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Let $Z$ be a linearly independent subset of a vector space $D$. Prove that if $W$ $\subseteq$ $Z$ then $W$ is also linearly independent.

What I tried: I tried to use the fact that $\alpha_1 z_1 + \alpha_2z_2 + ··· + \alpha_n z_n = \vec{0}$ where $\alpha_i \in \mathbb{R}$, $z_i \in Z$

but I had trouble relating that fact to W

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    $\begingroup$ Suppose that $W$ is not linearly independent. What does this say about the linear independence of element in $Z$? $\endgroup$
    – Roland
    Jan 19 '16 at 17:28
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The fact that you want to look at is:

$$ \alpha_1 z_1 + \alpha_2 z_2 + \cdots + \alpha_n z_n = 0 \Rightarrow \alpha_i = 0 \;\;\;\forall i $$

Without loss of generality let $W = \{z_i : i\leq k\}$. Now suppose that

$$ \alpha_1' z_1 + \alpha_2' z_2 + \cdots \alpha_k' z_k = 0. $$

Then if not all $\alpha_i'$ were zero then

$$ \alpha_1' z_1 + \alpha_2' z_2 + \cdots + \alpha_k' z_k + 0\cdot z_{k+1} + 0\cdot z_{k+2} + \cdots + 0\cdot z_n = 0 $$

contradicting the linear independence of $W$ since not all $\alpha'$ are zero.

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  • $\begingroup$ why would it contradict linear independence of $Z$ if not all $\alpha_i =0$? If $ \alpha_1 z_1 + \alpha_2 z_2 + \cdots \alpha_k z_k = 0 $, then that does not necessarily mean that $ \alpha_{n-k} z_{n-k} + \alpha_{n-k+1} z_{n-k+1} + \cdots \alpha_n z_n = 0$ $\endgroup$
    – Raton
    Jan 19 '16 at 17:46
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    $\begingroup$ I added a little bit to my answer. I think you are slightly confused about the definition of linear independence. The definition is the first line of my answer. $\endgroup$ Jan 19 '16 at 17:53
  • $\begingroup$ Okay, thanks, that clears a lot up. $\endgroup$
    – Raton
    Jan 19 '16 at 17:54

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