1
$\begingroup$

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 $rank({\bf A})=3$?

$\endgroup$
  • 2
    $\begingroup$ 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 $2$. $\endgroup$ – b00n heT Jun 29 '16 at 20:06
  • $\begingroup$ Isn't it sufficient to find three points that build a non-degenerated triangle? $\endgroup$ – Phil Jun 29 '16 at 20:08
1
$\begingroup$

If the triangle lies in a plane that goes through the origin, then the rank would be less than 3. More precisely, the rank would be 2. Otherwise the rank would be 3.

For the second matrix, we may consider the following equality

$(x+y+1)^2=1+2x+2y+x^2+2xy+y^2$

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 $6$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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