Eigenpairs Of Planar Rotation Matrices

Rotations and Complex Eigenvectors

A rotation of the plane usually has no nonzero real eigenvector: it turns every nonzero real vector to a different direction unless the rotation is through a special angle. Over \(\mathbb{C}\), however, a planar rotation has eigenvectors for every angle. The eigenvalues record the angle through their positions on the unit circle. We will compute the eigenpairs directly, verify them by multiplication, and distinguish the general case from the angles where the two eigenvalues are equal.

Let \(\theta\in\mathbb{R}\) be an angle measured in radians, and write \(c=\cos\theta\) and \(s=\sin\theta\). The standard matrix for counterclockwise rotation through \(\theta\) is

We regard this real matrix as acting on column vectors in \(\mathbb{C}^2\), using the same matrix entries and allowing complex coordinates. For a matrix \(\mathbf{A}\in\mathbb{C}^{2\times2}\), an eigenpair is a scalar \(\lambda\in\mathbb{C}\) and a nonzero vector \(\mathbf{v}\in\mathbb{C}^2\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The eigenvectors associated with \(\lambda\), together with the zero vector, form its eigenspace.

Finding the Eigenvalues and Eigenvectors

An eigenvalue must make the homogeneous system \((\mathbf{R}_{\theta}-\lambda\mathbf{I}_2)\mathbf{v}=\mathbf{0}\) have a nonzero solution. For a \(2\times2\) matrix this occurs exactly when its determinant is zero. Here, direct expansion gives the characteristic equation

where the last equality uses \(c^2+s^2=1\). The quadratic formula, or direct factorization, gives the two roots \(c+is\) and \(c-is\). By Euler's formula, these are \(e^{i\theta}\) and \(e^{-i\theta}\). To determine the eigenvectors, substitute each value into the matrix equation and row-reduce.

Proof. The determinant calculation above shows that any eigenvalue must be a root of \(\lambda^2-2c\lambda+1\), and its roots are \(c+is\) and \(c-is\). For \(\lambda=c+is\) and \(s\ne0\), subtract \(\lambda\mathbf{I}_2\) from \(\mathbf{R}_{\theta}\), then divide the matrix by the nonzero scalar \(s\). Row operations give

The first step multiplies row 1 by \(i\); the second replaces \(R_2\) by \(R_2-R_1\). The system is \(x-iy=0\), or \(y=-ix\), so its solutions are the scalar multiples of \(\begin{bmatrix}1\\-i\end{bmatrix}\). For \(\lambda=c-is\), the corresponding matrix after division by \(s\) is \(\begin{bmatrix}i&-1\\1&i\end{bmatrix}\). Multiplying row 1 by \(-i\), then subtracting row 1 from row 2, gives \(\begin{bmatrix}1&i\\0&0\end{bmatrix}\). Thus \(x+iy=0\), or \(y=ix\), and the eigenspace is the scalar multiples of \(\begin{bmatrix}1\\i\end{bmatrix}\).

These vectors are nonzero, so each root is indeed an eigenvalue; this also verifies existence directly. If \(s=0\), then \(\theta=k\pi\) for some integer \(k\), and \(c=(-1)^k\). The matrix is \((-1)^k\mathbf{I}_2\). Its only eigenvalue is \((-1)^k\), and every vector in \(\mathbb{C}^2\) satisfies the eigenvector equation for that scalar. In particular, the two roots coincide and the eigenspace is all of \(\mathbb{C}^2\). \(\square\)

The condition \(\sin\theta\ne0\) matters when describing the eigenspaces as one-dimensional spans: at the exceptional angles \(k\pi\), the two displayed candidate vectors are still eigenvectors, but they do not describe every eigenvector. The full eigenspace is two-dimensional there.

Worked Eigenpair Computations

How the Angle Controls the Eigenvalues

The numbers \(e^{i\theta}\) and \(e^{-i\theta}\) lie on the unit circle because their moduli are both \(1\). Their arguments are \(\theta\) and \(-\theta\), with arguments understood modulo \(2\pi\). In this precise sense, the two complex eigenvalues encode the rotation angle with opposite signs. The angle is not encoded by two different values in every case: when \(\theta=k\pi\), the values agree, since \(e^{ik\pi}=e^{-ik\pi}=(-1)^k\).

The eigenvalues are nonreal exactly when \(\sin\theta\ne0\). In that case they are a conjugate pair: \(\overline{e^{i\theta}}=e^{-i\theta}\). This agrees with the general fact that a real matrix's nonreal eigenvalues occur in conjugate pairs. If \(\sin\theta=0\), both eigenvalues are real and equal to \(1\) or \(-1\), and the rotation matrix itself is that scalar times the identity.

Proof. The eigenvalues are \(c+is\) and \(c-is\), where \(c,s\in\mathbb{R}\). Either value is real exactly when \(s=0\), which is the condition \(\sin\theta=0\). The real zeros of sine are precisely \(\theta=k\pi\) for integers \(k\). At such an angle \(c=\cos(k\pi)=(-1)^k\), so \(\mathbf{R}_{\theta}=(-1)^k\mathbf{I}_2\). Multiplying any vector by this matrix gives \((-1)^k\) times that vector; consequently every nonzero vector is an eigenvector for the only eigenvalue. Conversely, if the eigenvalues are real, their imaginary parts \(s\) and \(-s\) vanish, forcing \(\sin\theta=0\), and therefore \(\theta=k\pi\). \(\square\)

Common Mistake: Pairing the Wrong Eigenvector

A frequent error is to switch the signs in the eigenvalue-eigenvector pairing. For the quarter-turn matrix, one might incorrectly claim that \(\begin{bmatrix}1\\i\end{bmatrix}\) is an eigenvector for \(i\). Testing the claim immediately shows the mismatch:

These columns are not equal, so the claimed eigenpair is wrong. The correction is to pair \(\begin{bmatrix}1\\i\end{bmatrix}\) with \(-i\), since \(-i\begin{bmatrix}1\\i\end{bmatrix}=\begin{bmatrix}-i\\1\end{bmatrix}\). The other vector, \(\begin{bmatrix}1\\-i\end{bmatrix}\), pairs with \(i\). Direct substitution is a reliable way to catch this sign error before using an eigenpair in a further calculation.

Another point to keep distinct is the repeated-eigenvalue case. At \(\theta=\pi\), the formula still gives the correct eigenvalue \(-1\), but reporting only the span of \(\begin{bmatrix}1\\-i\end{bmatrix}\) would omit other eigenvectors. Since \(\mathbf{R}_{\pi}=-\mathbf{I}_2\), every nonzero vector is an eigenvector. The special case is therefore about the size of the eigenspace, not a failure of the eigenvalue formula.

When computing an eigenpair, a complete check has two parts: confirm that the proposed eigenvector is nonzero, and multiply the original matrix by that vector to verify \(\mathbf{R}_{\theta}\mathbf{v}=\lambda\mathbf{v}\). The characteristic equation identifies possible eigenvalues; the vector equation confirms the associated eigenvectors and their sign pairing.