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Prove: a set of vectors $K$ is linearly dependent iff a vector is linear combination of the others.

Let: $\alpha_1k_1 + \alpha_2k_2+...\alpha_nk_n = 0$

Then, There must be $\alpha_i \ne 0$. Therefore,

$$\alpha_1k_1 + \alpha_2k_2 + \alpha_{i-1}k_{i-1} + \alpha_{i+1}k_{i+1}+...+ \alpha_nk_n = -\alpha_ik_i.$$

Then,
$${\alpha_1 \over -\alpha_i}k_1 + {\alpha_2 \over -\alpha_i}k_2+...+{\alpha_n \over -\alpha_i}k_n = k_i$$

Indeed, $k_i$ is a linear combination of the other vectors.

My question is:
Is that answering the iff condition?

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    $\begingroup$ This answers the if and not the only if. $\endgroup$ – user63181 Jan 17 '14 at 22:30
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    $\begingroup$ You need to show that if a vector is a linear combination of the others, then the collection is linearly dependent. $\endgroup$ – copper.hat Jan 17 '14 at 22:34
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    $\begingroup$ @copper.hat, isn't it just reading the current proof from bottom to top? $\endgroup$ – SuperStamp Jan 17 '14 at 22:35
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    $\begingroup$ @SuperStamp Yes, just read backwards. All your implications go both ways. $\endgroup$ – T.J. Gaffney Jan 17 '14 at 22:36
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    $\begingroup$ @SuperStamp: It depends on your audience. Instructing your reader to read the steps in reverse can work, but it is completely unambiguous and less commanding to just establish a sequence of equivalences. $\endgroup$ – copper.hat Jan 17 '14 at 22:43
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Assume that there exist $a_i \in \mathbb{K}$ for which $a_1{\bf v}_1+\cdots+a_n{\bf v}_n = {\bf 0}$, where not all $a_i$ are zero.

Then, as you say, there exists an $a_k \neq 0$, and we can write

$${\bf v}_k = -\frac{1}{a_k}(a_1{\bf v}_1+\cdots + a_{k-1}{\bf v}_{k-1}+a_{k+1}{\bf v}_{k+1}+\cdots+a_n{\bf v}_n)$$

Hence ${\bf v}_k$ is a linear combination of the other ${\bf v}_i$.


Assume that ${\bf v}_k$ is a linear combination of the other ${\bf v}_i$. In that case, there exist $\lambda_j \in \mathbb{K}$ for which $${\bf v}_k = \lambda_1{\bf v}_1 + \cdots + \lambda_{k-1}{\bf v}_{k-1}+\lambda_{k+1}{\bf v}_{k+1}+\cdots + \lambda_n{\bf v}_n$$

Putting ${\bf v}_k=\sum_{i \neq k} \lambda_i{\bf v}_i$ into $a_1{\bf v}_1+\cdots + a_n{\bf v}_n$ shows that $\lambda_j = -a_j$ gives

$$a_1{\bf v}_1+\cdots+a_n{\bf v}_n = {\bf 0}$$


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