Hey guys, I am stuck on this question so any help would be really nice. I have read it over and over and can't get it.

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- Oct 16th 2011, 07:51 AMshimaralinear operator and eigenvector
Hey guys, I am stuck on this question so any help would be really nice. I have read it over and over and can't get it.

Attachment 22591

Thanks - Oct 16th 2011, 08:13 AMFernandoRevillaRe: linear operator and eigenvector
(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. - Oct 16th 2011, 12:49 PMshimaraRe: linear operator and eigenvector
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? - Oct 16th 2011, 01:40 PMDevenoRe: linear operator and eigenvector
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. - Oct 16th 2011, 02:20 PMshimaraRe: linear operator and eigenvector
- Oct 16th 2011, 04:00 PMDevenoRe: linear operator and eigenvector
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).

- Oct 16th 2011, 04:22 PMshimaraRe: linear operator and eigenvector
- Oct 16th 2011, 04:31 PMDevenoRe: linear operator and eigenvector
that's just "one" eigenvalue. might not A and B^t have a bunch of different eigenvalues?

- Oct 16th 2011, 04:40 PMshimaraRe: linear operator and eigenvector
- Oct 16th 2011, 05:57 PMDevenoRe: linear operator and eigenvector
you want to consider all possible products λμ, where λ is an eigenvalue for A, and μ is an eigenvalue for B (B and B^t have the same eigenvalues).

- Oct 16th 2011, 06:01 PMshimaraRe: linear operator and eigenvector
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! - Oct 16th 2011, 06:17 PMDevenoRe: linear operator and eigenvector
i'm not sure what the set "R" is. it would be better to just say: where $\displaystyle \lambda_i$ is an eigenvalue of A, and $\displaystyle \mu_j$ is am eigenvalue for B.

- Oct 16th 2011, 06:19 PMshimaraRe: linear operator and eigenvector
oh ok, thanks again deveno.