Let $V$ be a vector space. Let $\boldsymbol v_{1}, \ldots, \boldsymbol v_{n}$ be [[linearly independent vectors]] of $V$. Let $x_{1}, \ldots, x_{n}$ and $y_{1}, \ldots, y_{n}$ be [[scalar|scalars]]. If the associated [[linear combination]] for each is equivalent, $ x_{1} \boldsymbol v_{1}+\cdots+x_{n} \boldsymbol v_{n}=y_{1} \boldsymbol v_{1}+\cdots+y_{n} \boldsymbol v_{n} $ Then the sequences of scalars would have to be identical $ x_{i}=y_{i} \text { for } i=1, \ldots, n $ #### Proof $\begin{aligned} x_{1} v_{1}-y_{1} v_{1}+\cdots+x_{n} v_{n}-y_{n} v_{n}&=0\\ \left(x_{1}-y_{1}\right) v_{1}+\cdots+\left(x_{n}-y_{n}\right) v_{n}&=0 \end{aligned} $ Which shouldn't happen for linearly independent vectors unless all of the differences are zero.