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Is is true that $$ z \in \mathbb{R}^n, \forall u,v \in \mathbb{R}^n, \langle u,z\rangle = \langle v,z\rangle \implies u = v $$ i.e. if two inner products with fixed vector $ z $ are equal so that $ u $ and $ v $ are equals.

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    $\begingroup$ Why don't you use a "cross" to denote the cross product? i.e $u \times z$ instead of $<u,z>$. It's not a suggestion ; I'm really wondering why you do that. $\endgroup$ – Patrick Da Silva Oct 21 '12 at 23:08
  • $\begingroup$ And a more subtle question: why $<u,z>$ instead of $\langle u,z \rangle$? ;-) $\endgroup$ – Hans Lundmark Oct 22 '12 at 10:16
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For cross products, the answer is "no".

However, based on your notation, and the fact that you're talking about $\mathbb{R}^n$ rather than $\mathbb{R}^3$ (cross product defined specifically for $n=3$), it seems you may actually be asking about the inner product.

In that case, the answer is still "no".

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  • $\begingroup$ You can provide counter-examples to support your claim. $\endgroup$ – Artem Oboturov Oct 22 '12 at 20:33
  • $\begingroup$ @Artem: For the cross product, let $z = (1,0,0)$ and $u = (a,1,0)$, $v = (b,1,0)$ for any $a \ne b$. For the dot product, let $z = (1,0,0)$ and $u = (1,a,b)$, $v = (1,c,d)$ for $(a,b)\ne(c,d)$. $\endgroup$ – user856 Oct 23 '12 at 19:36
  • $\begingroup$ @RahulNarain seems it is you who should get a credit for an answer. Are there any conditions to be imposed on $u$ and $v$ so that implication would be true? $\endgroup$ – Artem Oboturov Oct 23 '12 at 21:54
  • $\begingroup$ @Artem: Seeing as you still fail to clarify whether you mean the cross product or the inner product, I cannot answer your question. $\endgroup$ – user856 Oct 23 '12 at 22:27
  • $\begingroup$ @RahulNarain the inner product. I did the edit. $\endgroup$ – Artem Oboturov Oct 24 '12 at 5:46
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No, they may not equal. Since vectors have directions. If the angle between u and z is $\theta_1$ $\langle u, z\rangle=|u||z|cos(\theta_1)$ the angle between v and z is $\theta_2$ $\langle v, z\rangle=|v||z|cos(\theta_2)$

Any two vectors satisfy $|u|cos(\theta_1)=|v|cos(\theta_2)$ will satisfy $\langle u, z\rangle=\langle v, z\rangle$

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Yes.

If $\langle v,z \rangle = \langle u,z \rangle$ for all $z \in V$ then you can certainly pick $z=v$ or $z=u$. Plugging this to get:

$$ \langle v,v \rangle = \langle v,u \rangle = \langle u,u \rangle. $$

From here you can rearrange:

$$ \langle v,v-u \rangle = 0, $$

which is true only if $v=u$.

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