[Math] Non parallel, linear independent vectors

linear algebra

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?

Best Answer

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.

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