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I use following definition:

let be $v:\Bbb{N} \to T$ with $T\subseteq V $ and $V $ a $K-$vector space: $$ (v_i)_{i \in \Bbb{N}} \text{ is lin. ind. if } \\ \forall \alpha \in K^\Bbb{N}: \sum_{i \in \Bbb{N}} \alpha_i \, v_i=0 \wedge \exists m \in \Bbb{N}:\forall n>m: \alpha_n=0 \to \forall i \in \Bbb{N}: \alpha_i=0$$

Now I need a definition not for family of vectors but for subsets of $V $ . I thinked $T \subseteq V$ is lin ind if $\forall x\in T: x \notin $ but I have following problem:

the family of vectors $((1,2), (1,2)) $ ist not lin ind, but the set $\{(1,2)\} $ ist lin ind...

Are there therefore two definitions of lin ind, one for families and one for sets?!

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    You are wrestling with a definition that has almost no English words. Try this definition and see if you can solve: a collection of vectors (vectors could be repeated, so not calling this collection a set) $v_1,v_2,\ldots, v_n$ is linearly independent if there is NO way, other than all 0, to find scalars $a_i$ such that the sum $a_1v_1+a_2v_2+\cdots a_nv_n$ will be zero vector. – 2017-02-11
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    What do you mean with "collection"? Why do you use a finite "collection"? In my question I have a countable subset!!! – 2017-02-11
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    What is a collection? Well here is an Illustration, instead of a definition: . a real polynomial of degree $n$ has *only a collection* of $n$ (possibly complex) roots, not always set of $n$ roots. About infinite sets, in algebra there is no concept of infinite sum, or convergence. So infinite collection of vectors is said to be linearly independent iff every *finite subcollection* is so. – 2017-02-11
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    @P Van, you are wrestling with definitions that have only English words! Collection is set, family is not set, a family is a function ( https://en.m.wikipedia.org/wiki/Indexed_family)... a polynomial ist function and see https://en.m.wikipedia.org/wiki/Multiplicity_(mathematics)#Multiplicity_of_a_root_of_a_polynomial and https://en.wikipedia.org/wiki/Linear_independence#Infinite_dimensions – 2017-02-11
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    @Marios, your definition for sets ist good!!! In addition, [$(v_i) $ is lin ind $\to$ $\{v_i\} $ ist lin ind] is true, but [$\{v_i\} $ is lin ind $\to$ $(v_i)$ is lin ind] is generally false and you have an example ;) . The question is interesting because for example $<(v_i)>=<\{v_i|i \in \Bbb{N}\}> $ is true!! – 2017-02-11

1 Answers 1

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By definition singleton sets are always linearly independent.

Let $\{\alpha \neq 0\}$ be a singleton set. $a_1 \alpha = 0 \implies a_1 = 0 $.

Now $a_1(1,2) + a_2 (1,2) = (0,0) \implies (a_1 +a_2, 2a_1+2a_2) = (0,0) \implies a_1 = -a_2 $. Thus the family of vectors $((1,2), (1,2)) $ is not linearly independent.

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    What definition for singleton? – 2017-02-11
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    See my edit.... I understand your answer but it is not my question – 2017-02-11
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    @user8795, you use two definitions for linear independence... – 2017-02-11