linear combination of linearly independent vectors is unique.
proof: assume, there exist 2 different linear combinations of linearly independent vectors $v_1$ and $v_2$ for generating vector $b$.
$c_1v_1 + c_2v_2 = b$ ..................... (1)
$d_1v_1 + d_2v_2 = b$ ...................... (2)
from equation (1) and (2),
$c_1v_1 + c_2v_2 = d_1v_1 + d_2v_2$
$c_1v_1 - d_1v_1 + c_2v_2 - d_2v_2 = 0$
$(c_1 - d_1)v_1 + (c_2 - d_2)v_2 = 0$
take $t_1 = c_1 - d_1$ and $t_2 = c_2 - d_2$
$t_1v_1 + t_2v_2 = 0$
if any one of them $t_1$ or $t_2$ is non-zero then, we can express them like this (assuming only $t_1$ is non-zero and it works also if both are non-zero),
$v_1 = -1/t_1.(t_2v_2) = -t_2/t_1.(v_2)$ ............................(3)
and this is a contradiction as from equation (3) we proved $v_1$ and $v_2$ are linearly dependent.
that means $t_1 = 0$ and $t_2 = 0$.
putting values of $t_1$ and $t_2$.
$c_1 - d_1 = 0 \Rightarrow c_1 = d_1$
$c_2 - d_2 = 0 \Rightarrow c_2 = d_2$
that means if $v_1$ and $v_2$ are linearly independent, there exist unique $c_1, c_2$ such that $c_1v_1 + c_2v_2 = b$.
hence, we can say that linear combination of linearly independent vectors is unique.
