Assume that ST = TS. Prove that the operators S and T have a common eigenvector.
Let V be a complex (i.e. F = R) finite dimensional vector space. Let S, T be elements of
L(V ) (set of operators on V).
Assume that ST = TS.
Prove that the operators S and T have a common eigenvector.
these are the steps:
a) Explain why T has at least one eigenvalue. We will denote this eigenvalue by λ.
Let W be the set of all eigenvectors of T which have the eigenvalue λ, together with the zero vector, 0 element of V .
b) Prove that W is a subspace of V .
c) Prove that W is invariant under S.
d) Prove that there is an eigenvector for S which belongs to W.
e) Finish the proof of the claimed result.
(a) To find the eigenvalue of T, we need to solve the equation det(xI-T)=0. Because V is a complex
finite dimensional vector space, then det(xI-T)=0 always has a solution in complex field. Thus T
has at least one eigenvalue.
(b) To show that W is a subspace of V, we consider any two vectors u,v in W and any two complex ...
Eigenvectors and eigenvalues are investigated. The solution is detailed and well presented. The response was given a rating of "5/5" by the student who originally posted the question.