linear dependence of a family of vectors

Question

I think I recall to have somewhere read the following:

Let $v_1, \dots, v_n$ let be vectors. Then they are linear dependent if and only if there is a natural number $1\leq i\leq n$ with $v_i = \sum_{j\not = i} \lambda_j v_j$, where $\sum_{j\not = i} \lambda_j v_j$ is a linear combination of the vectors $\{v_1, \dots, v_n\}\setminus \{v_i\}$.

1. Is this correct?
2. If $v_1, \dots, v_n$ are linear dependent, is it possible to express every vector $v_i\in\{v_1, \dots, v_n\}$ as a linear combination of the others? Or is this only possible for (at least) one vector but not necessarily for all?

Answer 1

I think this would help you:

The following are equivalent :

(1) $\{v_1,..,v_n\}$ independent.

(2) for each $1 \le i \le n $ $v_i \notin span\{v_1 , \dots , v_{i-1} \}$

(3) for each $1 \le i \le n $ $v_i \notin span(\{v_1 , \dots , v_{n} \} -\{v_i\})$

Answer 2

$\Rightarrow$:

If the vectors $v_j$ are linear dependent we have a vector $c = (c_j) \ne 0$ with $$ \sum_{j=1}^n c_j v_j = 0 $$ As $c\ne 0$ it has at least one non-zero component $c_i$ so we can solve for $v_i$: $$ 0 = c_i v_i + \sum_{j\ne i} c_j v_j \iff \\ v_i = \sum_{j\ne i} \underbrace{\left(- \frac{c_j}{c_i}\right)}_{\lambda_j} v_j $$ $\Leftarrow$:

$$ v_i = \sum_{j\ne i} \lambda_j v_j \iff \\ 0 = (-1) \cdot v_i + \sum_{j\ne i} \lambda_j v_j = \sum_{j=1}^n c_j v_j $$ The coefficient vector $c$ is not the null vector, because $c_i = -1 \ne 0$. So the $n$ vectors $v_j$ are linear dependent.