Let Z be a cycle of generalized eigenvectors of a linear operator T on V that corresponds to the eingenvalue lambda.Prove that span(Z) is a T-invariant subspace of V.

Yulia 2021-02-15 Answered
Let Z be a cycle of generalized eigenvectors of a linear operator T on V that corresponds to the eingenvalue lambda.Prove that span(Z) is a T-invariant subspace of V.
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Clelioo
Answered 2021-02-16 Author has 88 answers

Here, we are given that Z is the cycle of generalized eigenvectors of a linear operator T on V that corresponds to the eigenvalue λ.
Let T be a linear operator on a vector space , and let be an eigenvalue of . Suppose that
Z=v1,v2,,vn
Or if p is the smallest positive integer such that (TλI)pv=0,Z={(TλI)p1v,(TλI)p2v,,(TλI)v,v},
cycles of generalized eigenvectors of corresponding to .
Step 2
Now, we need to show that span(Z) is a T-invariant subspace of V.
Let us take a vector w from span(Z).
So, we can be written as linear combination of v1,v2,,vn.
i.e., w=α1v1+α2v2++αnvn where at least one αi0.
Now, it can be observed that if we can show that the vectors are T invariant then w is also T invariant.
T(vi)=T((TλI)kv), for some k (by the definition)
=(TλI+λI)((TλI)kv)
=(TλI)(TλI)kv+λI(TλI)kv
=(TλI)k+1v+λI(TλI)kv
=(TλI)k+1v+(TλI)k(λv)
And note that (TλI)k+1vand(TλI)k(λv) are also in span(Z), i.e., T-invariant.
Now, since w was taken randomly, span(Z) is a T-invariant subspace of V.

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