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bluesky314

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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)