What is the relationship between two vectors in terms of rotation and scaling?

[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: $$\\ \frac{\vec{v}}{\vec{u}} = M$$

 

[Jilang]

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

 

[Jhenrique]

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

 

[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.

 

[Jhenrique]

But my answer is valid too, correct?

 

[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.

 

[Jhenrique]

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

 

[pasmith]

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

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

 

[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.

 

[Jhenrique]

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.

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

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

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

 

[micromass]

So: $$\\ \frac{\vec{v}}{\vec{u}} = M$$

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

 

[WWGD]

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: $$\\ \frac{\vec{v}}{\vec{u}} = M$$

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.