# linear dependence and independence

Let $u_1$,$u_2$,$u_3$,$u_4$ be vectors in $R^2$ and $$u=\sum t_ju_j, 1\le j\le4$$:$t_j>0$ and $$\sum_{j=1}^4 t_j=1$$. Then Three vectors $v_1,v_2,v_3 \in R^2$ may be chosen from {$u_1,u_2,u_3,u_4$} such that $$u=\sum_{j=1}^3 s_jv_j, s_j\ge0,\sum_{j=1}^3 s_j=1$$.

Now since At max $2$ out of $4$ vectors in $R^2$ can be linearly independent, We can always write the vector as a sum of three vectors. But the condition on the sum of the coefficients is not satisfied. I am unable to proceed further.

Also is there any geometrical significance which I am missing?

• I see just one question (the last sentence). Was there another question before that? Commented Sep 29, 2013 at 2:34
• To find the solution Commented Sep 29, 2013 at 2:36

A key term you'll want to consider here is "convex hull". I see Calvin Lin has just given a brief answer, but it might be worth saying more what it means. I will use informal and intuitive language to get the idea across. (To avoid some slight awkwardness in language, I'll assume that no three points are collinear.)

It will help to draw a picture. Draw four points in the plane. Either we have that one of the points is inside the triangle formed by the other three, or the four points form a convex quadrilateral. In either case, the convex hull is the set of points inside or on this triangle or quadrilateral. (More formally, it is the smallest set $C$ that contains the four points and such that, if $x, y \in C$, then also $tx + (1-t)y \in C$ for any $t \in [0, 1]$. In other words, $C$ contains all line segments between points in $C$.)

It is easy to show that the convex hull of a set of points $S$ consists of all finite linear combinations of the form $t_1 x_1 + \ldots + t_n x_n$ where all the $x_i$ belong to $S$ and the $t_i$ satisfy the convexity conditions: $t_i \geq 0$ and $\sum_i t_i = 1$.

In the first triangle case, one point (say $u_4$) lies in the convex hull of the other three, so we have $u_4 = s_1 u_1 + s_2 u_2 + s_3 u_3$ where the $s_i$ satisfy the convexity conditions. If we then have

$$u = t_1 u_1 + t_2 u_2 + t_3 u_3 + t_4 u_4$$

then we may substitute in for $u_4$. We get a new linear combination

$$u = (t_1 + t_4 s_1)u_1 + (t_2 + t_4 s_2)u_2 + (t_3 + t_4 s_3)u_3$$

but these new coefficients satisfy the convexity conditions, as is trivially checked. That takes care of the triangle case.

The quadrilateral case is slightly trickier, but what we can do is divide the quadrilateral into the two triangles determined as the convex hulls of (in one case) $u_1, u_2, u_3$ and (in the other case) $u_2, u_3, u_4$. Now $u$, being inside the quadrilateral, is contained in one or the other triangle. That in turn means that $u$ will be a convex linear combination of either $u_1, u_2, u_3$ or of $u_2, u_3, u_4$.

That in any case is the geometric picture. The algebraic workload, which I will not carry out here, is in verifying the obvious geometric picture that the quadrilateral is a union of convex hulls of two sets of three points.

Your question is not really clear, but I believe that you are asking about Cartheodory's theorem.

It tells us that if $u$ is a vector in the convex hull of $u_1, u_2, u_3, u_4$, then it must be in the convex hull of 3 points.

Hence, we have $u= \sum s_i v_i$, where one of the $s_i$ is 0.
The reason why $\sum s_i = 1$ and $s_i \geq 0$ is because it is in the convex hull of the 3 points.