# Inner products equality for one of vectors fixed

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

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.
@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