# Proving Linear Independent Vector Space

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

• Suppose that $W$ is not linearly independent. What does this say about the linear independence of element in $Z$? Jan 19 '16 at 17:28

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.

• 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 $\alpha_{n-k} z_{n-k} + \alpha_{n-k+1} z_{n-k+1} + \cdots \alpha_n z_n = 0$ Jan 19 '16 at 17:46
• 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. Jan 19 '16 at 17:53
• Okay, thanks, that clears a lot up. Jan 19 '16 at 17:54