For 5 months! I have been struggling to solve the following equations analytically without numeric method (i,e, 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 $R(M^2,\mathbf{x})$:
$$R(M^2,\mathbf{x}):=\cfrac{\mathbf{x^{\text{T}}}M^2\mathbf{x}}{\mathbf{x^{\text{T}}}\mathbf{x}}$$
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.
Furthermore, same question is already asked on math overflow. Then answerers provided worthful information which may be solution to clue.
You were already told the answer on MO. Let $M^2$ have eigenvalues $\lambda_j$ with normalized eigenvectors $v_j$, and $v_j^T 1 = d_j$. Let $x = \sum_j c_j v_j$, where you want $$ \eqalign{x^T x = \sum_j c_j^2 &= 1 \cr x^T 1 = \sum_j c_j d_j &= 0\cr}$$ and $r = x^T M^2 x = \sum_j c_j^2 \lambda_j $. Then you want $ (M^2 - r E) x = 1$, which says $$ c_j (\lambda_j - r) = d_j$$ Thus (ignoring the possibility that some $\lambda_j = r$) $$c_j = \dfrac{d_j}{\lambda_j - r}$$ which means you need
$$\eqalign{\sum_j \dfrac{d_j^2}{(\lambda_j - r)^2} &= 1\cr \sum_j \dfrac{d_j^2}{\lambda_j - r} &= 0 \cr}$$ Note that the equation for $r$: $$\sum_j \dfrac{\lambda_j d_j^2}{(\lambda_j - r)^2} = r $$ follows from those two.
Since there are two (apparently independent) algebraic equations for the single variable $r$, there will usually be no solution. When there is, it will be the root of a complicated polynomial which in general has no expression in radicals.