# exterior product definition

i have question from vector mathematics,i know that if there is given two vector, for instance $a=\{a_1,a_2,a_3\}$,$b={b_1,b_2,b_3}$; then so called exterior product is determined as $a\wedge b=\frac{1}{2}(ab-ba)$, but I am confused. My question is what is $ab$ or $ba$? is it scalar? in this case $ab=ba$ and $a\wedge b=0$ which does not have any meaning or $ab=(b_1-a_1,b_2-a_2,b_3-a_3)$ or vector? thanks

Update:

consider for example following equation $$[a,b][X,Y]=a*x+b*y=c$$
then solution is given by following equalities $$[ X,Y]=\frac{1}{a\wedge b}[c\wedge b,a\wedge c]$$

• Where are you getting your definitions? The $1/2$ in your formula leads me to believe that you're doing differential geometry and working with the algebra of alternating multilinear maps. – Dylan Moreland Jul 20 '11 at 16:28
• the name of branch from which i am asking is called exterior algebra – dato datuashvili Jul 20 '11 at 16:32
• I thought that you might be reading a book. I can tell you how I would write down $a \wedge b$, but if it doesn't match up with your sources then I don't know how helpful that will be. – Dylan Moreland Jul 20 '11 at 16:39
• yes i am not reading book but it is also called as wedge multiplication – dato datuashvili Jul 20 '11 at 16:40
• Here's a simple example: suppose I have a two-dimensional vector space $V$, with a basis $\{v_1, v_2\}$. I can write $a = a_1v_1 + a_2v_2$ and $b = b_1v_1 + b_2v_2$. Then, using only the relations imposed upon $\wedge$, it is easy to see that $a \wedge b = (a_1b_2 - a_2b_1)(v_1 \wedge v_2)$. You can do this for a $3$-dimensional vector space almost as easily. – Dylan Moreland Jul 20 '11 at 16:50

## 2 Answers

I suspect that what you mean is $a\wedge b = 1/2(a\otimes b-b\otimes a)$, in which case $a\otimes b$ is the tensor product of $a$ and $b$.

$a\otimes b$ is an element of a new vector space that is higher-dimensional than the vector space containing $a$ and $b$. You can think about it as just a formal construction (basically it is mostly like the pair $(a,b)$), it cannot be simplified further, and satisfies some linearity properties like $(a+b)\otimes c = a\otimes c + b \otimes c$.

I'm struggling to interpret your equation, especially as you wrote down two three-dimensional vectors. Here's what I think is meant: you are working in a two-dimensional space. You have unknown scalars $x, y$, and column vectors $a, b, c$. You want to solve $A\begin{pmatrix}x \\ y\end{pmatrix} = c$, where the $2 \times 2$ matrix $A$ has columns $a$ and $b$. All you're doing is inverting $A$ (when that is possible) to get $\begin{pmatrix}x \\ y\end{pmatrix} = A^{-1}c$. If $a = \begin{pmatrix}a_1 \\ a_2\end{pmatrix}$ and $b = \begin{pmatrix}b_1 \\ b_2\end{pmatrix}$, then defining a scalar (see my comment above) $$a \wedge b = \frac{1}{2}(a_1b_2 - b_1a_2) = \frac{1}{2}\det A$$ makes this work. You'll also see that the $\frac{1}{2}$ plays no role in solving the system. Your formula is essentially Cramer's rule, using the fact that $\bigwedge^n (\mathbf{R}^n)$ is $1$-dimensional.

If solving matrix equations is your goal then thinking about the exterior product may just confuse things.