In my linear algebra course we defined a free / linearly independent family like this :

A family of vector of a $K$-vector space $\left(A_i\right)_{i\in I}$ is free / linearly independent iff $$\forall J \subset I \text{ such that }\operatorname{Card}(J) \in \mathbb{N}, \forall j \in J, \forall k \in J, \forall e_k \in A_j,e_k\notin \operatorname{Span}\left(A_j\setminus e_k\right),$$ with $\setminus$ defining the setminus (removing an element from a set)

Let us prove this is equivalent to the commonly used definition :

$$\forall J \subset I \text{ such that }\operatorname{Card}(J) \in \mathbb{N}\forall j \in J, \forall e_j \in (A_j)_{j \in J} \\ \sum_{j\in J} \lambda_j e_j=0 \implies \forall k \in J, \lambda_k=0$$

where $(\lambda_i)_{i \in I}$ is a family of scalars of $K$

Proof :

Let $J$ be a finite subfamily of $I$

Under the standard definition :

if $(A_i)_{i \in I}$ is not free, we have : $$ \forall j \in \!J,\forall e_j \in A_{j}, \sum_{j \in J} \lambda_j e_j=0 \wedge \exists k \in J, \lambda_k\neq 0 \\ \Leftrightarrow \exists k\in J \forall j \in J \forall e_j \in A_j,\sum_\limits{\substack{j \in J \\j \neq k }}\lambda_je_j+\lambda_ke_k=0\,\wedge \lambda_k\neq 0 \\ \Leftrightarrow\exists k\in J \forall j \in J \forall e_j \in A_j,\sum_\limits{\substack{j \in J \\j \neq k }}\lambda_je_j = -\lambda_k e_k \,\wedge \lambda_k\neq0\\ \exists k\in J \forall j \in J \forall e_j \in A_j,\sum_\limits{\substack{j \in J \\j \neq k }}-\frac{\lambda_j}{\lambda_k}e_j=e_k \, \wedge \lambda_k\neq 0 \\ \Leftrightarrow \exists k\in J \forall j \in J \forall e_j \in A_j, e_k \in \operatorname{Span}(A_j \setminus e_k) $$ I wonder if there is any way to prove the following theorem using this definition :

Let $F$ a free family of a vector space $S$, and $v$ a vector of $S$, then $ F\cup \lbrace v\rbrace$ is free iff $v \notin \operatorname{Span}(F)$


Let us rewrite what it means for $F\cup \lbrace v\rbrace$ to be free :

$F\cup \lbrace v\rbrace$ is free iff $$ \forall e \in F\cup \lbrace v\rbrace, e \notin\operatorname{Span}(F\cup \left\lbrace v\right\rbrace \setminus \lbrace e\rbrace )$$

So, in particular, for $v$ which is trivially an element of $F\cup \lbrace v\rbrace$ : $$v \notin\operatorname{Span}(F\cup \left\lbrace v\right\rbrace \setminus \lbrace v\rbrace )$$

but we have $(F\cup \left\lbrace v\right\rbrace) \setminus \lbrace v\rbrace=F$

so that $$v \notin \operatorname{Span}(F)$$

Now we will supose that $v$ belongs to this set, and prove that $F \cup\lbrace v\rbrace$ is not free. (this is the converse of the other implication)

$\forall v\in F, v \in \operatorname{Span}(F)$ as $v=1v$

as $F= (F\cup \left\lbrace v\right\rbrace) \setminus \lbrace v\rbrace$ it is self-evident that $$v \in\operatorname{Span}(F\cup \left\lbrace v\right\rbrace \setminus \lbrace v\rbrace )$$

so that $$ \exists u \in F, u \in\operatorname{Span}(F\cup \left\lbrace u\right\rbrace \setminus \lbrace u\rbrace )$$ this is the negation of "$\operatorname{Span}(F\cup \left\lbrace v\right\rbrace \setminus \lbrace v\rbrace )$ is free".

this statement is true for $u=v \in F$, and so, the proof is done.

Note : Free and linked are synonyms of linearly independent and linearly dependent in the above proof.


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