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?

  • $\begingroup$ 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. $\endgroup$
    – angryavian
    Oct 25, 2020 at 22:57
  • $\begingroup$ Yes i know @angryavian so the set can be linearly dependent $\endgroup$
    – ESCM
    Oct 25, 2020 at 23:16

1 Answer 1


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 $$v = (\alpha_1 + \beta_1) s_1 + \cdots + (\alpha_n + \beta_n) s_n.$$


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