Prove that UT and TU has the same eigenvalues.
So far I have and I need to prove they are the same. Should I take the inverses?
First, suppose that is a nonzero eigenvector of UT, so for some nonzero vector v (and hence also ). Then . So Tv is an eigenvector of TU, with eigenvalue . Hence every nonzero eigenvector of UT is also an eigenvalue of TU, and similarly the other way round.
The case 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.