Nonlinear matrix equation

Question

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 $-1$?

Answer 1

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.