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A Bit About Rotations
Rotations operate by transferring values from one axis to another.
2D Rotations
This is the 2D rotation equation:
\( \textbf{R}_\theta \left[ \matrix{ x \\ y} \right] = \left[ \matrix{ cos\theta & -sin\theta \\ sin\theta & cos\theta } \right] \left[ \matrix{x \\ y} \right] \)
This matrix has a similarity Euler's formula for imaginary exponents:
\( exp(i\theta) = cos\theta + i sin\theta \)
Observing multiplication operations on this exponential map of imaginary numbers reveals:
\( exp(i\theta) \textbf{x} \)
\( = exp(i\theta)(x+iy) \)
\( = (cos\theta + i sin\theta)(x+iy) \)
\( = x cos\theta + i(x sin\theta + y cos\theta) + i^2 y sin\theta \)
\( = (x cos\theta - y sin\theta) + i (x sin\theta + y cos\theta) \)
Rewriting this as a linear space with basis \(1\) and \(i\) gives us the following:
\( exp(i\theta) \textbf{x} \)
\( = exp(i\theta) \left[ \matrix{x\\y} \right] \)
\( = \left[ \matrix{ cos\theta & -sin\theta \\ sin\theta & cos\theta } \right] \left[ \matrix{x\\y} \right] \)
Take note that any complex number, \(a+ib\), can be treated as a 2x2 matrix reprsented as \( \left[ \matrix{a & -b \\ b & a} \right] \).
From this we can see that, in complex space, \( exp(i\theta) \) operates in the same way that rotations do in a 2D vector space.
Now let's look at the motion of the vector as it performs the rotation.
Let's watch \(\textbf{v}\) as the angle \(\theta\) changes. Consider the derivative \( {{d\textbf{R}_\theta \textbf{v}}\over{d\theta}}\):
\( {d\over{d\theta}} ( \textbf{R}_\theta \textbf{v} ) \)
\( = {{d\textbf{R}_\theta}\over{d\theta}} \textbf{v} \)
\( = {d\over{d\theta}} \left[ \matrix{ cos\theta & -sin\theta \\ sin\theta & cos\theta } \right] \textbf{v} \)
\( = \left[ \matrix{ -sin\theta & -cos\theta \\ cos\theta & -sin\theta } \right] \textbf{v} \)
\( = \left[ \matrix{ 0 & -1 \\ 1 & 0 } \right] \left[ \matrix{ cos\theta & -sin\theta \\ sin\theta & cos\theta } \right] \textbf{v} \)
\( = \left[ \matrix{0 & -1 \\ 1 & 0} \right] \textbf{R}_\theta \textbf{v} \)
Evaluate this at \( \theta = 0 \) and we get the following general rule for rotations:
\( \frac{d\textbf{v}}{d\theta} = \left[ \matrix{0 & -1 \\ 1 & 0} \right] \textbf{v} \)
Notice that the derivative of a rotating vector is that same vector multiplied by a 90' rotation matrix.
That is, the derivative to a vector is always at a right angle to that vector.
We can see this unfold much more quickly when considering complex rotations.
\( \frac{d}{d\theta} (exp(i\theta) \textbf{v}) \)
\( = \frac{d}{d\theta} exp(i\theta) \textbf{v} \)
\( = i exp(i\theta) \textbf{v} \)
In matrix form with linear basis \(1\) and \(i\) this looks strikingly similar:
\( = \left[ \matrix{0 & -1 \\ 1 & 0} \right] exp(i\theta) \textbf{v} \)
Keep note of the analogy between \( \textbf{R}_\theta \) and \( exp(i\theta) \).
Now let's look at the eigen-decomposition of this 90' rotation matrix:
\( \left| \matrix{ -\lambda & -1 \\ 1 & -\lambda } \right| = 0 \)
\( \lambda^2 + 1 = 0 \)
\( \lambda = \pm i \)
The corresponding eigenvectors are:
for \( \lambda = i \):
\( \left| \matrix{-i & -1 \\ 1 & -i} \right| = 0 \)
\( \left| \matrix{-i & -1 \\ i & 1} \right| = 0 \)
\( \left| \matrix{-i & -1 \\ 0 & 0} \right| = 0 \)
This is true for the eigenvector \( \xi_i = \left[ \matrix{1 \\ -i} \right] \).
Solving for the other gives us the eigenvector \( \xi_{-i} = \left[ \matrix{1 \\ i} \right] \).
Here we see a connection between complex eigenvalue conjugate pairs and planar rotations.
Performing the eigen-decomposition on our rotation matrix \( \textbf{R}_\theta \) gives us the following:
\( \left| \matrix{ cos\theta - \lambda & -sin\theta \\ sin\theta & cos\theta - \lambda } \right] = 0 \)
\( (cos\theta - \lambda)^2 + (sin\theta)^2 = 0 \)
\( (cos\theta)^2 - 2 \lambda cos\theta + \lambda^2 + (sin\theta)^2 = 0 \)
\( 1 - 2 \lambda cos\theta + \lambda^2 = 0 \)
\( \lambda = {1 \over 2} \left( 2cos\theta \pm \sqrt{2 (cos\theta)^2 - 4} \right) \)
\( \lambda = cos\theta \pm \sqrt{(cos\theta)^2 - 1} \)
\( \lambda = cos\theta \pm \sqrt{-(sin\theta)^2} \)
\( \lambda = cos\theta \pm i sin\theta \)
