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Math Help - Composition linear operator eigenvalues

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    Composition linear operator eigenvalues

    Prove that UT and TU has the same eigenvalues.

    So far I have  UT(v)= \lambda _1 v and  TU(v)= \lambda _2 v I need to prove they are the same. Should I take the inverses?
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    Quote Originally Posted by tttcomrader View Post
    Prove that UT and TU has the same eigenvalues.

    So far I have  UT(v)= \lambda _1 v and  TU(v)= \lambda _2 v I need to prove they are the same. Should I take the inverses?
    You can't assume that it's the same v in both those equations. (The eigenvalues may be the same, but that doesn't imply that the eigenvectors are the same.)

    First, suppose that \lambda is a nonzero eigenvector of UT, so UTv=\lambda v for some nonzero vector v (and hence also Tv\ne0). Then TU(Tv) = T(UTv) = \lambda Tv. So Tv is an eigenvector of TU, with eigenvalue \lambda. Hence every nonzero eigenvector of UT is also an eigenvalue of TU, and similarly the other way round.

    The case \lambda=0 has to be handled differently. Notice that if 0 is an eigenvalue of a matrix then the matrix is not invertible (so its determinant is 0), and conversely. But if det(TU) = 0 then also det(UT)=0.
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