show that M can only be 0 if either X or Y is 0, since M = XY^t. (that is, the entry mij of M is (xi)(yj)).
but neither X nor Y can be 0, because they are ____________s.
λ can't be an eigenvalue of T, unless 1 is an eigenvaule of B^t. similar reasoning applies for μ.
T takes M and "combines A and B (via M)". you should expect the eigenvalues to "combine" in a similar way.
Fernando showed you that M = XY^t is an eigenvector for T, with eigenvalue λμ. the problem is already telling you the eigenvalues for A and B are given (the λ's and the μ's). so...what are the eigenvalues of T? (you're over-thinking this, the answer isn't that complicated).