Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the matrix

\begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{matrix}

what effect does $({x_1},{y_1})$,$({x_2},{y_2})$,$({x_3},{y_3})$ being collinear have on the rank of the above matrix ?

share|cite|improve this question
I think the OP is talking of points, not vectors. Any two points in euclidean space are always collinear, but three need not... – DonAntonio Mar 14 '13 at 2:13
@DonAntonio Right. That was a very poor comment... – 1015 Mar 14 '13 at 2:36
up vote 2 down vote accepted

$$(x_i\,,\,y_i)\;,\,\,i=1,2,3\;,\;\;\text{are collinear}\;\;\iff \frac{y_3-y_2}{x_3-x_2}=\frac{y_3-y_1}{x_3-x_1}=\frac{y_2-y_1}{x_2-x_1}\iff$$

$$\iff (x_2-x_1)(y_3-y_2)=(y_2-y_1)(x_3-x_2)\\\;\;\;\;\;\;\;\;\;\;(x_3-x_1)(y_2-y_1)=(x_2-x_1)(y_3-y_1)\\\;\;\;\;\;\;\;\;\;\;(x_3-x_2)(y_2-y_1)=(x_2-x_1)(y_3-y_2)$$

If one of the denominators vanishes (and thus all of them), then $\,x_1=x_2=x_3\,$ , and we have two columns linearly dependent and the determinant in then zero, otherwise and using the above equality:

$$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix}=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=$$



so again zero. Do it slowly...!

share|cite|improve this answer

Elementary row/column operations yield $$ \left(\matrix{x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1}\right)\sim \left(\matrix{x_1&y_1&1\\x_2-x_1&y_2-y_1&0\\x_3-x_1&y_3-y_1&0}\right)\sim\left(\matrix{0&0&1\\x_2-x_1&y_2-y_1&0\\x_3-x_1&y_3-y_1&0}\right). $$ So the rank of your matrix is the same as the rank of the latter, which is $1$ plus the rank of the lower left $2\times 2$ block.

Now your three points are collinear, if and only if the two vectors $$ \vec{P_1P_2}=(x_2-x_1,y_2-y_1)\qquad\mbox{and}\qquad \vec{P_1P_3}=(x_3-x_1,y_3-y_1) $$ are collinear. And this is equivalent to $$ \det \left(\matrix{x_2-x_1&y_2-y_1\\x_3-x_1&y_3-y_1}\right)=0. $$ Finally, the latter is equivalent to the fact that the rank of this $2\times 2$ matrix is $0$ or $1$.

So the rank of your initial matrix is $1$ or $2$ when the three points are collinear. And it is $3$ when they are not collinear.

share|cite|improve this answer

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.