Suppose $V$ is a vector sapce and $S \subseteq V$. Then $S$ is linearly independent if and only if for each $v\in S, v \notin span(S-\{v\})$
Suppose for each $v\in S, v \notin (S-\{v\})$
then $v $ is not linear combination of elements of $(S-\{v\})$ i cant go for further and also get no idea for if part
On
First, suppose that $S$ is linearly independent and suppose that there exists $v\in S$ such that $v\in \text{span}(S\setminus\{v\})$. By definition, $v=\lambda_1x_1+\dots+\lambda_nx_n$ where $\lambda_k\in F$ (the field) and $x_k\in S$. Thus, $0=-v+\lambda_1x_1+\dots+\lambda_n x_n$ This is a contradiction because is a linear combination of the zero vector where not all the scalars are zero.
Now, if we suppose that for all $v\in S$ holds that $v\notin \text{span}(S\setminus\{v\})$. By contradiction, suppose that there exists non zero scalars and different vectors in $S$ such that $0=\lambda_1x_1+\dots\lambda_nx_n$. Suppose, w.l.g., that $\lambda_1\neq 0$, then, $x_1=\frac{\lambda_2}{\lambda_1}x_2+\dots+\frac{\lambda_n}{\lambda_1} x_n$. Contradiction because this means that $x_1\in\text{span}(S\setminus\{x_1\})$
If $S$ is linearly independent and $v\in S$, then $v\notin\langle S\setminus\{v\}\rangle$, because otherwise you could express $v$ as a linear combination $a_1v_1+\cdots+a_nv_n$ of elements of $S$, which would mean that $-v+a_1v_1+\cdots+a_nv_n=0$. This is impossible, since $S$ is linearly independent and not all coefficients of this linear combination are $0$.
And if $S$ is linearly dependent, there are $v_1,\ldots,v_n\in S$ and scalars $a_1,\ldots,a_n$, not all of which are $0$, such that $a_1v_1+\cdots+a_nv_n=0$. You can assume without loss of generality that $a_1\neq0$. Then $v_1=-\left(\frac{a_2}{a_a}v_2+\cdots+\frac{a_n}{a_1}v_n\right)$ and therefore $v_1\in\langle S\setminus\{v_1\}\rangle$.