# Existence of linear combination

My book proves the following theorems:

Three distinct points $$A,B,C$$ are collinear if and only if there exist 3 numbers, $$\lambda_1,\lambda_2,\lambda_3$$, all different from zero, such that $$\lambda_1 \boldsymbol a +\lambda_2 \boldsymbol b +\lambda_3 \boldsymbol c =0,\space \space \space \space \space \lambda_1+\lambda_2+\lambda_3=0$$ If no three of the points of $$A,B,C,D$$ are collinear, they will be coplanar if and only if there exist 4 numbers, $$\lambda_1,\lambda_2,\lambda_3,\lambda_4$$, different from zero, such that $$\lambda_1 \boldsymbol a +\lambda_2 \boldsymbol b +\lambda_3 \boldsymbol c +\lambda_4 \boldsymbol d=0,\space \space \space \space \space \lambda_1+\lambda_2+\lambda_3+\lambda_4=0$$

A logical extension of this is to ask the following:

Given that no four points of A,B,C,D,E are coplanar, will there always exist 5 numbers $$\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5$$ sll different from zero, such that $$\lambda_1 \boldsymbol a +\lambda_2 \boldsymbol b +\lambda_3 \boldsymbol c +\lambda_4 \boldsymbol d+\lambda_5 \boldsymbol e=0,\space \space \space \space \space \lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=0$$

If we assume that these points lie in (3 dimensional) space, then this is true. Indeed, if we consider the four-dimensional vector space $$V=\mathbb R^4$$ composed of vectors $$\lambda=(\lambda_1,\lambda_2,\lambda_3,\lambda_4)$$ with an implied $$\lambda_5=-(\lambda_1+\lambda_2+\lambda_3+\lambda_4),$$ the map $$\phi:V\to \mathbb R^3$$ sending $$\lambda\to\lambda_1\boldsymbol a+\lambda_2\boldsymbol b+\lambda_3\boldsymbol c+\lambda_4\boldsymbol d+\lambda_5\boldsymbol e$$ must have kernel dimension at least $$1$$.
If, however, the points lie in $$4$$ or higher dimensions, then your statement does not hold, for the same reason the analogous statements do not hold in lower dimensions.