Proving UT and TU Have Same Eigenvalues

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Linear operators U and T on a vector space V can be shown to have the same eigenvalues for the products UT and TU. The proof involves analyzing the characteristic polynomial of both operators, demonstrating that they yield the same roots. By considering the eigenvalue equations for UT and TU, one can establish that if λ is an eigenvalue of UT, it must also be an eigenvalue of TU, and vice versa. The discussion emphasizes the importance of the commutation relation and properties of determinants in the proof. Ultimately, the conclusion is that UT and TU share identical eigenvalues, reinforcing the fundamental nature of linear transformations in vector spaces.
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Let U, T be linear operators on a vector space V. Prove that UT and TU have the same eigenvalues.

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