Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This problem arises from my research in computer vision, specifically projective homography:

I have $n$ unknown variables, represented by an $n\times 1$ vector $\mathbf{x}$. There is a system of $n$ equations of the form $\mathbf{x}^{T}\mathbf{A_i}\mathbf{x} + \mathbf{b_i}^{T}\mathbf{x} + c_i = 0$, where $\mathbf{A_i}$ is an $n\times n$ symmetric matrix, $\mathbf{b_i}$ is an $n\times 1$ vector and $c_i$ is a scalar for $i=1\ldots n$, that $\mathbf{x}$ has to satisfy.

It is also known that every entry of the solution $\mathbf{x}$ is small, i.e. comparatively smaller than their respective coefficients.

Is there a closed-form or iterative method to solve for $\mathbf{x}$?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Newton's method converges fast, when it converges. In Newton's method, you start from an initial estimate $\bf x$, and solve a linear equation for $\bf h$ to get the next iterate $\bf x+h$. In this case this seems to give $$ {\bf x}^T({\bf A}_i{\bf x}+{\bf b}_i)+(2{\bf A}_i{\bf x}+{\bf b}_i)^T{\bf h}+c_i=0 $$ for $i=1$ to $n$. It is obtained by dropping the term which is quadratic in ${\bf h}$.

share|improve this answer

sometimes I think the most difficult part is not the method itself, but the guess of initialized value. Does anyone know how to give a better initialized value?

share|improve this answer
This is not an answer to the question. If you have a new question, please use the "Ask Question" link near the top right of the page. –  robjohn Jan 20 '13 at 16:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.