Finding the Zeros of a Matrix-Vector Equation

Here's another matrix algebra question, sorry if I'm coming at these incorrectly, but this kind of thing really isn't my forte :(

Lets say we have the equation:

$0 = -2 \mathbf{u}^{T} \mathbf{Z} \mathbf{v} + \mathbf{v}^{T} \left( \mathbf{Z} \mathbf{\Sigma} + \mathbf{\Sigma} \mathbf{Z} \right) \mathbf{v}$

The goal is to solve this equation for the matrix $\mathbf{Z}$. Right now I just get bogged down at

$\left(2 \mathbf{u}^{T} - \mathbf{v}^{T} \mathbf{\Sigma} \right)\mathbf{Z} = \mathbf{v}^{T} \mathbf{Z} \mathbf{\Sigma}$

and I'm stuck because you can't exactly take the inverse of a vector to move it to the other side.

The only piece other information to go on is that both $\mathbf{Z}$ and $\mathbf{\Sigma}$ are symmetric matrices, however they are not commutative, $\mathbf{Z \Sigma} \neq \mathbf{\Sigma Z}$

What is the correct analytical approach to this problem? Is there one?

Thanks!

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The original equation is just one single scalar equation, isn't it? So it's far from being able to uniquely determine $Z$ which has $N(N+1)/2$ entries. –  Luboš Motl Jun 2 '11 at 20:53

The equation as you've written it does not have a single solution. There are $\frac{1}{2}n(n+1)$ degrees of freedom in an $n\times n$ matrix, and your equation, being a scalar equation, only places one constraint on the matrix. There is a unique solution only when $\frac{1}{2}n(n+1) = 1$; and you can check that $n=1$ is the only positive solution to this equation.
Your second line of algebra is incorrect (unless we are in the case $n=1$). If you have two matrices $A,B$ that satisfy $Av=Bv$ for some vector $v$, you can't conclude that $A=B$.