Complex eigenvalues

A matrix that turns every arrow keeps no line at all. Watch a point step round a spiral, and its eigenvalue sit beside it.

A turn has no real eigen-line

On Eigenvectors the quarter turn $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ kept no line: every arrow comes out at a right angle to itself. Its characteristic polynomial, from Finding eigenvalues, is $\lambda^2+1$, which no real number makes zero.

But the complex numbers $j$ and $-j$ do, since $j^2=-1$. Allow complex entries in the eigenvector too, and the equation works. For $\lambda=j$, take $\mathbf{v}=(1,-j)$:

So the quarter turn does have eigenvectors. They just aren’t arrows in the real plane. The matrix is not broken; its eigenvalue $j$ is the turn itself, as multiplying by $j$ turns a complex number a quarter turn in Complex numbers for signals.

Conjugate pairs

A real matrix has a characteristic polynomial with real coefficients. If a complex number $\lambda$ is a root, so is its conjugate $\lambda^*$, the same number with the sign of its imaginary part flipped: conjugating the whole equation leaves the real coefficients alone. So complex eigenvalues of a real matrix come in pairs, $a\pm jb$, and their eigenvectors are conjugates too.

Size and angle

The simplest matrices with complex eigenvalues have the pattern

It sends $(x,y)$ to $(ax-by,\;bx+ay)$. That is exactly what multiplying the complex number $x+jy$ by $a+jb$ does. So by the rotate-and-scale rule for complex numbers, the matrix stretches every arrow by the size $\lvert\lambda\rvert$ and turns it by the angle $\theta$ of $\lambda$:

For the spiral at the top of the page, $\lambda=0.72+0.54j$: its size is $\sqrt{0.72^2+0.54^2}=0.9$ and its angle 36.87°, which is 0.6435 rad. Each step shrinks the point to 0.9 of its distance and turns it by 36.87°. After 10 steps it is $0.9^{10}=0.349$ of the way out, and it has turned 368.7°: one full turn and 8.7° more.

• $\lvert\lambda\rvert<1$: the spiral winds in.
• $\lvert\lambda\rvert=1$: the point goes round a circle.
• $\lvert\lambda\rvert>1$: the spiral winds out.

Drag the eigenvalue across the unit circle in the picture and watch the spiral change from one to the next.

A tilted spiral

Most matrices with complex eigenvalues don’t have that tidy pattern. Take

Its trace is 2 and its determinant $0.75+0.5=1.25$, so $\lambda^2-2\lambda+1.25=0$ and $\lambda=1\pm0.5j$. The size is $\sqrt{1.25}=1.118$ and the angle 26.57°: each step grows by 11.8 % and turns by 26.57°.

The eigenvector for $1+0.5j$ is $\mathbf{v}=(1+j,\;1)$. Its real part is $(1,1)$ and its imaginary part $(1,0)$, and the matrix moves them into each other:

So on the slanted grid of $(1,1)$ and $(1,0)$, the matrix is a plain stretch-and-turn with the pattern above. In the usual grid the spiral looks squashed, like an ellipse; on its own grid it is round.

Oscillation

The same eigenvalues make things swing. A spring with light damping obeys, for example, $u''+u'+9.25\,u=0$. Written as a system, as on Linear differential equations, its matrix has characteristic polynomial $\lambda^2+\lambda+9.25$, with roots

Each part of the solution grows like $e^{\lambda t}=e^{-0.5t}\,e^{\pm3jt}$, and $e^{3jt}=\cos3t+j\sin3t$. The two conjugate parts combine into a real solution:

The real part of $\lambda$, $-0.5$, is how fast the swing dies. The imaginary part, 3, is how fast it swings, in radians per second. One swing takes $2\pi/3=2.094$ s, and in that time the size shrinks to $e^{-0.5\times2.094}=0.351$ of what it was.

Worked example

1. The quarter turn. $\lambda^2+1=0$ gives $\lambda=\pm j$, with eigenvector $(1,-j)$ for $j$.

2. The spiral. $\lambda=0.72+0.54j=0.9\,e^{j\,0.6435}$, an angle of 36.87°. After 10 steps the size is $0.9^{10}=0.349$, and the turn is 368.7°, which is 8.7° past a full turn. A full turn takes $360/36.87\approx9.8$ steps, by which time the spiral has shrunk to about 0.36.

3. The tilted matrix. $\begin{bmatrix}1.5&-1\\0.5&0.5\end{bmatrix}$: $\lambda=1\pm0.5j$, size 1.118, angle 26.57°. Its square, $\begin{bmatrix}1.75&-2\\1&-0.25\end{bmatrix}$, has trace 1.5 and determinant 1.5625. Those match $\lambda^2=0.75+j$: twice its real part is 1.5, and $\lvert\lambda^2\rvert^2=1.25^2=1.5625$.

4. The damped spring. $\lambda=-0.5\pm3j$: period 2.094 s, and the swing shrinks to 0.351 of its size each period.

Try these before you read the answers.

• Eigenvalues of a 90° turn? $\pm j$.
• Size of $0.6+0.8j$? 1: a pure turn, no shrinking.
• A real matrix has the eigenvalue $2+3j$. What other eigenvalue must it have? $2-3j$.
• An eigenvalue of size 1.1: does the spiral wind in or out? Out.

Where you’ll meet this

Anything that rings. A tuning fork, a lightly damped RLC circuit and a guitar string all swing while they die away. Each is a system with a complex pair of eigenvalues.

Resonant filters. A digital filter with a complex pair of poles rings, turning by the poles’ angle each sample and shrinking by their size each sample: the spiral of this page, sampled.

Noisy rhythms. A time series model whose next value is a fixed mix of the last two, plus noise, swings irregularly when its matrix has a complex pair of eigenvalues. Sunspot counts and some economic cycles are described this way.

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