Generalized eigenspace invariant?

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The discussion centers on whether the generalized eigenspace is invariant under a finite-dimensional linear operator T on complex numbers. The generalized eigenspace for an eigenvalue y is defined as the set of vectors v in V such that there exists a j≥1 where (T - yI)^j(v) = 0. Participants seek clarification on the meaning of "invariant" in this context and request a proof of the invariance property. The conversation emphasizes the need for a deeper understanding of these mathematical concepts. Overall, the inquiry aims to establish the relationship between generalized eigenspaces and linear operators.
mind0nmath
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Hey,
Is the generalized eigenspace invariant under the operator T? Let T be finite dimensional Linear operator on C(complex numbers).
My understanding of the Generalized Eigenspace for the eigenvalue y is:
"All v in V such that there exists a j>=1, (T-yIdenitity)^j (v) = 0." plus 0.
thanks
 
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What does invariant mean? Try to prove it.
 
got it. thanks.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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