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In my Linear Algebra class we define Linear dependence as follows:

If $F$ is a field and $V$ is a vector space over the field $F$. The set $A = {\lbrace v_1,v_2,...,v_k \rbrace}$ where $v_1,v_2,...,v_k \in V $ is linearly dependent if $\exists$ a linear combination:

$a_1 v_1 + a_2 v_2 + ... +a_k v_k = 0$ where $ a_1, a_2 ,...,a_k\in F$

such that not all of the coefficients are zero

Then we are told to verify the following:

a set $A = {\lbrace v_1,v_2,...,v_k \rbrace}$ is linearly independent if and only if $a_1 v_1 + a_2 v_2 + ... +a_k v_k = 0 $ $\Rightarrow a_1=a_2=...=a_k =0$

Note: I know that some textbooks use this as a definition of linear independence. We have not defined linear independence at all.

In my opinion the proof is trivial in both directions, However I am concerned that what I have in mind is not rigorous enough.

My Idea/Attempt:

$(\Rightarrow)$ Suppose $A$ is linearly independent then by the converse of the definition above there can not exist a set of coefficients where not all of them are zero.

$(\Leftarrow)$ Suppose $\exists$ a set of coefficients $\lbrace a_1,a_2,...,a_k\rbrace $ such that $a_1v_1+a_2v_2+..a_kv_k =0 $ if some $a_i \neq 0 $ then of course $A$ is linearly dependent

What do you think of my idea?

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  • $\begingroup$ I suggest you post what you have in mind, and then we can comment on it. $\endgroup$ – Calle Feb 22 '14 at 12:57
  • $\begingroup$ I imagine that you are supposed to take the definition that a set of vectors is linearly independent if it is not linearly dependent? $\endgroup$ – hmmmm Feb 22 '14 at 13:04
  • $\begingroup$ @hmmmm I think that is the idea. $\endgroup$ – Mathman Feb 22 '14 at 13:07
  • $\begingroup$ @Calle What do you think? $\endgroup$ – Mathman Feb 22 '14 at 13:26
  • $\begingroup$ Well, "the converse of the definition" is a slightly odd phrase (a definition is not an implication). Just say "if not all of the coefficients are zero, this would be a contradiction". It would probably also be good to explicitly state that you take "linearly independent" to mean "not linearly dependent". The $\Rightarrow$ direction could be made clearer by stating that you prove the contrapositive (there exists $a_i$, not all zero, hence linearly dependent). $\endgroup$ – Calle Feb 22 '14 at 13:35
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$$\textrm{$A$ is linearly independent}$$ $$ \textrm{if and only if}$$ $$ \textrm{$A$ is NOT linearly dependent}$$ $$\textrm{if and only if}$$ $$\neg(\exists a_1\ldots\exists a_k(a_i\neq 0 \textrm{ some i } \wedge a_1v_1+\ldots +a_kv_k=0))$$ $$\textrm{if and only if}$$ $$\forall a_1\ldots a_k ( \forall i (a_i=0) \vee a_1v_1+\ldots +a_kv_k\neq 0))$$ $$\textrm{ if and only if}$$ $$\forall a_1\ldots a_k (a_1v_1+\ldots +a_kv_k=0\rightarrow \forall i(a_i=0))$$

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  • $\begingroup$ Beautiful. Simple and elegant. $\endgroup$ – Mathman Feb 22 '14 at 13:40

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