Re-checking my linear algebra, using Lax textbook *Linear algebra and its applications*.

I want to know if my attempt of proving the following theorem it's ok (it has been a while since I attended this lecture).

> 
> Show that if the vectors $x\_1, x\_2,\dots,x\_j$ are linearly independent, then
>  none of them is the zero vector.
> 
> 
> 

As far as I understand, this is a proof by contradiction. I have to exhibit a
nontrivial linear combination that fulfills

$$
k\_1x\_1+k\_2x\_2+\dots+k\_jx\_j=0,
$$

where $k\_j\in K$ are elements in the field $K$.

Without loss of generality, suppose that $x\_1$ is the zero vector. Then

$$
k\_1x\_1+\dots+k\_jx\_j=0,
$$

where all the $k\_i$ must be zero. We can sum on both sides the term $\tilde{k}x\_1$ (where $\tilde{k}\in K$), and by Eq. (10): $0x\_1=0$. Therefore

$$
(\tilde{k}+k\_1)x\_1+\dots+k\_jx\_j=\tilde{k}x\_1+\dots+k\_jx\_j=0,
$$

and I have showed that there is a non-trivial linear combination with the zero vector, which is a contradiction.