Transform a vector of positive and negative values to sum up to 0

Question

Is there a transformation that produces a vector with sum $0$?

There are positive and negative values and the transformation does not need to be preserve the weights.
E.g.: $f(x_1, x_2, x_3) = (x_1', x_2', x_3')$ s.t. $$\sum_{i=1}^3x_i' = 0$$

Answer 1

I will go with centering where I subtract the mean from each $x_i$: $$f(x) = x_i - \bar{x}, \text{ with }\ \bar{x} = \frac{1}{3} \sum_i^3 x_i$$

This way the absolute distances between the $x_i$ is preserved and the $x'_i$ add up to 0.

Answer 2

Intrigued, i started looking for a linear map, $$ f \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1' \\ x_2' \\ x_3' \end{pmatrix} $$ Imposing that $x_1'+x_2'+x_3'=0$, we get that $$(a_1+b_1+c_1)x_1+(a_2+b_2+c_2)x_2+(a_3+b_3+c_3)x_3=0$$ So one example of a linear map that produces vectors with sum of the component $0$, is the map given by the matrix $$ \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} \quad \text{with} \quad \begin{cases} a_1+b_1+c_1=0\\ a_2+b_2+c_2=0\\ a_3+b_3+c_3=0 \end{cases} $$ and i think you can produce infinitely many examples.