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prove or give a counterexample: if $v . w = 0$ for all $v$, then $w= 0$. v,w are vectors

I gave a counterexample but i think my answer is wrong.

Let $v= (v_{1},v_{2},v_{3})$ and $u= (1/v_{1}, -2/v_{2}, 1/v_{3})$

im pretty sure my answer is wrong.

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    $\begingroup$ This isn't a counterexample, since the statement is "for a fixed $w$, if for all $v$, $v\cdot w=0,$ then $w$ must be the $0$ vector." In your example, you're supposing $w$ depends on $v.$ Also, if any of the $v_{i}=0,$ your vector is not defined. $\endgroup$ Sep 14, 2017 at 4:00

3 Answers 3

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The problem states that the relation must hold for all $\mathbf v$ not just for a fixed $\mathbf v=(v_1,v_2,v_3)$. Instead we fix $\mathbf w=(w_1,w_2,w_3)$ and show that $w_1=w_2=w_3=0$ by varying our choice of $\mathbf v$. Consider $\mathbf v_a=(1,0,0)$, $\mathbf v_b=(0,1,0)$ and $\mathbf v_c=(0,0,1)$. Since $\mathbf v \cdot \mathbf w=0$ must hold for any $\mathbf v$ we get that $$\mathbf v_a \cdot \mathbf w=w_1=0$$ $$\mathbf v_b \cdot \mathbf w=w_2=0$$ $$\mathbf v_c \cdot \mathbf w=w_3=0$$ therefore $\mathbf w=(0,0,0).$

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You're right that your counter example is incorrect. The reason it doesn't work is that given your $v$, you have to show that $u\cdot v=0$ for any $u$. You are showing it only for a particular $u$.

In fact if $u\cdot v=0$ for all $u$, then $v$ must be equal to zero.

Hint: Is there any value of $u$ you can use such that $u\cdot v=v_i=0$?

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  • $\begingroup$ the zero vector $\endgroup$
    – NAT
    Sep 14, 2017 at 4:07
  • $\begingroup$ Not quite. $0\cdot v=0$ but you are already assuming that $u\cdot v=0$ for anything. Forgetting about the zero part for a minute is there any vector such that $v\cdot u=v_1$? $\endgroup$
    – CEH
    Sep 14, 2017 at 4:09
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If $v\cdot w=0$ for all vectors $v$ then in particular $|w|^2=w\cdot w=0$, and this implies $w=0$.

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