Division between vectors

1. Jun 15, 2014

Jhenrique

I want find a value M such that given v and u, satisfies the equation v=Mu.

Well, the vector u has a modulus u and a direction α wrt x-axis; the vector v has another modulus v and another direction β wrt x-axis. What is happened is a change of magnitude and direction of the vector u, therefore, M needs scalar the vector u and change its direction, so that:

$$\\ M = \frac{v}{u} \begin{bmatrix} \;\;\;\cos(\beta-\alpha) & -\sin(\beta-\alpha)\\ +\sin(\beta-\alpha) & \;\;\;\cos(\beta-\alpha)\\ \end{bmatrix}$$
So: $$\\ \frac{\vec{v}}{\vec{u}} = M$$

2. Jun 15, 2014

Jilang

Sorry, is this a question? Only it looks like the answer.

3. Jun 15, 2014

Jhenrique

It's a question! I just followed the tamplate of the homework section, where is necessery to post a tentative.

4. Jun 15, 2014

pasmith

The basic linear operations available in $\mathbb{R}^2$ are scaling, rotation, and shears.

There are infinitely many linear maps $A \in \mathrm{M}_2(\mathbb{R})$ such that $\vec v = A\vec u$; to determine $A$ uniquely you also need to specify the image under $A$ of some vector which is linearly independent of $\vec u$.

For example, if $\vec w \cdot \vec u = 0$ then you can have $A\vec w = 0$ or $A \vec w = \vec w$ or $A \vec w = \vec v$ or $A \vec w = \vec u$ or ..., and in each case it will still be the case that $A\vec u = \vec v$. This is why the notation $\vec v / \vec u$ doesn't make sense.

5. Jun 15, 2014

Jhenrique

But my answer is valid too, correct?

6. Jun 15, 2014

WWGD

No; the point was that if you have only a vector space structure, you cannot talk about division; you need at least a ring structure on your vectors. In a vector space alone, vectors cannot be divided.

7. Jun 15, 2014

Jhenrique

But if I have a problem like v= Mu, where u and v are known but M not, my solution isn't valid?

8. Jun 15, 2014

pasmith

You have exhibited an $M$ such that $\vec v = M\vec u$; the problem is that it isn't unique.

9. Jun 15, 2014

WWGD

Jhenrique: It would be easier to answer your question if you specified what v,u,M are, and where they "live", i.e., are v,u elements in a vector space, ring, etc? Is M just any matrix, or there specific coefficients you are using, specific dimension, etc.

10. Jun 15, 2014

Jhenrique

2D, M is a tensor and u and v are vectors.

The possible other answers will be just a change of basis. You still wants another answer better than formula in my first post?

11. Jun 15, 2014

micromass

Division of vectors is not defined. And if it were, it sure isn't going to be a matrix.

12. Jun 15, 2014

WWGD

The way I understand this is that you have a rotation matrix M, and you're then rotating u, to get v. So , you can rotate the plane by the necessary angle , and then rescale by the quotient of the norms of u, v ( and Dupree ). Then, yes, given vectors u,v in the plane, you can always get u from v by a combination of rotation and scaling : rotate by the difference from their angles , and then rescale by the quotient of their norms.