Eigenvalues & Eigenvectors

Example 1:    ■

Example 2: A rotation matrix in two-dimensions is of the form (frame fixed and matrix acts from left on column vectors); \[ {\bf R}(\theta ) = \begin{bmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \phantom{-}\cos\theta \end{bmatrix} , \tag{2.1} \] which represents a proper counterclockwise (= positive) rotation by an angle θ in the x–y plane. Consider the eigenvalue/eigenvector problem \[ {\bf R}(\theta )\, {\bf x} = \lambda\,{\bf x} . \tag{2.2} \] Since R(θ) rotates the vector x by an angle θ, we conclude that for θ ≠ 0 (mod π), there are no real eigenvectors that are solutions to Eq. (2.2). This can be easily checked by an explicit calculation as follows. \begin{align*} \det \left( \lambda\mathb{I} - {\bf R} (\theta ) \right) &= \det \begin{bmatrix}\lambda - \cos\theta & \sin\theta \\ -\sin\theta & \lambda - \cos\theta \end{bmatrix} \\ &= \left( \lambda - \cos\theta \right)^2 + \sin^2 \theta \\ &= \lambda^2 - 2\lambda\,\cos\theta + 1 . \end{align*} The roots of the characteristic equation det(λI &mimus; R) = 0 are complex conjugate numbers \[ \lambda_{1,2} = \cos\theta \pm \sqrt{\cos^2 - 1} = \cos\theta \pm {\bf j} \sqrt{1 - \cos^2 \theta} = \cos\theta \pm {\bf j} \,\sin\theta = e^{{\bf j}θ} , \] where j is the imaginary unit on the complex plane ℂ, so j² = −1. This yields eigenvalues λ_1,2 to which correspond eigenvectors \[ \mathbf{x}_{1,2} = \left[ \pm{\bf j}, \ 1 \right] . \] Mathematica confirms: For the special cases of <>R(0) = I and R(π) = −I, corresponding to θ = 0 and π, respectively, we obtain real eigenvalues as expected. Hee I is the identity matrix, as usual. In particular, the case of θ = π corresponds to a two dimension inversion x → −x, which implies that the eigenvalue of R(π) is doubly degenerate and equal to −1.

Let us consider a composition of rotation and reflection, called the improper rotation: \begin{align*} \mathbf{M} (\theta ) &= \begin{bmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \phantom{-}\cos\theta \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 &-1 \end{bmatrix} \\ &= \begin{bmatrix} \cos\theta & \phantom{-}\sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} . \end{align*} You can verify that matrix M(θ) acts on vectors as a pure reflection through a line that passes through the origin.

\[ \begin{bmatrix} \cos\theta & \phantom{-}\sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x\,\cos\theta + y\,\sob\theta \\ x\,\sin\theta - y\,\cos\theta \end{bmatrix} . \] So equation \[ \mathbf{M} \left( \theta \right) \mathbf{v} = \lambda \,\mathbf{v} \] has a nontrivial solution only when λ is a root of quadratic equation \[ \det \left( \lambda \mathbf{I} - {\bf M} \right) = 0 \qquad \iff \qquad \left( \lambda - \cos\theta \right) \left( \lambda + \cos\theta \right) = -\sin^2 \theta . \] This equation is equivalent to \[ \lambda^2 -1 = 0 , \] which yields the eigenvalues, λ = ±1.

The interpretation of this result is immediate. The matrix M(θ) when operating on a vector v represents a reflection of that vector through a line of reflection that passes through the origin. In the case of λ = 1 we have M(θ) v = v, which means that v is a vector that lies parallel to the line of reflection (and is thus unaffected by the reflection). In the case of λ = −1 we have M(θ) v = −v, which means that v is a vector that is perpendicular to the line of reflection (and is thus is transformed, v → −v, by the reflection).

One can determine the line of reflection by computing the eigenvector that corresponds to λ = 1, \[ \begin{bmatrix} \cos\theta & \phantom{-}\sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} . \tag{2.3} \] If θ = 0 (mod 2π), then any vector of the form [x, 0]^T is an eigenvector corresponding to the eigenvalue λ = 1. This implies that the line of reflection is the x-axis, which corresponds to the equation y = 0. In general (for any value of θ), the solution to Eq. (2.) is \begin{align*} x\,\cos\theta + y\,\sin\theta &= x . \\ x \,\sin\theta - y\,\cos\theta &= y . \end{align*} Since the determinant of this system of equations is zero, it is sufficient to use only one equation. Using trigonometric identities of double angles \[ 2x\,\sin^2 \left( \frac{\theta}{2} \right) - 2y\,\sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right) = 0 \] I θ ≠0 (mod 2π), then we can divide both sides of the latter byin(θ/2) to obtain \[ x\, \sin \left( \frac{\theta}{2} \right) - y\,\cos \left( \frac{\theta}{2} \right) = 0. \] We recognize this equation as one for a straight line that passes through the origin with a slope equal to tan(θ/2). Thus, we have demonstrated that the most general 2 × 2 orthogonal matrix with determinant equal to −1 given by M(θ) represents a pure reflection through a straight line of slope tan(θ/2) that passes through the origin.

Finally, it is worth noting that since M(θ) is both an orthogonal matrix, M(θ) M(θ)^T = I, and a symmetric matrix, M(θ)^T = M(θ), it follows that \[ \left[ \mathbf{M} \right]^2 = \mathbf{I}, \] which is property that must be satisfied by a reflection matrix since two consecutive reflections are equivalent to the identity operation when acting on a vector.    ■

Example 3: The most general three-dimensional proper (rotating in counterclockwise direction) rotation matrix, which we henceforth denote by R(n, θ), can be specified by an axis of rotation pointing in the direction of the unit vector n, and a rotation angle θ. Conventionally, a positive rotation angle corresponds to a counterclockwise rotation. The direction of the axis is determined by the right hand rule. Namely, curl the fingers of your right hand around the axis of rotation, where your fingers point in the θ direction. Then, your thumb points perpendicular to the plane of rotation in the direction of n. All possible proper rotations correspond to 0 ≤ θ ≤ π and the unit vector n pointing in any direction.

To learn more about the properties of a general three-dimensional rotation, consider the matrix representation R(n, θ) with respect to the standard basis α = {i, j, k}. We can define a new coordinate system in which the unit vector n points in the direction of the new axis, for instance, z-axis; the corresponding new basis will be denoted by β. The matrix representation of the rotation with respect to β is then given by \[ \mathbf{R} \left( {\bf k} , \theta \right) = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0&0&1 \end{\bmatrix} , \] where the axis of rotation points in the z-direction (i.e., along the unit vector k).    ■

1. Anton, Howard (2005), Elementary Linear Algebra (Applications Version) (9th ed.), Wiley International
2. Beezer, R.A., A First Course in Linear Algebra, 2017.