# 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.

• 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".

• 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 $(a,b)\ne(c,d)$. – user856 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. – user856 Oct 23 '12 at 22:27
• @RahulNarain the inner product. I did the edit. – Artem Oboturov Oct 24 '12 at 5:46

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

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 $$v=u$$.