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Suppose that $A=\{y_1,...,y_r\}$ is a subset of a vector space $V$ and that every vector $x \in V$ can be expressed uniquely as a linear combination of the vectors of $A$. Show that $A$ forms a basis of the span of $A$.

I am not sure whether this result is true or not because generally in order to show a basis, we show that every element can be written uniquely in terms of the basis vectors but here information about only one is given. How to approach this anyway?

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    $\begingroup$ Suppose there were two ways to create $y\in V$ via vectors in $A$. Consider the difference between these ways. Add that to the linear combination that produces $x$. You get a different linear combination which produces $x$, so it is not uniquely expressed, a contradiction. $\endgroup$ – Don Thousand Aug 17 at 5:00
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    $\begingroup$ This is just asking you to show that A is linearly independent, right? If it isn't, then some vector is A is a linear combination of the others. $\endgroup$ – saulspatz Aug 17 at 5:01
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It is enough to prove that $A$ is linearly independent. Let $x =\sum_{k=1} ^{r} b_k y_k$ be the unique representation of $x$. Suppose $\sum_{k=1} ^{r} a_k y_k=0$. Then $x=x+0=\sum_{k=1} ^{r} b_k y_k+\sum_{k=1} ^{r} a_k y_k=\sum_{k=1} ^{r} (a_k+b_k) y_k$. Since the representation of $x$ as a linear combination of $y_i$'s is given to be unique it follows that $a_k+b_k=b_k$ for each $k$ so $a_k=0$ for each $k$. This completes the proof.

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It suffices to show that $A$ is linearly independent. To this end, let $\mathbb{F}$ be the field over which $V$ is a vector space. If $\mathsf{T}: \mathbb{F}^r \longrightarrow V$ is defined by $$ x = \begin{bmatrix} x_1 \\ \vdots\\ x_r \end{bmatrix} \longmapsto \sum_{k=1}^r x_k y_k, $$ then $\mathsf{T}$ is linear.

By assumption, there is a unique $x \in \mathbb{F}^r$ such that $\mathsf{T}(x)=\mathbf{0}_V$. But we know that $\mathsf{T}(\mathbf{0})=\mathbf{0}_V$. Thus, $x = \mathbf{0}$ which gives us the result.

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