Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$x \in \mathbb{R}^{n}$ is a convex combination $C$ if there $p=p(x)\in \mathbb{N}$, $\lbrace \lambda_i\rbrace_{i=1}^{p} \subseteq [0,1]$ y $\lbrace x_i\rbrace_{i=1}^{p} \subseteq C$ such that $$ x=\sum_{i=1}^{p}\lambda_ix_i \ , \ \ \sum_{i=1}^{p}\lambda_i=1$$

For a triangle in $\mathbb{R}^{2}$, with vertices a, b, c. if x is a convex combination of {a,b,c} then $\lambda_1=\dfrac{\Vert x-a\Vert}{\Vert x-a\Vert +\Vert x-b\Vert +\Vert x-c\Vert }$? and so similarly for $\lambda_2,\lambda_3$??

share|improve this question
    
Otherwise you may give me a counterexample, please –  helmonio Dec 11 '12 at 19:11
    
When $x=a$ we want $\lambda_1 = 1$ but that is not true for your formula. –  GEdgar Dec 11 '12 at 19:23
    
Sure, but if $ C $ has dimension greater than one? –  helmonio Dec 11 '12 at 19:26
    
I was considering the question about a triangle in $\mathbb R^2$. –  GEdgar Dec 11 '12 at 19:29
add comment

1 Answer

up vote 3 down vote accepted

I am assuming that $a,b,c$ are not collinear.

Your answer cannot be correct. If $x=a$, your formula gives $\lambda_1 = 0$, when it should be $\lambda_1 = 1$. It is impossible for the formula to have any $\lambda_i = 1$.

However, it is straightforward to compute the multipliers:

You can show show that $A = \begin{bmatrix} 1 & 1 & 1 \\ a & b & c\end{bmatrix}$ in invertible, and that the multipliers $\lambda$ satisfy: $A \lambda = \begin{bmatrix} 1 \\ x\end{bmatrix}$, hence $\lambda = A^{-1} \begin{bmatrix} 1 \\ x\end{bmatrix} = A^{-1} \begin{bmatrix} 1 \\ 0\end{bmatrix}+A^{-1} \begin{bmatrix} 0 \\ x\end{bmatrix}$ . In particular, the multipliers are affine functions (ie, linear plus a constant) of $x$.

share|improve this answer
    
Thank you very much for the reply, you can generalize? –  helmonio Dec 11 '12 at 19:55
    
You are very welcome. –  copper.hat Dec 11 '12 at 19:55
    
Generalize in what way? If the points $a_k$ are affinely independent (ie, $\binom{1}{a_k}$ are linearly independent), then the same technique works for more points... –  copper.hat Dec 11 '12 at 20:26
add comment

Your Answer

 
discard

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.