Can someone explain symmetries to me? Specifically in group theory context if there is any confusion. And while you're at it, how do I do this problem: Consider the matrix (1 0; 0 -1) [2x2 matrix]. Is it in the group O(2)? How about SO(2)? What effect does it have on a vector A?

Question

irishboy123 · Answer

@ikram002p

ikram002p · Answer

That is long topic to do... I'll see it tonight =).

kainui · Answer

This really comes down to definitions. What does the group O(2) or SO(2) even mean?

O(2) means all the Orthogonal 2x2 matrices and SO(2) means all the Special Orthogonal 2x2 matrices, which are orthogonal matrices with determinant 1.

Ok great, but what's an orthogonal matrix and why do we care? Orthogonal matrices satisfy this property: Their transpose is their own inverse. This seems weird on its own, but it turns out this is how all rotation and reflection matrices behave, they have to! (more on this later)

So for example, this is an arbitrary orthogonal matrix, $A$, so you can see its transpose is its inverse:
$$A^T A =I$$

So now you have enough knowledge to answer the first part of your question. The next part of your question, just do it. Take some arbitrary column vector $\begin{pmatrix}x \ y \end{pmatrix}$ and multiply it by your matrix and see what happens. Graph a few arbitrary points and their image after the transformation until you figure it out.

So, uhh, I guess ask questions and/or give it your best shot.

nuttyliaczar · Answer

Okay so then for the first part would the inverse be the same as the starting matrix?

kainui · Answer

Yeah it is

nuttyliaczar · Answer

Alright so then it does belong in O(2) right?

nuttyliaczar · Answer

To clarify, because the transpose is the inverse

kainui · Answer

Haha yes exactly, I was just about to ask for clarification, good job.

nuttyliaczar · Answer

Thank you ^^ and then since the determinant is not 1, is it not in group SO(2)?

nuttyliaczar · Answer

Oh and one more question about significant orthogonals. I'm guessing that it's a subset of O(2)? Meaning that it must be in O(2) for the matrix to even have a chance to be in SO(2)

nuttyliaczar · Answer

Woops meant to say "special" not significant

kainui · Answer

Yeah, SO(n) is a subgroup of O(n) more specifically, since multiplying matrices in SO(n) is closed, quick check, if A and B are in SO(n) then so is their product because:

$$\det(AB) = \det(A)\det(B) = 1$$

kainui · Answer

I guess I should have just said it more straightforward, if it's in SO(n) it's in O(n) you're right.

nuttyliaczar · Answer

I gotcha, now I'm just trying to understand that determinant rule you mentioned

nuttyliaczar · Answer

Well maybe that can wait till later, linear algebra is important but I just need it to the extent that physics asks for it. So to continue, it's not in SO(2). What is meant by "a vector A"? Is it just some vertical matrix then?

kainui · Answer

Ah, the determinant is essentially the area (volume, etc for higher dimensions) that gets stretched by in the transformations, yeah let's not worry too much about linear algebra for now.

What they mean is take some 2x1 column vector and multiply it out like this to figure out what the transformation is. |dw:1457023296712:dw| 

And just say what happens to it. Basically throwing an arbitrary vector at a matrix to test it out lets you understand what kind of transformation the matrix represents since you can see the before and after. I don't know if that quite answers your question or not.