This must be very elementary, but I am really stuck after much reading.
Given a vector v and a basis A, to find v's coordinates in a new basis A' which is a rotation of A by angle θ, I understand that you must proceed as follows:
a) Find the coordinates in the original basis A of the unit vectors of the new basis (say e1' and e2'), which happen to be as shown in this picture:
b) To get coordinate v1' in the new basis A', make the dot product between v (coordinates in A) and e1' (also as per coordinates in A); whereas for v2', you would need dot product with again v but e2' in this case.
My problem is with a), second part. I see that a projection of e1' over X axis gives out cosθ as x coordinate and sinθ as y coordinate. But if we repeated the same projection with e2 over X axis, we would get -cosθ and sinθ, which is not the right answer.
I tend to think that my initial mistake, in the second problematic case, was projecting over X. After all, here the rotation starts from e2, not from e1, even if it is the same number of degrees… So I hypothesized that I should project e2' over e2 and then somehow convert the result into the language of the initial angle for the sake of homogeneity. Thus for the first coordinate I got a cos, which would be equivalent of a sine in terms of the first angle, but why then the negative sign, so I am lost…
Best Answer
This exercise involves the two complementary ideas of active and passive transformation: (a) rotating a vector, and obtaining its new coordinates in a fixed coordinate system, and (b) rotating the coordinate system, and obtaining the new coordinates of a fixed vector.
For part (a) the coordinates of the rotated basis vectors $\mathbf{e}_1'$ and $\mathbf{e}_2'$ are already correctly marked on the diagram: they are respectively $(\cos\alpha,\sin\alpha)$ and $(-\sin\alpha,\cos\alpha)$. (In your description, $\theta$ is the same as $\alpha$). I can't quite follow why you are having difficulty with the second of these: if you take the point at the end of the rotated vector $\mathbf{e}_2'$ and project (vertically) onto the original $x$ axis, the intercept is $-\sin\alpha$, while if you project (horizontally) onto the original $y$ axis, the intercept is $\cos\alpha$.
For part (b), the procedure is exactly as you describe: for some arbitrary fixed vector $\mathbf{v}$ you need to work out the projections onto the new axis system. One way to do this is to compute the scalar products with the new basis vectors, and you can do this if you express $\mathbf{v}$, $\mathbf{e}_1'$ and $\mathbf{e}_2'$ in the original coordinate system. But, if I understand you correctly, you don't have a problem with that.
Both active and passive rotations can be expressed as the product of a vector with a rotation matrix of the form $$ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \qquad \text{or} \qquad \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} $$ Knowing which one to use, and the sign convention to adopt for clockwise and anticlockwise rotation angles, are always sources of confusion. The best way to avoid this is to work through examples of exactly the kind that you are doing.