# An alternative proof of a lemma in linear algebra

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.