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Let $S$ be a finite set in a vector space $V$ with every $x$ in $V$ a unique representation as a linear combination of vectors in $S$. Show that $S$ is a basis for $V$.

My approach:

Since every $x$ in $V$ has a unique representation as a linear combination of vectors in $S$, the vector space $V$ is spanned by $S$.

How can we show the linear independence of $S$ given only this information?

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Here's a direct proof to complement the proof by contradiction above.

Let the vectors in $S$ be $v_1,\dots, v_n$.

Consider the zero vector: $0=0v_1+\cdots+0v_n$.

If we have a linear combination $a_1v_1+\cdots+a_nv_n=0$, then the fact that every vector, including the zero vector, has a unique representation implies that $a_1=0, a_2=0, \dots, a_n=0$.

Thus, the $v_i$ are linearly independent.

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Denote $S = \{s_1, \ldots, s_n\}$, it suffice to show $s_1, \ldots, s_n$ are linearly independent. Suppose $$\alpha_1 s_1 + \cdots + \alpha_n s_n = 0.$$

On the other hand, obviously, $$0 = 0s_1 + \cdots 0s_n$$

Since the representation of $0 \in V$ is unique, we conclude that $$\alpha_1 = \cdots = \alpha_n = 0$$ hence $s_1, \ldots, s_n$ are linearly independent.

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Let $S\equiv\{s_1,\ldots,s_n\}$ for some $n\in\mathbb N$. First, suppose that $n\geq 2$. If the vectors in $S$ were linearly dependent, then one would have $$\sum_{i=1}^n\alpha_is_i=0\tag{$\clubsuit$}$$ for some numbers $\alpha_1,\ldots,\alpha_n\in\mathbb R$ not all zero. Without loss of generality, suppose that $\alpha_1\neq 0$. Then, letting $\beta_i\equiv -\alpha_i/\alpha_1$ for $i\in\{2,\ldots,n\}$, one would have $$s_1=0\times s_1+\sum_{i=2}^n\beta_is_i.$$ However, one has also that $$s_1=s_1+\sum_{i=2}^n0\times s_i,$$ so the representation of $s_1$ is not unique.


Now, if $n=1$, then ($\clubsuit$) can hold only if $s_1=0$. But then $s_1=1\times s_1$ and $s_1=2\times s_1$, so the representation of $s_1$ is not unique in this case, either.

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Since every vector is in the span of $S$, in particular the zero vector lies in the span as well. Since $$\mathbf{0}=0s_1+0s_2+ \dotsb + 0s_k$$ Now by the uniqueness of the representation we know that the only solution to $$\sum c_is_i =\mathbf{0}$$ Is when all the coefficients are $0$. Thus the set is linearly independent.

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Say, $S=\{s_1, \ s_2, \ \dots \ , \ s_n \}$.

Definition of basis can be found here- wikipedia.

To show $S$ is a basis for $V$, we have to prove that-

  1. $S$ spans $V$.
  2. All the vectors in $S$ are linearly independent.

The condition-$1$ is already given to us because any vector $x \in V$ can be represented as a linear combination of the vectors in $S$. So, $S$ spans $V$.

So, we have to check for the condition-$2$.

Let assume that the vectors in $S$ are not linearly independent. So, we can write- $$ \sum_{i=1}^n \alpha_i s_i = 0, \ \alpha_i \neq 0 \text{ for at least one } i.\ \ \ \dots(1) $$

Again for any vector $x \in V$, we can write- $$ \sum_{i=1}^n \beta_i s_i = x. \ \ \ \dots(2) $$ Now adding $(1)$ and $(2)$, we will get- $$ \sum_{i=1}^n \alpha_i s_i + \sum_{i=1}^n \beta_i s_i = x, $$ $$ \sum_{i=1}^n (\alpha_i + \beta_i) s_i = x. \ \ \ \dots(3) $$

But we are given that every vector $x$ in $V$ has a unique representation as a linear combination of vectors in $S$. So, equation $(2)$ and $(3)$ both are correct representation of $x$ only when- $$\alpha_i + \beta_i= \beta_i, \ \forall \ i=1,2,\dots ,n$$ $$ \alpha_i = 0, \ \forall \ i=1,2,\dots ,n $$ which contradicts our assumption.

Hence we can say every vector in $S$ are linearly independent.

So, the condition-$2$ is also satisfied. Thus, it is proved that $S$ is a basis for $V$.

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