I am trying to formally prove this result:
Let $V$ be a vector space and $S$ be a finite non-empty subset, i.e., $S\subset V$, such that $S$ is LI set. Then, show that:
$w\notin L[S]\implies S\cup\{w\}$ is LI set.
$\underline{Terminology}:$
LI = Linear Independent
$L[S]$ = linear span of vectors from set $S$.
The proof shows that size of a linear independent set can be grown while maintaining its linear independence, if we append a vector $w$ that is not in the span of remaining vectors in the set.
My approach uses contradiction:
(contradiction approach): Suppose $T=S\cup\{w\}$ is LD set.
$\implies\exists u\in T$, such that $u\in L[T\setminus\{u\}]$,
i.e., for some $u\in T, u$ belongs to span of remaining vectors of $T$.
As given, $w\notin L[T\setminus\{w\}]$, so $u\neq w$.
$\implies\exists u\in S$, such that $u\in L[T\setminus\{u\}]$.
How to proceed further ?
I should perhaps convert expression $u\in L[T\setminus\{u\}$ into $u\in L[S\setminus\{u\}$, because $u\neq w$ and $u\in T\setminus\{w\}$.
Please help!
Assume the that $S$ is a linearly independent finite set and $w\not\in L[S]$ but, for the sake of contradiction, that $S\cup \{w\}$ is a linearly dependent set. Since $S$ is finite, it has $n$ vectors, i.e. $S=\{s_1,s_2,\ldots,s_n\}$ for some $n\in\mathbb{N}$. Furthermore, these vectors are nonzero since $S$ is linearly independent. We also have that $w\neq 0$ because $0\in L[S]$. Since $S\cup\{w\}$ is linearly dependent, we have that $\alpha_1s_1+...+\alpha_ns_n + \beta w=0$ for some scalars $\alpha_1,\ldots,\alpha_n,\beta$, not all of which are $0$. We can see that $\beta\neq 0$ by the linear independence of $S$, because if $\beta=0$ then at least one of the other scalars $\alpha_1,\ldots,\alpha_n$ would have to be nonzero and then we would have $\alpha_1s_1+\ldots\alpha_ns_n=0$ without $\alpha_i=0 \forall i\in\{1,\ldots,n\}$ which contradicts $S$ being a linearly independent set. Since $w\neq 0$ and $\beta\neq 0$ we have that $\alpha_i\neq 0$ for at least one $i\in\{1,\ldots,n\}$. Then, $w=\frac{\alpha_1}{-\beta}s_1+\ldots+\frac{\alpha_n}{-\beta}s_n$ which contradicts that $w\not\in L[S]$.