Complex Eigenvalues

A rotation by $90^\circ$ turns every direction, so it has no real eigenvectors. Its characteristic equation $\lambda^{2} + 1 = 0$ still has solutions: $\lambda = \pm i$. Complex eigenvalues are the algebra's way of saying "this matrix rotates". The polar form of a complex number tells us how much.

A real polynomial's complex roots come in conjugate pairs, so a real $2 \times 2$ matrix with a negative discriminant has eigenvalues $a \pm bi$ with $b \neq 0$.

Worked example: Complex eigenvalues

Find the eigenvalues of $A = \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix}$.

$\det(A - \lambda I) = \lambda^{2} - 4\lambda + 5$ (trace $4$, determinant $3 + 2 = 5$).
$\lambda$ $= \dfrac{4 \pm \sqrt{16 - 20}}{2}$ $= \dfrac{4 \pm 2i}{2}$ $= 2 \pm i$.

Answer: $2 + i$ and $2 - i$

Rotation and scaling

The matrix $C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ has eigenvalues $a \pm bi$ (its characteristic polynomial is $(a - \lambda)^{2} + b^{2}$). Write $a + bi = r(\cos\varphi + i\sin\varphi)$ with $r = \sqrt{a^{2} + b^{2}}$. Then $$C = r\begin{bmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{bmatrix},$$ a rotation by $\varphi$ followed by scaling by $r$. It does to vectors in the plane exactly what multiplying by $a + bi$ does to complex numbers.

Worked example: Reading off r and φ

$A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$. Describe its action.

It has the form $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ with $a = 1$, $b = 1$: eigenvalues $1 \pm i$.
$r = |1 + i| = \sqrt{2}$ and $\varphi = 45^\circ$.
So $A$ rotates by $45^\circ$ and stretches by $\sqrt{2}$. Repeating it makes points spiral outward.

Answer: Rotate $45^\circ$, scale by $\sqrt{2}$.

With complex eigenvalues there are no real eigenvector lines.

Multiplying by $z = 1 + i$ rotates by $45^\circ$ and scales by $\sqrt{2}$: the same action as the matrix.

Why it works

Why does every real $2 \times 2$ matrix with eigenvalues $a \pm bi$ behave like a rotation-scaling? Take a complex eigenvector $\mathbf{v}$ for $\lambda = a - bi$ and split it into real and imaginary parts. With $P = [\,\operatorname{Re}\mathbf{v}\ \ \operatorname{Im}\mathbf{v}\,]$, one can show $A = PCP^{-1}$ with $C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$. So in the right (slanted) coordinates, $A$ is a rotation-scaling: points move on ellipses, spiraling out if $r > 1$, in if $r < 1$, and staying on closed loops if $r = 1$.

Worked example: A complex eigenvector

Find an eigenvector of $C = \begin{bmatrix} 2 & -3 \\ 3 & 2 \end{bmatrix}$ for $\lambda = 2 - 3i$.

$C - \lambda I = \begin{bmatrix} 3i & -3 \\ 3 & 3i \end{bmatrix}$.
Row 1 with $\mathbf{v} = (1, y)$: $3i - 3y = 0$, so $y = i$.
Check row 2: $3 + (3i)(i) = 3 - 3 = 0$. ✓ The eigenvector for the conjugate $2 + 3i$ is the conjugate vector $(1, -i)$.

Answer: $(1, i)$

The same idea appears in differential equations: a system whose matrix has eigenvalues $a \pm bi$ produces oscillations with frequency $b$ that grow or decay like $e^{at}$.

Common mistake

Concluding "no eigenvalues" when the discriminant is negative. Over the real numbers there are no eigenvectors, but the eigenvalues exist as complex numbers, and they carry real information: $|\lambda|$ is the scale factor and the angle of $\lambda$ is the rotation.

Key idea

Complex eigenvalues $a \pm bi$ of a real $2 \times 2$ matrix mean rotation: the matrix acts, in suitable coordinates, like multiplication by $a + bi$, rotating by $\arg(a + bi)$ and scaling by $|a + bi|$.

Complex Eigenvalues

Rotation and scaling

Practice

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