# Linear independence Proof 4

I have a Field K , a K-vectorspace V and n $\in {N_0}$and n-tupel ($s_1,...,s_n$)in V and ${U}=<s_1,....,s_n>$
is this always True ?

for all v $\in{V}$, ($s_1,...,s_n,v$) linear independent in ${V}$ $\Rightarrow$ ($s_1,...,s_n$) linear independent in ${V}$

I think it is not always true , because we have to check if v a lineare in combinattion ($s_1,...,s_n$) then it would be wrong . but i can not proof it.

• Any subset of a linearly independent set is itself linearly independent. May 14, 2017 at 11:13
• then the proof would be because ($s_1,...,s_n,v$) linear independent in V then $\sum_{i=1}^{n} a_ is_i +bv=0 \rightarrow a_i= and b=0 \rightarrow \sum_{i=1}^{n} a_ is_i =-bv =0$ and because all ai=0 then is ($s_1,...,s_n$) linear independent in V May 14, 2017 at 11:22
• Try proving that linear dependence of $s_1, \cdots, s_n$ implies the linear dependence of $s_1, \cdots, s_n, v$. May 14, 2017 at 11:25
• What does $U$ here? May 14, 2017 at 13:55

Assume that $(s_1,\dots,s_n)$ is linearly dependent set in V, that means there's such a member of the set that can be shown as a linear combination of other members, particularly let it be the first one that can be shown as a linear combination of those prior to him. Let it be $s_r$ Then $s_r$ is a linear combination of past members: $$s_r=\sum_{i=1}^{r-1}\alpha_is_i$$ But that means $$s_r=\sum_{i=1}^{r-1}\alpha_is_i + \sum_{i=r+1}^{k} 0\cdot s_i + 0\cdot v$$ which would imply that $(s_1,\dots,s_n,v)$ is linearly dependent since there's an element which can be shown as a linear combination of other elements, which is a contradiction with your original assumption.