Hi I want to solve the following exercise:
Given $v_1,...,v_6$ in $K^2$, and x = $\sum_{l=1}^6 \frac{l}{21} v_l$,
show that there exist $\{i,j,k\} \in \{1,...,6\}$ and $\lambda_i,\lambda_j,\lambda_k \geq 0 $ and $\lambda_i+\lambda_j+\lambda_k = 1 $ with x = $v_i\lambda_i+v_j\lambda_j+v_k\lambda_k$.
My approach so far:
We know that each vector in $K^2$ can be written based on the basis vectors, i.e. $x = x_1*e_1 + x_2*e2$
This gives us three equations:
$\lambda_i+\lambda_j+\lambda_k = 1 $
$x_1 = v_{i1}\lambda_i+v_{j1}\lambda_j+v_{k1}\lambda_k$
$x_2 = v_{i2}\lambda_i+v_{j2}\lambda_j+v_{k2}\lambda_k$
A solution for this system of equations only exists if the determinant of the corresponding matrix is not equal 0.
As an example, I would chose $v_i = (1,0)^T$, $v_j = (0,1)^T$, $v_k=(1,1)^T$
and:
$\lambda_i = 1-x_1 $
$\lambda_j = 1-x_2$
$\lambda_k = x - 1$
Is my approach so far correct? Any feedback appreciated.