Two questions inspired by the rotation by 90 degrees

Question 1: Rotating by 180 degrees has unlimited eigenvectors?

Eigenvector’s definition is that the original vector changes by a constant vector after a linear transformation is applied. So if a linear transformation causes the vector to turn 180 degree, which is np.array( [ [-1, 0], [0, -1] ] ), is there unlimited eigenvectors because all vectors actually just turn to the opposite direction, which is considered a change of the original vector by -1?

Question2: Do all non-180 degree rotations end up with zero eigenvectors? For ex, the code snip below is rotating by 45 degrees

###############################Code##############################
a=np.sqrt(2)/2

B_rotation = np.array([[a, -a], [a, a]])
B_ratation_eig = np.linalg.eig(B_rotation)

print(“Matrix B_rotation:\n”, B_rotation,
“\n\n Eigenvalues and eigenvectors of matrix B_rotation:\n”, B_ratation_eig)

##############################Output##############################
Matrix B_rotation:
[[ 0.70710678 -0.70710678]
[ 0.70710678 0.70710678]]

Eigenvalues and eigenvectors of matrix B_rotation:
(array([0.70710678+0.70710678j, 0.70710678-0.70710678j]), array([[ 0. -0.70710678j, 0. +0.70710678j], [-0.70710678+0.j , -0.70710678-0.j ]]))

Hi @Paige_Yang!

Thanks for your post and sorry for the delay in answering it. I will try to answer all your questions.

Question 1. I’d say you are partially correct. You are correct in the sense that in a rotation of 180 degrees every vector is an eigenvector, since a rotation by 180 degrees is just “stretching” any vector backwards. You can see this by noting that, in this case, the eigenvalue is \lambda = -1 and

\left[ \begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = - \left[ \begin{array}{c} x \\ y \end{array} \right]

is satisfied by every pair (x,y) \in \mathbb{R}^2. The “partially” comes from saying that it has unlimited eigenvectors. This is because, if a (non-zero) vector is an eigenvector, then every multiple is also an eigenvector, thus if a linear transformation has one (non-zero) eigenvectors, then it has infinite, thus an unlimited number of them.

Question 2. This is a good question. Geometrically you can see that any rotation that the resulting vector does not lie in the same direction that the initial vector cannot have eigenvectors. Just note that we are implicitly limiting the rotations to satisfy \theta \in [0, 360], otherwise any rotation that is a multiple of 180 has every vector as an eigenvector. To see that this is the case, let’s do some calculations.

The rotation matrix in the plane is given by

R_{\theta} = \left[ \begin{array}{cc} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end {array}\right]

So, the eigenvalues can be found by computing the determinant of

R_{\theta} = \left[ \begin{array}{cc} \cos \theta -\lambda & - \sin \theta \\ \sin \theta & \cos \theta - \lambda \end {array}\right]

Which is equal to (\cos \theta - \lambda)^2 + \sin^2 \theta = \cos^2 \theta - 2 \lambda \cos \theta + \lambda^2 + \sin^2 \theta. Using the fact that \sin^2 \theta + \cos^2 \theta = 1 the equation becomes \cos^2 \theta - 2 \lambda \cos \theta + \lambda^2 + \sin^2 \theta = 1 + \lambda^2 - 2\lambda \cos \theta. So the eigenvalues are such that this equation is equal to 0.

1 + \lambda^2 - 2\lambda \cos \theta = 0 \rightarrow \cos \theta = \dfrac{1 + \lambda^2}{2\lambda}

By analyzing the function above (I will not do this since it is quite lenghty, but you can take a look here to convince yourself), this function has only values greater or equal to 1 or less or equal to -1. Since \cos \theta \in [-1, 1], the only values that can occur (in the real plane) are \lambda = 1 or \lambda = -1. In the former, this implies \cos \theta = 1 \rightarrow \theta = 2k \pi for k \in \mathbb{Z} (this means \theta can be 0, 360, -360, 720, -720 and so on. Or \cos \theta = -1 which implies \theta = (2k + 1) \pi, for k \in \mathbb{Z} (this implies \theta = 180, -180, 540, -540, and so on).

I hope this answers yout questions!

Cheers,
Lucas