Let $N = {\{v_1,v_2,...,v_n}\}$ linearly dependent Prove that there are scalars all different from $0$ such that $x_1v_1+x_2v_2+..+v_nv_n=0$

Question

Let $N = {\{v_1,v_2,...,v_n}\}$ linearly dependent set in $V$ , every subset of $N$ and not equal to $N$ is linearly independent.

Prove that there are scalars all different from $0$ such that $x_1v_1+x_2v_2+..+v_nv_n=0$

I think this is wrong

my example is :

$N = {\{\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 1\end{pmatrix},\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}}\}$

so $N$ is linearly dependent becouse of $\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}$ but there are no scalars all different from $0$ such that:

$x_1\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}+x_2\begin{pmatrix}0\\ \:1\\ \:0\end{pmatrix}+x_3\begin{pmatrix}0\\ \:0\\ \:1\end{pmatrix}+x_4\begin{pmatrix}1\\ \:1\\ \:0\end{pmatrix}=0$

because the rank of $\begin{pmatrix}1&0&0&1\\ 0&1&0&1\\ 0&0&1&0\end{pmatrix}$ is 3

Is this correct , if not how to prove it ?

thanks .

Answer 1

Your example is not correct. In that example $v_1,v_2$ and $v_4$ are not linearly independent.

There exist $a_i$'s not all of them $0$ such that $ \sum\limits_{i=1}^{n}a_iv_i=0$. If some $a_j$is $0$ remove that term from the sum and you will get a contradiction to the fact that $\{v_i: i\neq j\}$ is linearly independent. Hence $a_i \neq 0$ for all $i$.