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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?

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    $\begingroup$ It should be written $$a_1x+b_1y+c_1z=0 \iff a_1=b_1=c_1=0$$ as an if and only if. $\endgroup$ – gimusi Mar 7 '18 at 8:52
  • $\begingroup$ @gimusi: nonsense! The other implication is always true in the most trivial way. $\endgroup$ – ancientmathematician Mar 7 '18 at 14:50
  • $\begingroup$ @ancientmathematician then why it should be a nonsense if it is always true? I prefer to express the condition by an if and only of to emphatize that it is the only case $\endgroup$ – gimusi Mar 7 '18 at 14:53
  • $\begingroup$ @gimusi: of course you are perfectly free to do it your way, but not to insist other people "should" do so. Some of us think definitions are best when they are minimal, and not embroidered with tautologies. $\endgroup$ – ancientmathematician Mar 7 '18 at 15:04
  • $\begingroup$ @ancientmathematician it was simply a suggestion, you are free to write whatever symbol you like! $\endgroup$ – gimusi Mar 7 '18 at 15:05
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$1(x-y)+1(y-z)+1(z-x)=0$, therefore system is clearly linear-dependant.

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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.

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    $\begingroup$ I have to vote for this one because it shows the mistake in the questioners reasoning. $\endgroup$ – David Stanley Mar 7 '18 at 15:37
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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.

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