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Posted

I have a question about something i read here.

 

Why did they choose not to define vector division exactly? They said something about non-uniqueness, and I don't follow.

 

To help me refresh my memory, suppose that we have two lines in an xy plane.

 

For the sake of reality, let the plane be a real plane in real space.

 

An equation for any line in the plane will have the following form:

 

Ax+By+C=0

 

Suppose we have two lines.

 

[math] A_1 x + B_1 y + C_1 = 0 [/math]

[math] A_2 x + B_2 y + C_2 = 0 [/math]

 

We can rewrite the equations as follows:

 

[math] A_1 x + B_1 y = - C_1 [/math]

[math] A_2 x + B_2 y = - C_2 [/math]

 

We can rewrite this in matrix form now.

 

[math] A \mathbf{v} = \left[ \begin{array}{cc}

A_1 \ B_1 \\

A_2 \ B_2 \\

\end{array} \right]

 

\left[ \begin{array}{c}

x \\

y \\

\end{array} \right]

 

=

 

\left[ \begin{array}{c}

-C1 \\

-C2 \\

\end{array} \right]

 

= B

 

[/math]

 

Where I am using notation found here.

 

Now the same site says, "vector division is not defined becase there is no unique solution to the matrix equation y=Ax UNLESS x is parallel to Y."

 

So in that matrix equation they are talking about, y is a column vector, and x is a column vector right?

Posted
What is division? it's the inverse operation to multiplication. so, what ways do you have of multiplying two vectors to give a vector?

 

I only know of one mathematical process which operates on two vectors, and outputs a vector... the cross product.

 

The dot product also operates on two vectors, but it outputs a scalar.

Posted

And if you take two non-zero vectors, u and v, then uxv may be zero. Thus you cannot formally invert multiplication for all vectors. (And this only works in R^3 anyway).

 

The point is that there is almost always no useful way to define a multplication operation on vectors that is invertible (ie there will always be two non-zero vectors that mulitply to give the zero vector) . The notable exception is R^2 where we can define (a,b)*(c,d)=(ac-bd,bc+ad), and we get a space that is isomorphic to the complex numbers.

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