(a) By hypothesis $\displaystyle AX=\lambda X$ and $\displaystyle B^tY=\mu Y$ . Choose $\displaystyle M=XY^t$ then,
$\displaystyle T(M)=AMB=A(XY^t)B=(AX)(B^tY)^t=$
$\displaystyle (\lambda X)(\mu Y)=\lambda \mu (XY^t)=\lambda \mu M$
As $\displaystyle M\neq 0$ (why?) then, $\displaystyle M$ is an eigenvector of $\displaystyle T$ .
(b) Try it.
Thanks fernando, for part a I don't understand why $\displaystyle M \ne 0$.
For part b, aren't the eigenvalues of A and B, $\displaystyle \lambda$ and $\displaystyle \mu$ respectively?
And these are also the eigenvalues of T?
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).
omg, you were right, it is really simple and I was way over thinking it. Thanks deveno.
So I would write it like:
the eigenvalues of T are $\displaystyle \lambda_i \mu_j$ for $\displaystyle i,j \in R$.
I will make sure to go over this and fully comprehend all your wonderful help!