Math Core

A rotation of the plane by $90^\circ$ moves every nonzero vector off its line, so it has no real eigenvectors at all. Yet its characteristic equation still has roots; they are just complex numbers. Allowing complex eigenvalues and eigenvectors completes the theory, and it also reveals a hidden structure: every real $2 \times 2$ matrix with complex eigenvalues is, in a suitable basis, a rotation combined with a scaling.

When the characteristic equation has no real roots

The rotation matrix $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ has $\operatorname{tr} A = 0$ and $\det A = 1$ , so its characteristic equation is $\lambda^2 + 1 = 0$ , with roots $\lambda = \pm i$ . There is no real solution of $A\mathbf{x} = \lambda\mathbf{x}$ , but if you let $\mathbf{x}$ have complex entries there is: $A(1, -i) = (i, 1) = i(1, -i)$ .

In general, the definitions from earlier in the unit carry over word for word to vectors in $\mathbb{C}^n$ and complex scalars. A complex scalar $\lambda$ is a (complex) eigenvalue of $A$ if $A\mathbf{x} = \lambda\mathbf{x}$ for some nonzero $\mathbf{x} \in \mathbb{C}^n$ . You find the eigenvalues from $\det(A - \lambda I) = 0$ and the eigenvectors by solving $(A - \lambda I)\mathbf{x} = \mathbf{0}$ with complex arithmetic.

For a real $2 \times 2$ matrix, the characteristic polynomial is $\lambda^2 - (\operatorname{tr} A)\lambda + \det A$ , and complex eigenvalues appear exactly when the discriminant $(\operatorname{tr} A)^2 - 4\det A$ is negative.

Complex eigenvalues come in conjugate pairs

If $A$ has real entries and $A\mathbf{x} = \lambda\mathbf{x}$ , then taking complex conjugates of both sides gives $A\overline{\mathbf{x}} = \overline{\lambda}\,\overline{\mathbf{x}}$ . So $\overline{\lambda}$ is also an eigenvalue, with eigenvector $\overline{\mathbf{x}}$ .

This saves half the work: once you have an eigenvector for $a - bi$ , conjugate every entry to get one for $a + bi$ .

Worked example: Eigenvalues and an eigenvector

Find the eigenvalues of $A = \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix}$ and an eigenvector for each.

$\operatorname{tr} A = 4$ and $\det A = 3 + 2 = 5$ , so $\lambda^2 - 4\lambda + 5 = 0$ and

\lambda = \frac{4 \pm \sqrt{16 - 20}}{2} = 2 \pm i.

For $\lambda = 2 - i$ :

A - (2 - i)I = \begin{bmatrix} -1 + i & -2 \\ 1 & 1 + i \end{bmatrix}.

Because $\lambda$ is an eigenvalue, this matrix is singular, so its two rows are multiples of each other and you only need one of them. The second row says $x_1 + (1 + i)x_2 = 0$ . Taking $x_2 = 1$ gives

\mathbf{v} = \begin{bmatrix} -1 - i \\ 1 \end{bmatrix}.

As a check, the first row gives $(-1 + i)(-1 - i) - 2 = (1 + 1) - 2 = 0$ . By conjugation, $\overline{\mathbf{v}} = (-1 + i, 1)$ is an eigenvector for $\lambda = 2 + i$ .

Tip

For a $2 \times 2$ matrix $A - \lambda I = \begin{bmatrix} p & q \\ r & s \end{bmatrix}$ that you know is singular, the vector $(-q, p)$ always solves the first equation, and therefore both. This avoids complex row reduction entirely (provided the first row is not all zero).

Rotation-scaling matrices

The model example of complex eigenvalues is a matrix of the form

C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}, \qquad a, b \text{ real, not both } 0.

Its characteristic polynomial is $(a - \lambda)^2 + b^2$ , so its eigenvalues are $\lambda = a \pm bi$ . Let $r = |\lambda| = \sqrt{a^2 + b^2}$ and let $\varphi$ be the angle with $\cos\varphi = a/r$ and $\sin\varphi = b/r$ . Then

C = r\begin{bmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{bmatrix}.

So $C$ rotates every vector by $\varphi$ and then scales it by $r$ . The eigenvalue $a + bi = re^{i\varphi}$ carries exactly this information: its modulus is the scale factor and its argument is the rotation angle.

Worked example: Reading off the rotation and scaling

Describe $C = \begin{bmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{bmatrix}$ geometrically.

Here $a = \sqrt{3}$ and $b = 1$ , so $r = \sqrt{3 + 1} = 2$ , $\cos\varphi = \sqrt{3}/2$ and $\sin\varphi = 1/2$ . Thus $\varphi = \pi/6$ : $C$ rotates by $30^\circ$ counterclockwise and doubles lengths. Its eigenvalues are $\sqrt{3} \pm i$ .

Iterating a rotation-scaling matrix produces a spiral. With $C = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$ ( $r = \sqrt{2}$ , $\varphi = 45^\circ$ ), starting from $\mathbf{x}_0 = (1, 0)$ gives $\mathbf{x}_1 = (1, 1)$ , $\mathbf{x}_2 = (0, 2)$ , $\mathbf{x}_3 = (-2, 2)$ and $\mathbf{x}_4 = (-4, 0)$ . No vector keeps its direction, which is why there are no real eigenvectors.

Repeatedly applying C = [1 −1; 1 1] rotates by 45° and stretches by √2 each time, so the iterates spiral outward.Open in grapher →

For $r > 1$ the iterates spiral outward, for $r < 1$ they spiral in toward the origin, and for $r = 1$ they circle forever at a fixed distance.

Every real 2 × 2 matrix with complex eigenvalues

The rotation-scaling picture is not special to matrices of the form $C$ . It describes every real $2 \times 2$ matrix with non-real eigenvalues, in the right coordinates.

Real factorization

Let $A$ be a real $2 \times 2$ matrix with a complex eigenvalue $\lambda = a - bi$ ( $b \ne 0$ ) and associated eigenvector $\mathbf{v}$ . Then

A = PCP^{-1}, \qquad P = \big[\operatorname{Re}\mathbf{v}\ \ \operatorname{Im}\mathbf{v}\big], \qquad C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}.

Here $\operatorname{Re}\mathbf{v}$ and $\operatorname{Im}\mathbf{v}$ are the real vectors formed from the real and imaginary parts of the entries of $\mathbf{v}$ . Note the sign convention: use the eigenvalue $a - bi$ (with the minus sign) to get $C$ in the form shown.

Worked example: Factoring A = PCP⁻¹

For $A = \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix}$ , the first example gave $\lambda = 2 - i$ with $\mathbf{v} = (-1 - i, 1)$ . So $a = 2$ , $b = 1$ , and

\mathbf{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} + i\begin{bmatrix} -1 \\ 0 \end{bmatrix}, \qquad P = \begin{bmatrix} -1 & -1 \\ 1 & 0 \end{bmatrix}, \qquad C = \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}.

Check $AP = PC$ instead of inverting $P$ :

AP = \begin{bmatrix} -3 & -1 \\ 2 & -1 \end{bmatrix}, \qquad PC = \begin{bmatrix} -3 & -1 \\ 2 & -1 \end{bmatrix}.

In the basis formed by the columns of $P$ , $A$ rotates by $\arctan(1/2) \approx 26.6^\circ$ and scales by $\sqrt{5}$ . In standard coordinates the orbits are ellipse-like spirals rather than circles.

Common mistake

Complex eigenvalues do not make a matrix "broken" or non-diagonalizable. $A$ above is diagonalizable over $\mathbb{C}$ (two distinct eigenvalues); it simply cannot be diagonalized using real matrices. Also remember that non-real eigenvalues of a real matrix always come in pairs, so a real $3 \times 3$ matrix always has at least one real eigenvalue.

Practice

Practice 1

The eigenvalues of $A = \begin{bmatrix} 1 & -5 \\ 2 & 3 \end{bmatrix}$ are $a \pm bi$ . Find the real part $a$ .