Let $\{v_1, . . . , v_k\}$ be a sequence of vectors from a vector space $V$ . Prove that the sequence is linearly dependent if and only if for some $j$, $1 ≤j ≤ k$, $v_j$ is a linear combination of $\{v_1, . . . , v_k\} \setminus \{v_j\}$.

I have to prove two directions. I am having troubles with the following one:

If the sequence of vectors is linearly dependent then $\exists v_j$ for$ 1 \leq j \leq k$ such that $v_j$ is a linear combination of $\{v_1,...,v_k\}\setminus \{v_j\}$


if the sequence of vectors is linearly dependent then, $$c_1v_1+...+c_kv_k=0 \text { not all $c_j=0$}$$ Assume $c_j \neq 0$, so $$c_1v_1+....+c_jv_j+....+c_kv_k=0$$ $$c_1v_1+...+c_{j-1}v_{j-1}+c_{j+1}v_{j+1}+...+c_kv_k=c_jv_j$$

how can I express $v_j$ without the constant as a linear combination of the other vectors.

  • $\begingroup$ $c_j \neq 0$, so you can divide both sides of the last equality by $c_j$ to get $\sum_{i \neq j} c_ic_j^{-1} v_i = v_j$ $\endgroup$ – Crostul May 22 '16 at 23:28
  • $\begingroup$ how do you know that it preserves the right field? $\endgroup$ – TheMathNoob May 22 '16 at 23:32
  • 1
    $\begingroup$ That is the definition of a field: every $c \ne 0$ has an inverse that is also an element of the same base field, and multiplying by this inverse results in a product also still in the same base field. Upvoted for the very clear statement of the problem, showing good effort and explaining clearly what the last obstacle was. Good example for how questions should be asked on the forum. $\endgroup$ – mathguy May 22 '16 at 23:37

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