my basis, then Tu does not scale it so it is mapped to some other span. Therefore I'm guessing U is a set.
Any other comments about (T-𝜆I)v would be appreciated. How to think of it in the context of our original T? (I know the null space is the eigenvector)

 

[Stephen Tashi]

But if I had chosen a non-eigenvalue for my basis, then Tu does not scale it so it is mapped to some other span.

The range of a linear operator $L$ from a vector space $V$ to a vector space $W$ is the set $\{u\in W:$ such that $Lx =u$ has a solution $x\in V\}$. The range is also a subspace of $W$.

Simplifying the right hand side of the equation $Tu = (T - \lambda)u + \lambda u$ shows it is equivalent to the identity $Tu = Tu$.

The proof does not assume that $u$ is an eigenvector of $T$ or of $(T - \lambda)$, so your objection is unclear. The assumption $u \in U$ is equivalent to assuming the existence of $x$ such that $(T-\lambda)x = u$. I don't know whether you could that expression to formulate a different proof that $U$ is invariant under $T$.