Non parallel, linear independent vectors I have came across a problem in my notes that I am looking for some help with. Let 
$$ \text{vector 1} = \begin{bmatrix}
1   
\\2
\\3
\\4
\end{bmatrix} \qquad\qquad \text{vector 2} =  \begin{bmatrix}
1
\\4
\\6
\\8
\end{bmatrix}$$ 
I am using these vectors combined with the standard basis vectors to form a basic in $\mathbb R^4$. So I understand that they need to have a $rank=4$ and they must be linearly independent. In my notes it says that these two vectors above are non-parallel and they are linearly independent. Just looking to clarify why this is non parallel and linearly independent. I understand the rest, just not these two terms. Are these columns independent because they have non trivial parts?
 A: Apply the definition of linear independence to two vectors $v$ and $w$.

Two vectors $v$ and $w$ are linearly independent if for any scalars $a$ and $b$, the condition $av+bw=0$ implies $a=b=0$.

Also, since you seem to be confused about the definition of parallel,

Two vectors $v$ and $w$ are parallel if there exists a scalar $c$ such that $v=cw$.

Now, if $v=cw$ (i.e. $v$ and $w$ are parallel), then we have $v-cw=0$. This implies $v$ and $w$ are not linearly independent. The converse follows a similar argument. So in the case of two vectors, linear independence is the same as being non-parallel.

For your example, it is easy to see that the first vector is not a multiple of the other, so they are non-parallel and linearly independent.
A: Parallel and linearly dependent is one and the same in this (euclidean) case: none of $\;v_1,\,v_2\;$ is a multiple scalar of the other one. Because of this they are non-parallel = they are linearly independent.
A: First, you cannot write $v_1$ as a linear combination(scaling) of the other: There is no $\lambda$ to satisfy $v_1=\lambda v_2$. If you have other vectors, you shouldn't be able to find $\mathbf{\lambda}=\{\lambda_1, \lambda_2, \lambda_3\}$ s.t.:
$$
v_1 = \lambda_1 v_2 + \lambda_2 v_3 + \lambda_3 v_4
$$
This is easy to show. Two (hyper-)planes will be parallel if one of the vectors (i.e. the plane parameters) can be written as a scaling of the other:
$$
\lambda v_1 = v_2
$$
