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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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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. – Patrick Da Silva Oct 21 '12 at 23:08
And a more subtle question: why $<u,z>$ instead of $\langle u,z \rangle$? ;-) – Hans Lundmark Oct 22 '12 at 10:16

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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You can provide counter-examples to support your claim. – Artem Oboturov Oct 22 '12 at 20:33
@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 $(a,b)\ne(c,d)$. – Rahul Oct 23 '12 at 19:36
@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? – Artem Oboturov Oct 23 '12 at 21:54
@Artem: Seeing as you still fail to clarify whether you mean the cross product or the inner product, I cannot answer your question. – Rahul Oct 23 '12 at 22:27
@RahulNarain the inner product. I did the edit. – Artem Oboturov Oct 24 '12 at 5:46

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