# Proof relation between linearly independent sets and unique scalars of the linear combination

I have the following statement to prove or disprove:

Let $$v$$ a vector of vectorial space $$V$$ different of $$0_{\text{V}}$$ and $$S= \{s_k\}_{k=1}^n$$ a subset of $$V$$ and $$v \in span(V)$$ prove ir disprove that:

The linear combination of $$S$$ that forms $$v$$ have unique scalars $$\iff$$ $$S$$ is linearly independent

My attempt was:

First, prove the $$\impliedby$$ direction:

If a vector $$v \neq 0_v$$ can be written as a linear combination of a linearly independent subset $$S$$ of a vectorial subspace $$V$$ with some scalars $$\{\alpha_k\}_{k=1}^n$$ implies that these scalars are unique.

Proof: If $$v$$ can be written as a linear combination of a linearly independent subset $$S=\{s_k\}_ {k=1}^n$$ this will be: $$v=\alpha_1s_1+...+a_ns_n$$.

Suppose scalars are not unique, so we also have: $$v=\beta_1s_1 + ...+\beta_ns_n$$

Equating $$\alpha_1s_1+...+a_ns_n=\beta_1s_1 + ...+\beta_ns_n \implies (\alpha_1-\beta_1)s_1+(\alpha_n+\beta_n)s_n=0$$ and since $$S$$ is linearly independent $$\alpha_k-\beta_k=0 \iff a_k = b_k$$ for every $$k \in [1, n]$$.

Second, disprove the $$\implies$$ direction:

Let $$S= \{s_k\}_{k=1}^n$$ and since $$v$$ is a linear combination of $$S$$ with unique scalars $$\{\alpha_k\}_{k=1}^n$$ we have:

$$v=\alpha_1s_1+\alpha_2s_2 + ... + \alpha_ns_n$$ for unique $$a_k$$.

Suppose $$S$$ is linearly dependent.This will implies that we have at least some vector $$s_k$$ in $$span(S - \{s_k\})$$, w.l.o.g lets say $$s_k$$ is $$s_1$$ so we have $$s_1 = \beta_2s_2 + ... +\beta_ns_n$$

And rewriting $$v$$ as $$v = \alpha_1(\beta_2s_2 + ... +\beta_ns_n) + \alpha_2s_2 + ... + \alpha_ns_n= 0\cdot s_1 + (\alpha_1\beta_2 + \alpha_2)s_2 + ... + (\alpha_1\beta_n+\alpha_n)s_n$$

and since scalar are unique we have that $$\alpha_1 = 0$$ and $$\alpha_1\beta_k=0$$ for every $$k$$, and since $$\alpha_1$$ is $$0$$ it holds for every $$\beta_k \in \mathbb{R}$$ and therefore the $$\implies$$ direction is false. This can be interpreted as: If we have unique scalars doesn't not means that is linearly independent because we can set all scalars of the "repeated"(that can be written as linear combination of the others) vectors of a linearly dependent set to $$zero$$ and the scalars will be unique.

Is my proof correct?

• In the proof of the $\implies$ direction, if for example $v = s_2$ (so that $\alpha_2 = 1$ and all other $\alpha_i$ are zero), you have not really obtained a contradiction. Oct 25, 2020 at 22:57
• Yes i know @angryavian so the set can be linearly dependent
– ESCM
Oct 25, 2020 at 23:16

To prove the $$\implies$$ direction by contraposition:
Suppose $$S$$ is linearly dependent. That means there exist constants $$\beta_1, \ldots, \beta_n$$ not all zero such that $$\beta_1 s_1 + \cdots + \beta_n s_n = 0$$. Then for representation of $$v$$ in terms of $$S$$ $$v = \alpha_1 s_1 + \cdots + \alpha_n s_n$$ you can obtain a different representation with $$v = (\alpha_1 + \beta_1) s_1 + \cdots + (\alpha_n + \beta_n) s_n.$$