5.5 Complex eigenvalues

Big idea

Rotate the plane a quarter turn. No line through the origin survives, so no real eigenvector exists, and the characteristic polynomial has no real root. Over the reals the matrix has no eigenvalues at all, which is an awkward place to stop.

Allowing complex scalars removes the obstruction. The characteristic polynomial of an $n \times n$ matrix always has $n$ complex roots counted with multiplicity, so every square matrix has a full set of eigenvalues once you accept complex ones. The arithmetic of eigenvectors is unchanged; only the entries are now complex numbers.

The payoff for a real matrix is geometric. A complex eigenvalue $a \pm bi$ of a real $2 \times 2$ matrix says the matrix is a rotation combined with a uniform scaling, seen in some skewed coordinate system. The modulus $\sqrt{a^2 + b^2}$ is the scaling factor, the argument is the rotation angle, and the real and imaginary parts of a complex eigenvector give the coordinate change that reveals it.

Decoder

If $\lambda = a - bi$ with $b \neq 0$ is an eigenvalue of a real $2 \times 2$ matrix $A$ with eigenvector $\mathbf{v}$, then $A = PCP^{-1}$ where $P = [\,\operatorname{Re}\mathbf{v} \;\; \operatorname{Im}\mathbf{v}\,]$ and $C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.

Unpacked: take the complex eigenvector, split it into a real part vector and an imaginary part vector, and stand those two real vectors up as the columns of $P$. Then $A$ looks like $C$ in the coordinate system $P$ sets up, and $C$ is a rotation by the angle of $\lambda$ scaled by the size of $\lambda$. It is the diagonalization pattern with $C$ standing in for $D$, because a real matrix will not become diagonal without complex entries but will become this.

Two details cause trouble. Use the eigenvalue with the negative imaginary part, $a - bi$, to match the signs in $C$. And the columns of $P$ come from the same eigenvector, not from two different ones.

Definitions and results

Complex eigenvalues and eigenvectors. For a square matrix $A$ with real or complex entries, a nonzero complex vector $\mathbf{v}$ with $A\mathbf{v} = \lambda\mathbf{v}$ for a complex $\lambda$ is an eigenvector. Every step of the real procedure carries over: eigenvalues are the roots of $\det(A - \lambda I) = 0$, and eigenvectors span the null space of $A - \lambda I$, computed with complex arithmetic.

Existence. Over the complex numbers every polynomial of degree $n$ has $n$ roots counted with multiplicity, so every $n \times n$ matrix has $n$ eigenvalues counted with multiplicity. No real matrix is left without eigenvalues.

Conjugate pairs. If $A$ has real entries and $\lambda$ is a non-real eigenvalue with eigenvector $\mathbf{v}$, then $\overline{\lambda}$ is also an eigenvalue, with eigenvector $\overline{\mathbf{v}}$, the vector of entrywise conjugates. Non-real eigenvalues of real matrices come in pairs, so an odd-sized real matrix always has at least one real eigenvalue.

Modulus and argument. Write $\lambda = a + bi$. Its modulus is $|\lambda| = \sqrt{a^2 + b^2}$ and its argument $\varphi$ is the angle from the positive real axis, with $\cos\varphi = a/|\lambda|$ and $\sin\varphi = b/|\lambda|$. Powers follow the polar rule: $\lambda^k$ has modulus $|\lambda|^k$ and argument $k\varphi$.

Rotation-scaling matrices. The matrix

$$ C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} $$

has trace $2a$ and determinant $a^2 + b^2$, so its characteristic polynomial is $\lambda^2 - 2a\lambda + (a^2+b^2)$ and its eigenvalues are $a \pm bi$. With $r = \sqrt{a^2+b^2}$ it factors as $r$ times a rotation matrix through the angle $\varphi$ of $a + bi$. Acting on the plane, $C$ rotates by $\varphi$ and scales by $r$.

The real factorization. Every real $2 \times 2$ matrix with non-real eigenvalues is similar to such a $C$, by the recipe in the decoder above. So up to a change of coordinates, there is only one kind of real $2 \times 2$ matrix with complex eigenvalues: rotate and scale.

Behaviour of powers. Since $A^k = PC^kP^{-1}$ and $C^k$ is a rotation through $k\varphi$ scaled by $r^k$, iterating $A$ spirals. Outward when $r > 1$, inward when $r < 1$, and around a closed orbit in the $P$-coordinates when $r = 1$. That is exactly the reading of $|\lambda|$ you already use for real eigenvalues, with a rotation added.

Larger matrices. For a real $n \times n$ matrix, each conjugate pair contributes a two-dimensional plane on which the transformation rotates and scales, and each real eigenvalue contributes a line on which it scales. Complex eigenvalues never mean something new is happening, only that the movement is circular rather than along a line.

Worked examples

A quarter turn

Let

$$ A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $$

The trace is $0$ and the determinant is $1$, so the characteristic polynomial is $\lambda^2 + 1$, with roots $\pm i$. No real root, so no real eigenvector, which matches the picture: this matrix rotates the plane by ninety degrees and leaves no line in place.

Find an eigenvector for $\lambda = i$. The shifted matrix is

$$ A - iI = \begin{bmatrix} -i & -1 \\ 1 & -i \end{bmatrix} $$

The second row says $x_1 = i x_2$. Take $x_2 = -i$, so $x_1 = i(-i) = 1$ and $\mathbf{v} = (1, -i)$. Check directly: $A(1,-i) = (0 \cdot 1 + (-1)(-i), \; 1 \cdot 1 + 0) = (i, 1)$, and $i \mathbf{v} = (i, \, i(-i)) = (i, 1)$. They agree. The conjugate $(1, i)$ is then an eigenvector for $-i$, as the pairing rule promises. Here $|\lambda| = 1$ and $\varphi = 90$ degrees: pure rotation, no scaling.

Extracting the rotation and the scaling

Let

$$ A = \begin{bmatrix} 3 & -2 \\ 1 & 1 \end{bmatrix} $$

The trace is $4$ and the determinant is $3 + 2 = 5$, so the characteristic polynomial is $\lambda^2 - 4\lambda + 5$ and the eigenvalues are $2 \pm i$.

Work with $\lambda = 2 - i$. The shifted matrix is

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

The second row gives $x_1 = (1 - i)x_2$. Take $x_2 = 1$, so $\mathbf{v} = (1 - i, \, 1)$. The first row confirms it: $(1+i)(1-i) - 2 = (1 + 1) - 2 = 0$.

Split the eigenvector: $\operatorname{Re}\mathbf{v} = (1,1)$ and $\operatorname{Im}\mathbf{v} = (-1, 0)$. So

$$ P = \begin{bmatrix} 1 & -1 \\ 1 & 0 \end{bmatrix}, \qquad C = \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix} $$

Check by computing $PC$ and comparing with $AP$. First $AP$: the columns are $A(1,1) = (1,2)$ and $A(-1,0) = (-3,-1)$, so $AP$ has columns $(1,2)$ and $(-3,-1)$. Now $PC$: its first column is $2(1,1) + 1(-1,0) = (1,2)$, and its second is $-1(1,1) + 2(-1,0) = (-3,-1)$. They match, so $AP = PC$ and hence $A = PCP^{-1}$.

Read the geometry off $C$. Here $r = \sqrt{4+1} = \sqrt{5}$, about $2.236$, and the rotation angle satisfies $\tan\varphi = 1/2$, so $\varphi$ is about $26.6$ degrees. Iterating $A$ spirals outward, turning roughly $26.6$ degrees and growing by a factor of about $2.236$ each step, in the coordinates the columns of $P$ define.

A power made easy by the angle

Let

$$ S = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} $$

This is already in rotation-scaling form with $a = b = 1$, so its eigenvalues are $1 \pm i$, its modulus is $\sqrt{2}$, and its angle is $45$ degrees. Eight steps turn through $360$ degrees and scale by $(\sqrt{2})^8 = 16$, so $S^8$ should be $16I$.

Verify with repeated squaring. First

$$ S^2 = \begin{bmatrix} 1-1 & -1-1 \\ 1+1 & -1+1 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} $$

which is $2$ times the quarter-turn matrix, as expected after two $45$ degree steps. Squaring again gives $S^4 = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix}$, a half turn scaled by $4$. One more squaring gives $S^8 = 16I$. The prediction holds, and no entry of the eight-fold product ever had to be computed directly.

Practice

Get the eigenvalues from the characteristic polynomial. These drills stay with real roots, but the polynomial step is identical when the discriminant turns negative: solve it anyway and keep the two complex roots.

Then the eigenvectors. Again the drills are real; the complex case is the same row reduction with complex arithmetic, and the worked examples above show each step.

Videos

Watch the part where a rotation is applied and no vector stays on its own line. The narration then names the complex eigenvalues and connects the angle of the eigenvalue to the angle of the rotation, which is the content of this section.

Quiz

Five items on eigenvalues from the characteristic polynomial and eigenvectors from the shifted matrix.