2
$\begingroup$

I translated this to an alternative statement: if $x\neq 0$ in $V$, $cx = 0$ iff $c = 0$.

My attempt by contradiction:

$cx = 0$

$cx + 0 = 0$

$cx + cx = cx$

$2cx = cx$

But so far all I'm getting are circular statements like this, and I feel that this isn't the right direction, but I'm at a lost for what theorem to use

EDIT: so I divided $c$ from both sides,

$2cx = cx$

$2x = x$

and then say that since $x$ is not zero, this is not satisfied, which means the assumption that $cx = 0$ (leading to this conclusion is wrong) Though I still feel that this logic is quite circular.

$\endgroup$
2
$\begingroup$

Assume $0\neq x\in V$ and $0\neq c\in F$ where $V$ is a vector space over field $F$. Suppose, for a contradiction, $cx=0$. Since $c\neq 0$, then it's invertible, yielding $x= c^{-1}cx = c^{-1}0 = 0$, a contradiction.

Explanation: the reason this works is due to the axioms of a vector space.

  1. Vector spaces are structures that are made up of what we call "vectors" and there are operations we can do with them. We can, among other things, multiply vectors with scalars (the field elements). In every vector space $$\forall f,g\in F\quad\forall x\in V\qquad f(gx) = (fg)x$$
  2. A field is a group under multiplication. Therefore every non-zero element is invertible i.e $$\forall f\in F\setminus\{0\}\quad\exists f^{-1}\in F\qquad ff^{-1} = f^{-1}f = 1_F, $$ where $1_F =: 1$ is the uniquely determined identity element.
$\endgroup$
1
$\begingroup$

Hint: Do a proof by contradiction. Assume that $x$ is not the zero vector, $cx$ is the zero vector, and $c$ is not zero. There's a particular property of fields that you used in your edit that is helpful here, specifically regarding multiplication.

$\endgroup$
  • $\begingroup$ Could you point out why my statement isn't justified? $\endgroup$ – L to the V Oct 4 '17 at 3:13
  • $\begingroup$ @LtotheV Honestly, you're pretty much there after the edit, but you need to talk about the multiplicative identity. If $2x=x$, then $2x-x=1x=0$. Then because we have a multiplicative identity, we must have $x=0$. You also skip over the assumption that $c$ is nonzero, which is why you can divide by $c$ in the first place. $\endgroup$ – Kevin Long Oct 4 '17 at 3:17
  • $\begingroup$ In that case, could I say since 2x = x, then -x is the additive inverse of 2x, but we know that cx + 0 = cx, thus 0 is the additive identity in this case, and 2x - x = 1x, and 1x isn't 0, thus leading to a contradiction? $\endgroup$ – L to the V Oct 5 '17 at 18:01
  • $\begingroup$ "but we know that cx + 0 = cx, thus 0 is the additive identity in this case" Everything else you said makes sense (same as my comment), but I don't see why you need this line. $\endgroup$ – Kevin Long Oct 5 '17 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.