Let $A\in\mathbb{R}^{n\times n}$ be a positive semidefinite matrix, $b\in\mathbb{R}^n$, $k>0$, and $g:\mathbb{R}^n\rightarrow\mathbb{R}$ be a positive function. Consider the system of nonlinear equations $$ (1) \quad Ax=-k\frac{x}{g(x)}-b. $$ Let $A^+$ be the Moore–Penrose pseudoinverse of $A$. For every $q\in\mathbb{R}^n$, consider the system of nonlinear equations $$ (2)\quad x=-kA^+\frac{x}{g(x)}-A^+b+(I-A^+A)q. $$ Find $q\in\mathbb{R}^n$ such that (1) is equivalent to (2).

My attempt

1) Observe that if $A$ is non-singular then (1) is equivalent to (2) for all $q\in\mathbb{R}^n$. Indeed, since $A$ is non-singular, $A^+=A^{-1}$ and so (2) is rewritten as
$$ x=-kA^{-1}\frac{x}{g(x)}-A^{-1}b+(I-A^{-1}A)q $$ which is equivalent to (1).

2) For each soluion $x$ of (1) there exists $q\in\mathbb{R}^n$ such that $x$ is the solution set of (2). Indeed, suppose that $x$ is a solution of (1). Then $$ Ax=-k\frac{x}{g(x)}-b. $$ Multiplying both sides of the above equation with $AA^+$ we obtain $$ AA^+Ax=-AA^+k\frac{x}{g(x)}-AA^+b. $$ Since $AA^+A=A$, the above equation is rewritten as $$ A\left(x+kA^+\frac{x}{g(x)}+A^+b\right)=0. $$ Therefore, there exits $q\in\mathbb{R}^n$ such that $$ x+kA^+\frac{x}{g(x)}+A^+b=(I-A^+A)q $$ or $$ x=-kA^+\frac{x}{g(x)}-A^+b+(I-A^+A)q. $$ Hence, $x$ is the solution of (2).

Thank you for all kind help and comments.

  • $\begingroup$ It is my pleasure if someone gives comments or hint in my question. $\endgroup$
    – Blind
    Jul 6, 2015 at 16:29

1 Answer 1

  1. We may assume that $A=D=diag((\lambda_i)_i)$ where $\lambda_i\geq 0$. Indeed $A=PDP^T$, where $P$ is orthogonal, and let $x=Py,b=Pc$. Then the equation (1) can be rewritten $Dy=-ky/g(Py)-c=-ky/h(y)-c$ where $h>0$.

In the sequel $A=D$ and, consequently, for every $i$, $x_i=\dfrac{-b_i}{\lambda_i+k/g(x)}$. Note that $D^+=diag((\mu_i)_i)$ where $\mu_i=1/\lambda_i$ if $\lambda_i\not=0$ and $\mu_i=0$ if $\lambda_i=0$.

  1. Consider the equation (2). When $\lambda_i\not=0$, we obtain the same relations as in equation (1) and $q_i$ is arbitrary. When $\lambda_i$=0, we obtain $x_i=q_i$, that is $q_i=\dfrac{-b_i}{\lambda_i+k/g(x)}$. In particular, $q_i$ depends on $x$, that implies that the equation (2) is very complicated!!
  • $\begingroup$ Thank for your comments and helping. Here, $A$ is not symmetric. How can we diagonalize $A$? $\endgroup$
    – Blind
    Jul 9, 2015 at 13:52

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