Generalized eigenspace invariant?

[mind0nmath]

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

 

[matt grime]

What does invariant mean? Try to prove it.

 

[mind0nmath]

got it. thanks.