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 $-1$?



Nonlinear matrix equation
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.


