Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

1 Answer

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".

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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