Linear Transformation involving pi/2

Click For Summary
The discussion focuses on understanding the matrix representation of linear transformations, specifically a counterclockwise rotation by an angle β, which is defined as π/2 in this case. The transformation of the point (1,0) to (cos(β), sin(β)) and (0,1) to (-sin(β), cos(β)) is explained through trigonometric identities. The matrix for a counterclockwise rotation by an arbitrary angle β is given as [[cos(β), -sin(β)], [sin(β), cos(β)]], and for β=π/2, it simplifies to [[0, -1], [1, 0]]. The discussion highlights the confusion around identifying the sine and cosine values and their signs based on the angle of rotation. Overall, the conversation clarifies the relationship between rotation matrices and trigonometric functions.
dwn
Messages
165
Reaction score
2
Resource: Linear Algebra (4th Edition) -David C. Lay

I understand that there are identities associated with transformations, but what I don't understand is when the transformation is rotated about the origin through an angle β. I believe β in this case is \frac{}{}\pi/2

\left[1,0\right] into [cos(\beta) , sin(\beta)]
\left[0,1\right] into [-sin(\beta), cos(\beta)]

Can someone please explain to me why this is the case? Why do these values suddenly translate to trig identities?

Thanks!
 
Physics news on Phys.org
I don't understand the question. "...β in this case..." What case?

Are you asking why (1,0) rotated counterclockwise by an angle of β is (cos β,sin β)? This is a very common way to define sin and cos. Do you want to use another definition?
 
I suppose my questions reiterates my confusion...haha.

There is something I'm not grasping in the definition of this counterclockwise rotation. How am I suppose to know the positive/negative values of the matrix and whether they're sin or cosine..? What if this type of rotation is not \pi/2?
 
\begin{pmatrix}\cos\beta & -\sin\beta\\ \sin\beta & \cos\beta\end{pmatrix} is the matrix representation of a counterclockwise rotation by an arbitrary angle β. When ##\beta=\pi/2##, we have ##\sin\beta=1## and ##\cos\beta=0##, so the matrix representation of a counterclockwise rotation by ##\pi/2## is
\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}
 
Sometimes its difficult "to see the wood for the forest". That's all I will say.

Thanks for your clarifying this point for me.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 26 ·
Replies
26
Views
878
  • · Replies 3 ·
Replies
3
Views
1K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K