# Explicit solution to a Rayleigh quotient equation

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.

• Since $M^2$ is symmetric, it is diagonalizable with orthogonal eigenvectors. Also the eigenvalues are positive. Let $(v_i)$ be a basis – Anthony Quas May 10 '16 at 5:44

## 2 Answers

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.

• Can't some eigenvalues of $M$ be zero? – Dirk May 10 '16 at 9:05
• Yes. I guess so. (editing answer). – Anthony Quas May 10 '16 at 14:20

You may already know this, but your Main Equation is equivalent to the system $$\begin{cases} (M^2-\lambda E)\mathbf{x} = \mathbf{1} & (*) \\ \mathbf{x}^T (M^2 - \lambda E) \mathbf{x} = 0 & (**) \end{cases}.$$ The first constraint, $\mathbf{x}^T \mathbf{1} = 0$, follows from the equations, while the second one, $\mathbf{x}^T \mathbf{x} = u$. in general does not. An obvious remark then is that your equation may not have any solutions (unless the matrix $M$ is somehow special), because the number of equations exceeds the number of unknowns.

Here's a way to simplify the above system, under the assumption that $\lambda$ is not an eigenvalue of $M^2$. Let $W(\lambda) = (M^2-\lambda E)^{-1}$, which by hypothesis is invertible. Then $\mathbf{x} = W(\lambda) \mathbf{1}$, by $(*)$, and $\mathbf{1}^T W(\lambda) \mathbf{1} = 0$, by $(**)$. The last scalar equation, when multiplied by $\det (M^2-\lambda E)$ becomes a polynomial in $\lambda$ of degree $n$ and in general gives $n$ solutions. The constraint equation becomes $\mathbf{1}^T W(\lambda)^2 \mathbf{1} = u$. In general, two independent equations for $\lambda$ will give you no solutions.

It is also possible to examine the case when $\lambda$ is an eigenvalue of $M^2$, with eigenvector $\mathbf{e}$. Then, $(*)$ has solutions only if $\mathbf{e}^T \mathbf{1} = 0$. Lets assume that is the case and denote by $W^+(\lambda)$ the pseudo-inverse of $(M^2-\lambda E)$. Then, the general solution of $(*)$ is $\mathbf{x} = W^+(\lambda) \mathbf{1} + \mu \mathbf{e}$, which converts $(**)$ and the constraint to the system $$\begin{cases} \mathbf{1}^T W^+(\lambda) \mathbf{1} = 0 \\ \mathbf{1}^T W^+(\lambda)^2 \mathbf{1} + \mu^2 \mathbf{e}^T \mathbf{e} = u \end{cases} .$$ Now you have two equations for two unknowns $(\lambda,\mu)$, so in general you can expect to find non-trivial solutions.

• Exactly, your statement is correct. On the contrary, the origin is that this consideration and constraint equations generate main equation eventually. Therefore, main equation must satisfy these conditions absolutely. Nevertheless, is it possible that we solve like $\mathbf{x}=\cdots$ ? – Hayabusananji May 10 '16 at 3:07
• $\mathbf{x} = W(\lambda) \mathbf{1}$ is already of that form. Of course, you also need to solve a separate equation for $\lambda$. But I doubt that the solution could be made much simpler, since you need to solve a generic $n$-degree polynomial equation for $\lambda$. – Igor Khavkine May 10 '16 at 4:30
• Yes, $\mathbf{x}=W(\lambda)\mathbf{1}$ is established form. But this equation forces numeric calculation to get $\lambda$ because of higher degree equation. On the other hand, main equation is also a non-linear simultaneous cubic equation. Theoretically, this is solvable. Just, this is my theme question. – Hayabusananji May 10 '16 at 5:23