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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 $-1$?

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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 $1$? –  Federico Poloni Oct 16 '12 at 8:26
    
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. –  Minh Tran Oct 18 '12 at 7:10
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1 Answer 1

up vote 4 down vote accepted

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.

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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? –  Minh Tran Oct 18 '12 at 10:24
    
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. –  Federico Poloni Oct 18 '12 at 10:40
    
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. –  Minh Tran Oct 25 '12 at 3:31
    
You'd better ask a new question with this second problem. Is the text correct? It is unpleasingly asymmetric. –  Federico Poloni Oct 25 '12 at 9:22
    
Thank you for your advice. I will start a new question. There was indeed some typo anyway. –  Minh Tran Oct 27 '12 at 8:04
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