Understanding Vector Division in Algebraic Structures

[TheDestroyer]

How can We Devide 2 vectors? and what is the result? is it a scalar or another vector?

if we said:

$$\frac{\vec{E_1}}{\vec{E_2}}$$

What the result would be if the unit vectors or not equal or opposite? if they are then we'll get a scalar,

Can we say:

$$\frac{\vec{E_1}}{\vec{E_2}} = \vec{E_1} \cdot \frac{1}{\vec{E_2}}$$

Can this be a dot product? then we'll get a scalar also ! and if it's like this, what's the defenition of a vector inverse?

 

[matt grime]

If you want to define division then you need to define multiplication. So how do you define the product of two (arbitrary) vectors?

 

[cronxeh]

there are infinitely many answers, because there are that many possibile vectors, therefore vector division is not defined

 

[TheDestroyer]

How come? can you explain??

In electromagnetic we define the freenels coeffecients with the ratio of to electrical fields,

can you explain why there are infinite answers?

and about the definition of 2 dot product vectors simply:

a.b = ax.bx + ay.by + az.bz
or
a.b = ab cos(a,b)

 

[cronxeh]

http://www.mcasco.com/qa_vdq.html

 

[matt grime]

The dot product of two vectors isn't a vector so doesn't define a division operation.

If you're going to ahve a sensible notion of division then you need a sensible notion of multiplication. 1/b is *define* to be the (unique) element (of whatever objects you're thinking about) such that b*1/b=1. This requires a *well defined* mutliplication, and an identity element. You've so far got neither of these. well defined means something like a*b=a*c implies b=c.

 

[TheDestroyer]

I didn't like this ! or we can say this is not logic ! because as I said, some physical quantities are defined as the ratio of 2 vectors, that means it must have a value, even if it's not single,

I'll work on this a little, and i'll repost the result, thanks everybody for caring about answering,

TheDestroyer

 

[HallsofIvy]

WHAT physical quantities are defined as the ratio of two vectors? I've never seen such a thing! (The Fresnel coefficients, in particular, are NOT defined as the ratio of two vectors.)

 

[dextercioby]

The Fresnel coefficents are defined with absolute values of the 2 electric field vectors.

I guess one could say that euclidean vectors form a vector/linear space over R,and not an algebra...

Daniel.

 

[Data]

Well, in special cases, they do form a division algebra with an appropriate multiplication. For example, we can look at $\mathbb{R}^2$ and define the operation

$$(a, b)(c, d) = (ac-bd, ad+cb)$$

which does permit meaningful division using the definition

$$(a, b)/(c, d) = \left(\frac{ac+bd}{c^2+d^2}, \frac{cb-ad}{c^2+d^2}\right), \ \forall (c, d) \neq (0, 0)$$

Of course, these definitions just stem from the fact that $\mathbb{C}$ is a division algebra. The same thing can clearly be done for $\mathbb{R}^4$ and $\mathbb{R}^8$ using the multiplication rules for quaternions and octonions.

It doesn't work for $\mathbb{R}^3$, though.

 

[matt grime]

The Fresnel coefficents are defined with absolute values of the 2 electric field vectors.

I guess one could say that euclidean vectors form a vector/linear space over R,and not an algebra...

Daniel.

They do form an algebra in many non-isomorphic ways. Just almost never division algebra.

 

[dextercioby]

Okay.However,there are more definitions of a vector...

Daniel.

 

[matt grime]

Given any (finite dimensional) vector space I can create an algebra from it. And probably from most infinite ones too, I shouldn't wonder.