# Show that $A$ forms a basis of the span of $A$.

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?

• 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. – Don Thousand Aug 17 at 5:00
• 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. – saulspatz Aug 17 at 5:01

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.
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.