Show that system is linearly dependent

Question

We have that the system of vectors $$x, y, z$$ is linearly independent. Show that the system $$x-y, y-z, z-x $$ is linearly-dependent. Here is my try. As the first system is independent, it means that $$a_1x+b_1y+c_1z=0 => a_1=b_1=c_1=0$$ In the second system, writing the combination $$a(x-y) + b(y-z) + c(z-x) = ax-ay+by-bz+cz-cx = (a-c)x + (b-a)y+(c-b)z$$ if we put ($a_1 = a-c, b_1 = b-a, c_1 = c-b$), then from the first system's independence the second system is independent too.

Where is the mistake?

Answer 1

$1(x-y)+1(y-z)+1(z-x)=0$, therefore system is clearly linear-dependant.

Answer 2

Note that

$$ (a-c)x + (b-a)y+(c-b)z=0\iff a=b=c$$

which has solution also for $a=b=c\neq 0$.

Note that, as an example, the system $x,x+y,x+y+z$ is also linearly independent.

Answer 3

The mistake lies in the passage in which you jump from$$a-b=b-c=c-a=0\tag1$$to $a=b=c=0$. Note that if, say, $a=b=c=1$, then $(1)$ still holds.