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 '20 at 22:57
  • $\begingroup$ Yes i know @angryavian so the set can be linearly dependent $\endgroup$ Oct 25 '20 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 $$v = (\alpha_1 + \beta_1) s_1 + \cdots + (\alpha_n + \beta_n) s_n.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.