Hi jayshizwiz,

I think you haven't been given a reply until now because some of the parts of your question are hard to follow. You must ask your question so carefully that the reader doesn't have to guess what you mean: what do you mean by "a single solution"? and the hint surely should be: "Suppose that the above holds, but that $\displaystyle u_{1}, \ldots, u_{k}$ are linearly dependent", not merely "Let $\displaystyle u_{1}, \ldots, u_{k}$ be independent", because if we *define* these vectors to be independent, we cannot contradict ourselves! But if we *suppose* that they are dependent, then we can contradict the initial hypoethesis.

In this case, the proper statement of the problem should be: suppose that each $\displaystyle v\in\mathbb{R}^{n}$ has a *unique* representation as a linear multiple of the vectors $\displaystyle u_{1},\ldots,u_{k}.$ Then prove that $\displaystyle u_{1},\ldots,u_{k}$ are linearly independent.

**Proof **(I will get you started)

Suppose that the statement holds and that $\displaystyle u_{1}, \ldots, u_{k}$ are dependent, and let $\displaystyle v$ be written uniquely as

$\displaystyle v = \lambda_{1}u_{1}+\ldots+\lambda_{k}u_{k}$ **(1)**

By linear dependence, there exist scalars $\displaystyle \mu_{1},\ldots,\mu_{k}$ such that

$\displaystyle \mu_{1}u_{1}+\ldots+\mu_{k}u_{k} = 0$ **(2)**

with some $\displaystyle \mu_{j}\neq 0,\quad 1\leq j \leq k.$ Then rearranging **(2) **gives

$\displaystyle u_{j}=\frac{1}{\mu_{j}}\sum_{i\neq j}{\mu_{i}u_{i}}$

Now try to substitute this representation of $\displaystyle u_{j}$ into the original representation of $\displaystyle v$ given in **(1)**. Is this a different expression for $\displaystyle v$ as a linear combination of $\displaystyle u_{1},\ldots,u_{k}$?