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Federico Poloni
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Is it possible that we solve like $\mathbf{x}=\cdots$? Explicit solution to a Rayleigh quotient equation

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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 $R(M^2,\mathbf{x})$:

$$R(M^2,\mathbf{x}):=\cfrac{\mathbf{x^{\text{T}}}M^2\mathbf{x}}{\mathbf{x^{\text{T}}}\mathbf{x}}$$

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.

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 $R(M^2,\mathbf{x})$:

$$R(M^2,\mathbf{x}):=\cfrac{\mathbf{x^{\text{T}}}M^2\mathbf{x}}{\mathbf{x^{\text{T}}}\mathbf{x}}$$

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.

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 $R(M^2,\mathbf{x})$:

$$R(M^2,\mathbf{x}):=\cfrac{\mathbf{x^{\text{T}}}M^2\mathbf{x}}{\mathbf{x^{\text{T}}}\mathbf{x}}$$

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.

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Does the analytic solution or method exist Is it possible that we solve like $\mathbf{x}=\cdots$?

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