3
$\begingroup$

Solve the following nonlinear equations for $v$ and $w$

$Avv^TAw=\lambda_1v+\lambda_2w$

$Aww^TAv=\lambda_1w+\lambda_2v$

$v^Tw=w^Tv=0$

$v^Tv=w^Tw=1$

where $\lambda_1, \lambda_2, \lambda_3$ are real. $A$ is a symmetric matrix.

How would you generalize to the case

$Avv^TAw+Bvv^TBw=\lambda_1v+\lambda_2w$

$Aww^TAv+Bww^TBv=\lambda_1w+\lambda_2v$

Where both A and B are symmetric? Would it help if they are also similar and each of them has exactly $n/2$ eigenvalues equal to $+1$ and $n/2$ eigenvalues equal to $-1$?

$\endgroup$
2
  • 2
    $\begingroup$ Where does this problem come from? Could you provide more context, also to convince us that this is no homework? Also, why are there a $v^Tv$ and a $w^Tw$ in the equations when you know that they are both equal to $1$? $\endgroup$ Oct 16, 2012 at 8:26
  • $\begingroup$ Hi, Sorry for the typo. It should be $vv^T$ and $ww^T$. The problem may look simple as if it is a homework, but it's not, and I think it's not trivial, at least to me. This is part of my attempt to minimize $\sum_{\sigma}|v^{\dagger}\sigma w|^2$ with Lagrange multiplier. Here {\sigma} are tensor products of some Pauli matrices, and $v$, $w$ are two orthonormal pure state. It is needed to prove another conjecture for my research project in quantum entanglement. I don't even know if it holds although random test suggest it does. $\endgroup$
    – Minh Tran
    Oct 18, 2012 at 7:10

1 Answer 1

4
$\begingroup$

First of all, note that $w^TAv=v^TAw$ is a scalar.

Here is an idea that should greatly simplify the equation:

Your equations say that $Aw$ and $Av$ are both contained in $U=\operatorname{span}(v,w)$, therefore $U$ is an invariant subspace of $A$. You can get all two-dimensional invariant subspaces by taking $U=\operatorname{span}(x_1,x_2)$, where $x_1$ and $x_2$ are eigenvectors of $A$ (proof: consider $A$ restricted to the subspace $U$; it is a symmetric linear operator, so it has two eigenvalues which are also eigenvalues of $A$).

So all solutions must be of the form $v=\alpha x_1 +\beta x_2$ and $w=\gamma x_1+\delta x_2$, where $x_1$ and $x_2$ are two eigenvectors of $A$. Making this ansatz the problem becomes a $2\times 2$ one in $\alpha,\beta,\gamma,\delta$ and should be easy to solve explicitly.

$\endgroup$
5
  • $\begingroup$ Thank you for your answer. It is not so clear to me why the two eigenvalues of $A$ restricted to $U=span{v,w}$ are also eigenvalues of $A$. Can you explain a bit more? $\endgroup$
    – Minh Tran
    Oct 18, 2012 at 10:24
  • $\begingroup$ It's the same operator, just seen on a smaller vector subspace. The relation $Au=\lambda u$ still holds, does not depend on the ambient space. $\endgroup$ Oct 18, 2012 at 10:40
  • $\begingroup$ Thank you for you great idea. I would like to ask you some more. How would you extend your idea to the case where you have two matrices instead of one as above? I have spent me some time to think about it but really I don't see a way. $\endgroup$
    – Minh Tran
    Oct 25, 2012 at 3:31
  • $\begingroup$ You'd better ask a new question with this second problem. Is the text correct? It is unpleasingly asymmetric. $\endgroup$ Oct 25, 2012 at 9:22
  • $\begingroup$ Thank you for your advice. I will start a new question. There was indeed some typo anyway. $\endgroup$
    – Minh Tran
    Oct 27, 2012 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.