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So here is the link to my problem, which i've copied here:

A set is linearly independent $\iff$ $\det(W)\neq0$

Proof. First $\det(W)\neq 0 \implies$ linear independence:

By contradiction. Assume our vectors are linearly dependent. Then $\exists c_i \neq 0$ such that

$$\sum_{i=1}^k c_i\cdot y_i(x)=0$$

Taking derivatives:

$$\left. \begin{matrix} \sum_{i=1}^k c_i y_i=0 \\ \vdots \\ \sum_{i=1}^k c_i y_i^{(k-1)}=0\end{matrix} \right \} k~\text{equations}$$

$\exists \vec{c}\neq 0_v; W(x)\vec{x}=0_v \implies \det(W)=0$ So by the contrapositive: $\det(W)\neq 0 \implies$ linear independence.

On page 15 at the top between where it says "Taking derivatives: ..." and "So by the contrapositive..." I don't understand the line in between. Could someone please explain the logic in a bit more depth than the notes do?

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    I'm assuming this got downvoted because of the "I don't know how to write matrices in latex." So, rather than downvote, I'll pass along some helpful links for you: 1. A site that will help you to construct things like matrices very simply https://www.codecogs.com/latex/eqneditor.php 2. A link to the mathjax basics here on stack exchange http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-02-13
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    Thanks for the link David, I'll have a look now. Also big thank you to projectile for the edit! – 2017-02-13
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    @mrnovice You're welcome. Is it all correct? – 2017-02-13
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    Yes although it's been pointed out there is a typo in the 2nd last line in that it should be W(x)*c rather than W(x)*x. – 2017-02-13
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    @mrnovice Yes, the typo is also there on the pdf you've attached. – 2017-02-13
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    I think that is partially what has been confusing me so much – 2017-02-13

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There's a typo. You should red $$\exists\vec c\neq 0_v,\enspace W_x \color{red}{\vec c}=0_v$$ (I don't know what the $v$ subscript denotes; probably the vector space $\mathbf R^k$) In other words, for any $x$, this linear system has a non-trivial solution (actually the same for all $x$), which implies the determinant $\det W(x)$ is $0$ for all $x$.

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    Ok now I follow the proof, thank you – 2017-02-13