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Two vectors: $a=(a_1, a_2, a_3)$; $b=(b_1, b_2, b_3)$ I am looking for an operator that results in a third vector $c=(a_1b_1, a_2b_2, a_3b_3)$; Obviously it’s dot product nor cross product.

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There is no need to look anywhere except at your own question. You have defined a valid operation yourself. It is not an especially nice operation because the non zero vectors can multiply to give a zero vector. See Hadamard product for more details.

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  • $\begingroup$ Or pointwise product, which is in a sense a generalization of the Hadamard product. $\endgroup$ – Mario Carneiro Jun 9 '15 at 4:47
  • $\begingroup$ Thanks! Understanding why such operation is unpopular is much more difficult than thinking about the operation. $\endgroup$ – whitegreen Jun 9 '15 at 5:52
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That is an elementwise product of vectors. It is occasionally represented by $\odot$, though generally you'll want to specify what you're doing for the sake of clarity.

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This algebra on $\mathbb{R}^3$ is the direct product of $\mathbb{R}$ with itself three times. Recall for rings $R,S$ we have the product ring $R \times S$ with multiplication $(r_1,s_1)(r_2,s_2)= (r_1r_2,s_1s_2)$. We can do the same for finitely many rings (three for your operation).

Algebraically, we might wonder what this algebra is? Or, more precisely, to which other algebras is this algebra isomorphic. I think the isomorphism to $\mathbb{R} \times \mathcal{H}$ where $\mathcal{H} = \{ a+bj \ | \ a,b \in \mathbb{R} \ \ \& \ \ j^2=1\}$ are the hyperbolic numbers is interesting: $$ \Phi (a,b,c) = \left(a, \frac{b+c}{2}+j\frac{b-c}{2}\right) $$ The identity $(1,1+0j)$ in $\mathbb{R} \times \mathcal{H}$ corresponds to the identity $(1,1,1)$ in $\mathbb{R}^3$ with the direct product. Of course, you could also realize the first and second or first and third copies of $\mathbb{R}$ as hyperbolic numbers. In fact, there are infinitely many algebras which are isomorphic to this given direct three-fold product of $\mathbb{R}$.

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  • $\begingroup$ Thanks! The solution sounds very interesting, though I need some time to get it. $\endgroup$ – whitegreen Jun 10 '15 at 5:01
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This is also the matrix multiplication of the vectors as:

$A*(B*I_{n})$

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