Prove that UT and TU has the same eigenvalues.
So far I have $\displaystyle UT(v)= \lambda _1 v $ and $\displaystyle 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 $\displaystyle \lambda$ is a nonzero eigenvector of UT, so $\displaystyle UTv=\lambda v$ for some nonzero vector v (and hence also $\displaystyle Tv\ne0$). Then $\displaystyle TU(Tv) = T(UTv) = \lambda Tv$. So Tv is an eigenvector of TU, with eigenvalue $\displaystyle \lambda$. Hence every nonzero eigenvector of UT is also an eigenvalue of TU, and similarly the other way round.
The case $\displaystyle \lambda=0$ 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.