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 · Accepted Answer

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 (T−λI)pv=0,Z={(T−λI)p−1v,(T−λI)p−2v,…,(T−λI)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 αi≠0. 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((T−λI)kv), for some k (by the definition) =(T−λI+λI)((T−λI)kv) =(T−λI)(T−λI)kv+λI(T−λI)kv =(T−λI)k+1v+λI(T−λI)kv =(T−λI)k+1v+(T−λI)k(λv) And note that (T−λI)k+1vand(T−λI)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.

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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