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 is anonzeroeigenvector 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.