I Question about an eigenvalue problem: range space

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 𝐡 for 𝑉 such that the matrix of 𝑇 with respect to the basis 𝐡 is upper triangular.
In the proof, he defines U=range(T-πœ†I) (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-πœ†I)u + Tu and both (T-πœ†I)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-πœ†I)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

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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##.
 

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