Consider the equation: $$ c\boldsymbol{a}+d\boldsymbol{b}=\boldsymbol{0} $$ where $\boldsymbol{a}$ and $\boldsymbol{b}$ are vectors, $c$ and $d$ scalars (let's suppose they are not zero).
Why would $\boldsymbol{a}$ and $\boldsymbol{b}$ be dependent when it equals zero? I mean, I don't find the logic.
On
I'm not sure if you're asking why summing vectors to $\mathbf{0}$ makes them linearly dependent, or why the $\mathbf{0}$ vector is always linearly dependent, so I'll quickly answer both.
The first is exactly from the definition of linear dependence.
A set of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \cdots, \mathbf{v}_n\}$ is linearly dependent if you have some linear combination $a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+a_n\mathbf{v}_n=0$ where not every $a_1$ is $0$. So if you have a linear combination of vectors that sum to $\mathbf{0}$ when the scalars are not all $0$, you have a linearly dependent set of vectors.
The second is a direct result of this defintion.
Specifically, you only need $one$ of your scalars to be nonzero.
So if one of your vectors $\mathbf{v}_i$ is $\mathbf{0}$ you can let every $a_k$ be $0$ when $k\neq i$ and let $a_1$ be any nonzero number and you have your linear combination that shows linear dependence.
On
Assume that
$$a_1v_1+a_2v_2=0$$ with $a_2\ne 0.$ Then it is
$$v_2=-\frac{a_1}{a_2}v_1.$$ That is, $v_1,v_2$ are linearly dependent.
In general, assume that
$$a_1v_1+a_2v_2+\dots +a_nv_n=0$$ with $a_n\ne 0.$ Then it is
$$v_n=-\frac{a_1}{a_n}v_1+\dots -\frac{a_{n-1}}{a_n}v_{n-1}.$$ That is, $v_n$ is a linear combination of $v_1,\dots,v_{n-1}.$ In other words, $v_1,\dots,v_n$ are linearly dependent.
Because that's the way linear dependence is defined: the vectors $v_1,v_2,\ldots,v_k$ are linearly dependent if the null vector can be written as $\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_kv_k$, where $\alpha_1,\alpha_2,\ldots,\alpha_k$ are scalars and not all of them are equal to $0$.