# Rank of the matrix

Let ${\bf A}$ is a matrix that construct with coordinate of the $n$ distributed points in a two dimensional domain like follow: $${\bf A }=\begin{pmatrix} 1&x_1&y_1\\ 1&x_2&y_2\\ \vdots & \vdots & \vdots\\ 1&x_n&y_n \end{pmatrix}$$ Is there any condition for this kind of matrix and points that lead to the $rank({\bf A})=3$? If yes, How about the following matrix i.e. $rank({\bf A})=6$? $${\bf A }=\begin{pmatrix} 1&x_1&y_1&x_1^2&x_1y_1&y_1^2\\ 1&x_2&y_2&x_2^2&x_2y_2&y_2^2\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 1&x_n&y_n&x_n^2&x_n y_n&y_n^2 \end{pmatrix}$$

If there are just $3$ points and this points be vertices of a triangle, can we sure that $rank({\bf A})=3$?

• For the first: instead of thinking of $n$ points in $2-$dimensional space, think of $2$ points in $n-$dimensional space instead: then the matrix has rank $3$ if and only if the vector $(y_1,\dots,y_n)$ does not lie on the plane spanned by the vector $(1,\dots,1)$ and $(x_1,\dots,x_n)$. Of course here we assume that the vectors are pairwise independent, else it cannot have rank $2$. – b00n heT Jun 29 '16 at 20:06
• Isn't it sufficient to find three points that build a non-degenerated triangle? – Phil Jun 29 '16 at 20:08

$(x+y+1)^2=1+2x+2y+x^2+2xy+y^2$
The right hand side contains all the terms that appear in the rows of matrix $A$. So, if we find 6 pairs of $x$ and $y$ that satisfy $x+y+1=0$ and put them in matrix A, then we have a linear combination of columns that adds to the vector $0$. Therefore, we want to have at least one $(x,y)$ that does not satisfy $x+y+1=0$. Having the new condition and considering the fact that there are non-linear elements (for example, $x^2$) involved, I can see no problem with $A$ to be of rank $6$.