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Suppose that $\circ$ is an operation on $\Bbb R^2$ with the following properties:

  1. For any $\vec p,\vec q \in \mathbb{R}^2$, and $t \in \mathbb{R}$, $(t \vec p ) \circ \vec q = t(\vec p \circ \vec q)$ holds.
  2. For any $\vec p, \vec q, \vec r \in \mathbb{R}^2$, $\vec p \circ (\vec q + \vec r) = \vec p \circ \vec q + \vec p \circ \vec r$ holds.
  3. For any $\vec p, \vec q \in \mathbb{R}^2$, $\vec p \circ \vec q = -\vec q \circ \vec p$ holds.
  4. For any $\vec p, \vec q, \vec r \in \mathbb{R}^2$, $(\vec p \circ \vec q) \circ \vec r = (\vec p \cdot \vec r)\vec q - (\vec q \cdot \vec r)\vec p$ holds.

Why is it then true that $\vec p \circ \vec q = \vec 0$ all the time?

Cameron Buie
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veco
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  • Does a dot denote the usual scalar product? If so, it is easy to find three vectors such that the RHS of 4) is not zero, while you claim that the LHS must always be zero. – Andrea Mori Sep 23 '13 at 12:53
  • Dot denote dot(scalar) product. – veco Sep 23 '13 at 13:38
  • And $\circ$ is just an operator we are defining. ($\cdot$ is usually scalar product) – veco Sep 23 '13 at 13:45
  • Ok, then I claim that the rules 1--4 are incompatible. I cannot fully answer since the question has been put on hold. – Andrea Mori Sep 24 '13 at 00:54
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    @AndreaMori I believe that is the intention, that the rules are "incompatible" and hence the only possibility is that $\vec{p} \cdot \vec {q} = 0$. (I'm assuming it's the scalar 0, and not the 0 vector ...) We can trivially verify that this definition satisfies the conditions, hence the rules are not incompatible. – Calvin Lin Sep 24 '13 at 02:01

2 Answers2

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I assume that the operation $\circ$ is inner, i.e. $$ \circ:\Bbb R^2\times\Bbb R^2\longrightarrow\Bbb R^2 $$ and satisfies rules 1.--4. as stated (else, 4. would make no sense).

By antisimmetricity (rule 3.) $\vec v\circ\vec v=0$ for all $\vec v\in\Bbb R^2$.

Fix an orthonormal basis $\{\vec e_1,\vec e_2\}$. By bilinearity and what we just observed, the operation $\circ$ is left completely determined by the value $$ \vec w=\vec e_1\circ\vec e_2=a\vec e_1+b\vec e_2. $$ Now, applying 4. $$ 0=\vec w\circ\vec w=(\vec e_1\circ\vec e_2)\circ\vec w=(\vec e_1\cdot\vec w)\vec e_2-(\vec e_2\cdot\vec w)\vec e_1=a\vec e_2-b\vec e_1. $$ By linear independence of basis vectors, $a=b=0$, so that $\circ$ needs to be the constant zero function.

BUT: let now $\vec r=c\vec e_1+d\vec e_2\neq\vec0$, so that $(c,d)\neq(0,0)$. Rule 4. would yield $$ \vec 0=(\vec e_1\circ\vec e_2)\circ\vec r=c\vec e_2-d\vec e_1\neq\vec 0 $$ (again by the linear independence of basis vectors) which is obviously impossible.

THUS no such a function exists, not even the constant zero function.

Andrea Mori
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Hint: Take a look at the outer product in $\mathbb R^3$ and restrict one component to be constant. What do you see?

AlexR
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    This would only prove that one model (rule) of the given vector product always yield $\vec 0$. The task of then deciding whether there are other models which yield other results is just as hard as the original question. – Arthur Sep 23 '13 at 12:39
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    @Arthur that's true but it should help visualising the problem. – AlexR Sep 23 '13 at 13:54