How to proceed with proof that a finite set of vectors is a basis under a uniqueness condition. Claim: Suppose $B = \{b_{1},...,b_{k}\}$ is a set of vectors from a finite dimensional vector space $\mathbb{X}$. Show that if every $x \in \mathbb{X}$ can be written uniquely as 
$$x = \sum_{i=1}^{k} \phi_{i}b_{i}$$
i.e. a linear combination of the vectors in $B$, then $B$ must form a basis of $\mathbb{X}$.


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*Does non-uniqueness imply linear dependence?

*Can anyone give me a  push to complete the proof?
 A: If there are some $(a_{i}), (a'_{i})$ such that $a_{i} \neq a'_{i}$ for all $i$ and $\sum_{i}a_{i}b_{i} = \sum_{i}a'_{i}b_{i}$, then $\sum_{i}(a_{i}-a'_{i})b_{i} = 0$. So if $\{ b_{i} \}$ is linearly independent, then $a_{i} = a'_{i}$ for all $i$, a contradiction. So $\{ b_{i} \}$ is linearly dependent.
If every $x \in \mathbb{X}$ can be uniquely written as a linear combination of $\{b_{i}\}$, then the span of $\{ b_{i} \}$ is $\mathbb{X}$ by definition, and there is some unique $(a_{i})$ such that $0 = \sum_{i}a_{i}b_{i}$. If some $b_{i} = 0$, then $a_{i}$ admits more than one choice, a contradiction. So every $b_{i}$ is $\neq 0$. If $\{ b_{i} \}$ is linearly dependent, then some $b_{i}$ is such that $a_{i} \neq 0$, say $b_{1}$, so $0 = b_{1} + \frac{a_{2}}{a_{1}}b_{2} + \cdots + \frac{a_{k}}{a_{1}}b_{k}.$ But then $0 = -b_{1} -  \frac{a_{2}}{a_{1}}b_{2} - \cdots - \frac{a_{k}}{a_{1}}b_{k},$ so $0$ admits two representations, a contradiction. Hence $\{ b_{i} \}$ is linearly independent, and hence a basis for $\mathbb{X}$.
