Development of Linear combinations of vectors Suppose that $v_1 \in \operatorname{span}\{v_1, \ldots,v_{i-1}, v_{i+1}, \ldots, v_n  \}$. Then we can write $v_i$ in two different ways as a linear combination of the vectors $v_1,\ldots,v_n$:
$$v_i = r_1v_1 + \cdots + r_{i-1}v_{i-1} +  r_{i+1}v_{i+1} + \cdots + r_{n}v_n = $$
$$r_1v_1 + \cdots + r_{i-1}v_{i-1} + 0v_i + r_{i+1}v_{i+1} + \cdots + r_{n}v_n $$
and
$$v_i = 0v_1 + \cdots + 0v_{i-1} + 1v_i + 0v_{i+1} + \cdots + 0v_n$$
Suppose that $\vec{0} =  r_1v_1 + \cdots + r_nv_n$. if $r_i \neq 0$ then:
$$r_iv_i = (-r_1)v_1 + \cdots + (-r_{i-1})v_{i-1} +  (-r_{i+1})v_{i+1} + \cdots + (-r_{n})v_n$$
Can someone please help me understand how the last development i made? I cannot understand why all the $r_1,\ldots,r_n$ changes sign, to me the most logical thing would be if they all were multiplied with $r_i$...
Thank you!

I think that last step is inherently related to the vector equation equals to the zero vector (after your argument "suppose that") The terms changed sign because you brought them over to the other side of the equal sign, solving it for RiVi
