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

Quote:

---

Originally Posted by shimara

Attachment 22591

---

(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.

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.

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).

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 $\displaystyle \lambda \mu$. 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)

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

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 $\displaystyle \lambda \mu$ is only one of many eigenvalues of T? This confuses me more.

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).

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 $\displaystyle \lambda_i \mu_j$ for $\displaystyle i,j \in R$.
I will make sure to go over this and fully comprehend all your wonderful help!

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.

oh ok, thanks again deveno.