**For 5 months!** I have been struggling to solve the following equations analytically without numeric method (ie, Newton method):



> 
> **Main equation:**
> 
> 
> $$
>  \biggl(M^2-\cfrac{\mathbf{x^{\text{T}}}M^2\mathbf{x}}{\mathbf{x^{\text{T}}}\mathbf{x}}E\biggr)\mathbf{x}=\mathbf{1}
> $$
> 
> 
> **Constraint equations:**
> 
> 
> $$
> \begin{cases}
>  \mathbf{x^{\text{T}}1}=0 \\
> \\
> \mathbf{x^{\text{T}}x}=u 
> \end{cases} $$
> 
> 
> where $\{M,E\}\in\mathbf{R}^{n \times n}$ and $\{\mathbf{1},\mathbf{x}\}\in\mathbf{R}^n$ are defined, then $M$ is an arbitrary symmetric matrix, $E$
>  is an identical matrix,
>  $\mathbf{1}$ is all one vector, $\mathbf{x}$ is a
>  variable vector and $u\in\mathbf{R}$ is a scalar.
>  Furthermore, as a knowledge, the below equation form is called [Rayleigh
>  quotient](https://en.wikipedia.org/wiki/Rayleigh_quotient) $R(M^2,\mathbf{x})$:
> 
> 
> $$R(M^2,\mathbf{x}):=\cfrac{\mathbf{x^{\text{T}}}M^2\mathbf{x}}{\mathbf{x^{\text{T}}}\mathbf{x}}$$
> 
> 
> 


Now, we attempt to estimate the $\mathbf{x}$. Does the analytic solution or method exist? My ability is shortage but, I guess that this problem has a beautiful solution. Also, main equation is a simultaneous cubic equation. Theoretically, this is solvable. Just, this is my theme question.



Explicit solution to a Rayleigh quotient equation
Since $M^2$ is symmetric, it is diagonalizable with orthogonal eigenvectors. Also the eigenvalues are positive. Let $(v\_i)$ be an orthonormal basis of eigenvectors for $M^2$ with eigenvalues $\lambda\_i\ge 0$. Now express $\mathbf 1$ in terms of the eigenvectors as $\sum b\_iv\_i$ and write $\mathbf x=\sum a\_iv\_i$. The equations reduce to $a\_i(\lambda\_i-K)=b\_i$, where $K$ is the Rayleigh quotient $(\sum \lambda\_i a\_i^2)/(\sum a\_i^2)$. 


I would approach this by defining for each $t$, $\alpha\_i(t)=b\_i/(\lambda\_i-t)$ and then $F(t)$ to be the Rayleigh quotient for the corresponding family of $(\alpha\_i(t))$, that is 
$$
F(t)=\left(\sum \lambda\_i \frac {b\_i^2}{(\lambda\_i-t)^2}\right)\Big /
\left(\sum \frac{b\_i^2}{(\lambda\_i-t)^2}\right).
$$
If for some $t$, $F(t)=t$, then you have a solution to your equations (namely $a\_i=\alpha\_i(t)$). Multiplying through by the denominators on both sides, you see that $F(t)$ is a continuous function of $t$, bounded between $\min\lambda\_i$ and $\max\lambda\_i$. Hence there is always a solution to the equation (by the intermediate value theorem).


But I don't think you should expect to be able to find an analytic solution, as this seems to be essentially the same as solving a typical degree $2n$ polynomial equation.


