# Linear Convex Decomposition

Let $X=\{x_1,x_2,x_3,x_4\}$ be the four vertexes of a square in $\mathbb{R}^2$, and $H$ the convex hull of $X$.

Then, in general, the points $y\in H$ that cannot be written in a unique way as a convex combination of $\{x_1,x_2,x_3,x_4\}$, i.e. $$\exists y:\quad y=\sum_{i=1}^{4}\lambda_ix_i=\sum_{i=1}^{4}\lambda'_ix_i, \text{ with}\quad \lambda_i,\lambda'_i\geq0, \sum_i\lambda_i=\sum_i\lambda'_i=1,$$ but $\lambda_i\neq\lambda'_i$.

Question:

Is there a method make a choice among all possible convex coordinates of every $y\in H$ in a linear fashion?

More specifically, for every $y\in H$, give a method to obtain some convex coordinates $y=\sum_i\bar{\lambda}_i(y)x_i$ among all possibles, such that if $y=(1-b)y_1+by_2$, for $b\in[0,1]$ and whatever $y_1,y_2\in H$, we have that $$\bar{\lambda}_i(y)=(1-b)\bar{\lambda}_i(y_1)+b\bar{\lambda}_i(y_2)$$

No, this is not possible. Note that $$\bar\lambda_1(x_1) = 1$$ and $$\bar\lambda_1(x_2) = \bar\lambda_1(x_3) = \bar\lambda_1(x_4) = 0.$$ This cannot be achieved by a linear $\lambda_1$.