# linear operator and eigenvector

• Oct 16th 2011, 08:51 AM
shimara
linear 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, 09:13 AM
FernandoRevilla
Re: linear operator and eigenvector
Quote:

Originally Posted by shimara

(a) By hypothesis $AX=\lambda X$ and $B^tY=\mu Y$ . Choose $M=XY^t$ then,

$T(M)=AMB=A(XY^t)B=(AX)(B^tY)^t=$

$(\lambda X)(\mu Y)=\lambda \mu (XY^t)=\lambda \mu M$

As $M\neq 0$ (why?) then, $M$ is an eigenvector of $T$ .

(b) Try it.
• Oct 16th 2011, 01:49 PM
shimara
Re: linear operator and eigenvector
Thanks fernando, for part a I don't understand why $M \ne 0$.

For part b, aren't the eigenvalues of A and B, $\lambda$ and $\mu$ respectively?
And these are also the eigenvalues of T?
• Oct 16th 2011, 02:40 PM
Deveno
Re: 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, 03:20 PM
shimara
Re: linear operator and eigenvector
Thanks deveno, X and Y are non-zero because they are eigenvectors.

Quote:

Originally Posted by Deveno
T takes M and "combines A and B (via M)". you should expect the eigenvalues to "combine" in a similar way.

Sorry deveno, I don't really get this part. I know the eigenvalues will be combined like you say, but I don't understand how to find them.
• Oct 16th 2011, 05:00 PM
Deveno
Re: 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, 05:22 PM
shimara
Re: linear operator and eigenvector
Quote:

Originally Posted by Deveno
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).

Yeah, I am definitely over-thinking this, all I can stupidly think of at this time of day is $\lambda \mu$. Can you just put me out of my misery and tell me the answer?

And thanks for being patient with my stupidity at the moment. (Doh)
• Oct 16th 2011, 05:31 PM
Deveno
Re: linear operator and eigenvector
that's just "one" eigenvalue. might not A and B^t have a bunch of different eigenvalues?
• Oct 16th 2011, 05:40 PM
shimara
Re: linear operator and eigenvector
Quote:

Originally Posted by Deveno
that's just "one" eigenvalue. might not A and B^t have a bunch of different eigenvalues?

Because A and B are general complex matrices they can have many different eigenvalues, so $\lambda \mu$ is only one of many eigenvalues of T? This confuses me more.
• Oct 16th 2011, 06:57 PM
Deveno
Re: 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, 07:01 PM
shimara
Re: linear operator and eigenvector
Quote:

Originally Posted by Deveno
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).

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 $\lambda_i \mu_j$ for $i,j \in R$.
I will make sure to go over this and fully comprehend all your wonderful help!
• Oct 16th 2011, 07:17 PM
Deveno
Re: linear operator and eigenvector
i'm not sure what the set "R" is. it would be better to just say: where $\lambda_i$ is an eigenvalue of A, and $\mu_j$ is am eigenvalue for B.
• Oct 16th 2011, 07:19 PM
shimara
Re: linear operator and eigenvector
oh ok, thanks again deveno.