Let vectors $v_1,v_2,\dots, v_n$ be linearly independent.

Let $v$ be a vector s.t. $v,v_1,v_2,\dots,v_n$ are linearly dependent. How do I prove that $v$ is a linear combination of $v_1,v_2,\dots,v_n$?

  • $\begingroup$ What is the definition of "linear independence" you are using? What have you tried? $\endgroup$ – Andrés E. Caicedo Sep 7 '13 at 17:41
  • $\begingroup$ Write down what it means for $v,v_1,...v_n$ to be linearly dependent. Then use the other condition to show that the multiplier for $v$ is non-zero. Divide across by this constant. $\endgroup$ – copper.hat Sep 7 '13 at 17:49
  • $\begingroup$ To me it means is v1,v2 ... vn are linearly independent then a1v1 + a2v2...... + anvn = 0 has only trivial solution i.e. a1=a2 =.... an = 0 $\endgroup$ – Hashtag Sep 7 '13 at 17:52
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    $\begingroup$ There is a definition of linear independence: $v_k$ are linearly independet iff $\sum_k \alpha_k v_k = 0$ implies $\alpha_k = 0$. $\endgroup$ – copper.hat Sep 7 '13 at 17:55
  • $\begingroup$ Logically the answer is obvious but how can it be proven mathematically $\endgroup$ – Hashtag Sep 7 '13 at 18:11

If $v,v_1,\ldots ,v_n$ are linearly dependent over $k$ then $rv+r_1v_1+\cdots +r_nv_n=0$ with some $r,r_i\in k$ not all zero. Suppose that $r=0$. Then all $r_i=0$ because $v_1,\ldots ,v_n$ are linearly independent over $k$. This is a contradiction, because not all $r_i$ and $r$ are zero. Hence $r\neq 0$ and we have $$ v=-r^{-1}(r_1v_1+\cdots +r_nv_n). $$

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