Linear Algebra

Complex eigenvalues and rotation-scaling matrices

A real matrix can turn every arrow in the plane, so that no real direction is kept and the characteristic equation has no real root. What do its complex roots say about the motion?

You need

The characteristic equation and finding eigenvaluesLinear Algebra Polar form of a complex numberGrades 11 to 12

Where you are 5 in Eigenvalues and eigenvectors

Linear algebra6

Eigenvalues and eigenvectors5

The characteristic equation and finding eigenvalues

Complex eigenvalues and rotation-scaling matrices

Finding an eigenvector Diagonalization Matrix powers, dynamical systems and steady states

Why this exists

A leaf floats on a whirlpool in a river. Sangmin records its position \mathbf{x}_k, in meters from the center of the whirlpool, once a second, and finds that one matrix carries each position to the next:

\mathbf{x}_{k+1} = W\mathbf{x}_k, \qquad W = \begin{bmatrix} 1 & -1 \\ \tfrac12 & 0 \end{bmatrix}.

The leaf circles as it drifts inward, so no line through the center is carried to itself, and W has no real eigenvector. The The characteristic equation and finding eigenvalues agrees: \lambda^2 - \lambda + \tfrac12 = 0 has discriminant 1 - 2 = -1 < 0. With real numbers only, that ends the analysis. With complex numbers it does not: the roots \tfrac12 \pm \tfrac12 i carry the whole motion. Their modulus \displaystyle \tfrac{\sqrt2}{2} is the factor by which the leaf closes in each second, and their argument 45^\circ is how far it turns, once the plane is measured in axes suited to the whirlpool. This lesson finds those axes.

Definition

Let A be a real n \times n matrix. A complex number \lambda is an eigenvalue of A when A\mathbf{v} = \lambda\mathbf{v} for some nonzero vector \mathbf{v} with complex entries; \mathbf{v} is a complex eigenvector. The eigenvalues are exactly the complex roots of \det(A - \lambda I), and the eigenvectors are found by row reducing A - \lambda I as before, with complex arithmetic. A complex eigenvalue here means a non-real one, a + bi with b \neq 0.

For a complex vector \mathbf{v}, the conjugate \bar{\mathbf{v}} conjugates each entry, and \mathbf{v} = \operatorname{Re}\mathbf{v} + i\operatorname{Im}\mathbf{v} splits it into two real vectors. For example (1 - 2i, 1) = (1, 1) + i(-2, 0).

A rotation-scaling matrix is a real matrix

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

Write a + bi in polar form, a + bi = r(\cos\theta + i\sin\theta), with modulus \displaystyle r = \sqrt{a^2 + b^2} and argument \theta, the angle from the positive real axis. Then a = r\cos\theta, b = r\sin\theta, and

C = r\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} = rR_\theta,

a turn through \theta by the rotation matrix R_\theta, followed by stretching every length by r.

Intuition

Treat the point (x, y) as the complex number x + yi. Multiplying by a + bi gives

(a + bi)(x + yi) = (ax - by) + (bx + ay)i,

which is exactly C\begin{bmatrix} x \\ y \end{bmatrix}. So C is “multiply by a + bi” written as a real matrix, and multiplying by a complex number turns by its argument and scales by its modulus. Applied again and again, C moves a point around a round spiral, inward when r < 1:

Figure 1The rotation-scaling matrix C = \tfrac12\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} applied again and again. Each step turns by 45^\circ and multiplies length by \displaystyle \tfrac{\sqrt2}{2}, the argument and modulus of its eigenvalue \tfrac12 + \tfrac12 i.

A general real 2 \times 2 matrix with complex eigenvalues is a rotation-scaling matrix in disguise. There is a basis, the columns of a matrix P, in which it acts exactly as C. Measured in that basis the motion is a turn by \theta and a stretch by r; seen in the standard axes the circles are squashed into ellipses, which is why the leaf’s path is lopsided. The eigenvalue keeps the angle and the scale; the eigenvector supplies the basis.

Results

Conjugate pairs. If A is real and A\mathbf{v} = \lambda\mathbf{v}, then A\bar{\mathbf{v}} = \bar{\lambda}\bar{\mathbf{v}}. So the complex eigenvalues of a real matrix come in conjugate pairs a \pm bi, with conjugate eigenvectors.

Odd size. A real matrix of odd size has at least one real eigenvalue.

Rotation-scaling matrices. C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} has eigenvalues a \pm bi, and C = rR_\theta with r = |a + bi| and \theta the argument of a + bi.

Uncovering C. Let A be a real 2 \times 2 matrix with eigenvalue \lambda = a - bi, b \neq 0, and eigenvector \mathbf{v}. Then P = \begin{bmatrix} \operatorname{Re}\mathbf{v} & \operatorname{Im}\mathbf{v} \end{bmatrix} is invertible and

A = PCP^{-1}, \qquad C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}.

Powers. In that case A^k = PC^kP^{-1} with C^k = r^kR_{k\theta}, and \det A = r^2 = |\lambda|^2. For \mathbf{x} \neq \mathbf{0}, the points A^k\mathbf{x} spiral in to \mathbf{0} when r < 1, stay on one ellipse when r = 1, and spiral out without bound when r > 1.

Why it is true

Conjugate pairs. Conjugation respects sums and products, so the conjugate of A\mathbf{v} is \bar{A}\bar{\mathbf{v}}, which is A\bar{\mathbf{v}} because A is real. The conjugate of \lambda\mathbf{v} is \bar{\lambda}\bar{\mathbf{v}}, and \bar{\mathbf{v}} \neq \mathbf{0}. The same argument on the polynomial: p(\lambda) = \det(A - \lambda I) has real coefficients, so p(\bar{\lambda}) = \overline{p(\lambda)}, and one is 0 exactly when the other is.

Odd size. For odd n, p(\lambda) has degree n and leading coefficient (-1)^n = -1, so it is positive for large negative \lambda and negative for large positive \lambda. Being continuous, it crosses 0 at some real \lambda (the intermediate value theorem).

Rotation-scaling matrices. \det(C - \lambda I) = (a - \lambda)^2 + b^2, which is 0 exactly when a - \lambda = \pm bi, that is \lambda = a \mp bi. The factorization C = rR_\theta was shown in the Definition.

Uncovering C. Write \mathbf{v} = \mathbf{x} + i\mathbf{y} with \mathbf{x}, \mathbf{y} real. Then

\begin{aligned} A\mathbf{x} + iA\mathbf{y} &= (a - bi)(\mathbf{x} + i\mathbf{y}) \\ &= (a\mathbf{x} + b\mathbf{y}) + i(-b\mathbf{x} + a\mathbf{y}). \end{aligned}

Since A\mathbf{x} and A\mathbf{y} are real, the real and imaginary parts match separately: A\mathbf{x} = a\mathbf{x} + b\mathbf{y} and A\mathbf{y} = -b\mathbf{x} + a\mathbf{y}. These are the two columns of AP = PC, because by the column rule of matrix multiplication the columns of PC are P times the columns (a, b) and (-b, a) of C. It remains to see that \mathbf{x}, \mathbf{y} are linearly independent. If \mathbf{x} = \mathbf{0}, then \mathbf{v} = i\mathbf{y} and \mathbf{y} is a real eigenvector for a - bi; if \mathbf{y} = c\mathbf{x} with c real, then \mathbf{v} = (1 + ci)\mathbf{x} and \mathbf{x} is. Either way a nonzero real vector \mathbf{w} would satisfy A\mathbf{w} = (a - bi)\mathbf{w}, whose left side is real and whose right side has imaginary part -b\mathbf{w} \neq \mathbf{0}. So P is invertible, and A = PCP^{-1}.

Powers. In (PCP^{-1})^k each P^{-1}P in the middle cancels, leaving PC^kP^{-1}. Rotations compose by adding angles (Composing transformations as a matrix product), so C^k = r^kR_\theta^k = r^kR_{k\theta}. Also \det A = \det P \det C \det P^{-1} = a^2 + b^2. Put \mathbf{c} = P^{-1}\mathbf{x}, the coordinates of \mathbf{x} in the basis \operatorname{Re}\mathbf{v}, \operatorname{Im}\mathbf{v}. Then P^{-1}A^k\mathbf{x} = C^k\mathbf{c} has length r^k\lVert\mathbf{c}\rVert, since a rotation keeps lengths. Those coordinates shrink to \mathbf{0}, keep a fixed length, or grow without bound as r < 1, r = 1, r > 1, and P carries them back: a circle of coordinates becomes an ellipse in the plane (stated here without proof).

Worked examples

Example 1

Day 1: A rotation-scaling matrix read at sight

Find the eigenvalues of C = \begin{bmatrix} 3 & -4 \\ 4 & 3 \end{bmatrix}, its scale and angle, and a complex eigenvector.

Solution

This is a rotation-scaling matrix with a = 3 and b = 4, so its eigenvalues are 3 \pm 4i. As a check, \det(C - \lambda I) = (3 - \lambda)^2 + 16, which is 0 when 3 - \lambda = \pm 4i.

The modulus is \displaystyle r = \sqrt{9 + 16} = 5. The argument has \cos\theta = \tfrac35 and \sin\theta = \tfrac45, so \theta = \arctan\tfrac43 \approx 53.13^\circ, in the first quadrant. Thus C = 5R_\theta, where R_\theta = \tfrac15\begin{bmatrix} 3 & -4 \\ 4 & 3 \end{bmatrix} is a rotation: C turns every arrow by about 53.13^\circ and makes it five times as long.

For \lambda = 3 - 4i,

C - (3 - 4i)I = \begin{bmatrix} 4i & -4 \\ 4 & 4i \end{bmatrix}.

The first row says 4ix - 4y = 0, so y = ix, and x = 1 gives \mathbf{v} = (1, i). Check: C\mathbf{v} = (3 - 4i, 4 + 3i), and (3 - 4i)\cdot i = 4 + 3i, so C\mathbf{v} = (3 - 4i)\mathbf{v}. By the conjugate-pair rule, (1, -i) is an eigenvector for 3 + 4i. Here \operatorname{Re}\mathbf{v} = (1, 0) and \operatorname{Im}\mathbf{v} = (0, 1), so P = I: a matrix already in rotation-scaling form needs no change of basis.

Example 2

Day 1: A matrix in disguise

Find the eigenvalues of A = \begin{bmatrix} 2 & -5 \\ 1 & 0 \end{bmatrix} and an eigenvector for each.

Solution

The trace is 2 and the determinant is 0 + 5 = 5, so p(\lambda) = \lambda^2 - 2\lambda + 5. The quadratic formula gives

\lambda = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i.

For \lambda = 1 - 2i,

A - (1 - 2i)I = \begin{bmatrix} 1 + 2i & -5 \\ 1 & -1 + 2i \end{bmatrix}.

The matrix is singular, so its rows are complex multiples of each other, and one row is enough. The second row says x + (-1 + 2i)y = 0, so x = (1 - 2i)y, and y = 1 gives \mathbf{v} = (1 - 2i, 1). The first row agrees: (1 + 2i)(1 - 2i) - 5 = 5 - 5 = 0.

Check directly: A\mathbf{v} = (2(1 - 2i) - 5,\ 1 - 2i), which is (-3 - 4i,\ 1 - 2i), and (1 - 2i)^2 = 1 - 4i - 4 = -3 - 4i, so A\mathbf{v} = (1 - 2i)\mathbf{v}. The conjugate \bar{\mathbf{v}} = (1 + 2i, 1) is an eigenvector for 1 + 2i with no further work.

Example 3

Day 1: Uncovering the rotation-scaling matrix

For A = \begin{bmatrix} 2 & -5 \\ 1 & 0 \end{bmatrix} from the last example, find P and C with A = PCP^{-1}, and describe what A does.

Solution

Use \lambda = 1 - 2i, so a = 1 and b = 2, with \mathbf{v} = (1 - 2i, 1) = (1, 1) + i(-2, 0). Then

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

and \det P = 2 \neq 0. Rather than invert P, check AP = PC:

AP = \begin{bmatrix} -3 & -4 \\ 1 & -2 \end{bmatrix} = PC.

So A = PCP^{-1}. The modulus is \displaystyle r = \sqrt{1 + 4} = \sqrt5 \approx 2.236 and the argument is \theta = \arctan 2 \approx 63.43^\circ. In the basis (1, 1), (-2, 0), each application of A turns by about 63.43^\circ and stretches by \displaystyle \sqrt5, and \det A = 5 = r^2 as the Powers result says.

Example 4

Day 2: The leaf in the whirlpool

Find the eigenvalues of Sangmin’s whirlpool matrix W = \begin{bmatrix} 1 & -1 \\ \tfrac12 & 0 \end{bmatrix}, a complex eigenvector, and P and C with W = PCP^{-1}. How far does the leaf turn each second, and by what factor does it close in?

Solution

The trace is 1 and the determinant is \tfrac12, so \lambda^2 - \lambda + \tfrac12 = 0 and

\lambda = \frac{1 \pm \sqrt{1 - 2}}{2} = \tfrac12 \pm \tfrac12 i.

Take \lambda = \tfrac12 - \tfrac12 i, so a = b = \tfrac12. The first row of W - \lambda I is \left(\tfrac12 + \tfrac12 i,\ -1\right), so y = \tfrac{1 + i}{2}x, and x = 2 gives \mathbf{v} = (2, 1 + i). The second row checks: \tfrac12 \cdot 2 + \left(-\tfrac12 + \tfrac12 i\right)(1 + i) = 1 + (-1) = 0.

Splitting \mathbf{v} = (2, 1) + i(0, 1),

P = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}, \qquad C = \begin{bmatrix} \tfrac12 & -\tfrac12 \\ \tfrac12 & \tfrac12 \end{bmatrix},

and both WP and PC equal \begin{bmatrix} 1 & -1 \\ 1 & 0 \end{bmatrix}. The modulus is \displaystyle r = \sqrt{\tfrac14 + \tfrac14} = \tfrac{\sqrt2}{2} \approx 0.707 and the argument of \tfrac12 + \tfrac12 i is 45^\circ. In the coordinates given by (2, 1) and (0, 1), the leaf turns 45^\circ counterclockwise each second and its distance from the center is multiplied by about 0.707. Since r < 1, it spirals in.

Example 5

Day 2: Where the leaf is after four and eight seconds

The leaf starts at \mathbf{x}_0 = (2, 0). Find W^4, W^8, \mathbf{x}_4 and \mathbf{x}_8.

Solution

Four steps turn by 4 \cdot 45^\circ = 180^\circ and scale by r^4 = \tfrac14, so C^4 = \tfrac14 R_{180^\circ} = -\tfrac14 I. Then

W^4 = PC^4P^{-1} = -\tfrac14 PP^{-1} = -\tfrac14 I,

and squaring, W^8 = \tfrac1{16}I. So \mathbf{x}_4 = -\tfrac14\mathbf{x}_0 = \left(-\tfrac12, 0\right) and \mathbf{x}_8 = \tfrac1{16}\mathbf{x}_0 = \left(\tfrac18, 0\right).

Stepping one second at a time confirms it: \mathbf{x}_1 = (2, 1), \mathbf{x}_2 = (1, 1), \mathbf{x}_3 = \left(0, \tfrac12\right), \mathbf{x}_4 = \left(-\tfrac12, 0\right). Every eight seconds the leaf is back on the same ray, one sixteenth as far out.

Figure 2The leaf under W = \begin{bmatrix} 1 & -1 \\ \tfrac12 & 0 \end{bmatrix}, which has the same eigenvalues as C. Seen in the standard axes the turn and the shrink are squashed into an elliptical spiral; four steps take the leaf to -\tfrac14 of where it started.

Example 6

Day 2: Tuning the whirlpool

Henok models stronger and weaker whirlpools with W_k = \begin{bmatrix} 1 & -1 \\ k & 0 \end{bmatrix}, k > 0; Sangmin’s is k = \tfrac12. For which k does the leaf circle at all, and for which does it spiral in, stay on one closed path, or spiral out?

Solution

The trace is 1 and the determinant is k, so p(\lambda) = \lambda^2 - \lambda + k with discriminant 1 - 4k. The eigenvalues are non-real exactly when k > \tfrac14. For 0 < k \le \tfrac14 they are real and positive, and the leaf does not circle.

For k > \tfrac14 the eigenvalues are a conjugate pair, and their product is the constant term: \lambda\bar{\lambda} = |\lambda|^2 = k, so \displaystyle r = \sqrt k. The leaf spirals in for \tfrac14 < k < 1, stays on one ellipse for k = 1, and spirals out for k > 1.

At k = 1 the eigenvalues are \displaystyle \tfrac12 \pm \tfrac{\sqrt3}{2}i, with modulus 1 and argument 60^\circ, so W_1^6 = PR_{360^\circ}P^{-1} = I and the leaf returns to its start every six seconds. Directly, W_1^3 = -I.

Boundaries

The conjugate-pair rule needs a real matrix. \begin{bmatrix} i & 0 \\ 0 & 2 \end{bmatrix} has eigenvalue i but not -i.
The angle and the scale are read in the basis \operatorname{Re}\mathbf{v}, \operatorname{Im}\mathbf{v}, not in the standard axes. The leaf’s actual distance from the center does not shrink by r every second: \lVert\mathbf{x}_0\rVert = 2 but \displaystyle \lVert\mathbf{x}_1\rVert = \sqrt5. Only over whole turns, or in the new coordinates, is the factor exact.
The sign of \theta depends on which eigenvalue of the pair is used with which eigenvector. Using a + bi with \bar{\mathbf{v}} flips the turn to -\theta; the modulus is the same for both.
When b = 0 the eigenvalue is real, C = aI, and nothing turns.
The Uncovering result is stated for 2 \times 2 matrices. For a larger real matrix the same computation still shows that A carries the plane spanned by \operatorname{Re}\mathbf{v}, \operatorname{Im}\mathbf{v} to itself and acts on it as C. Splitting all of \mathbb{R}^n into such planes and real eigenvector lines, one block for each eigenvalue or pair, needs extra conditions, stated here without proof: for instance that all the eigenvalues are distinct.

Where this goes next

Matrix powers, dynamical systems and steady states studies the long-run behavior of \mathbf{x}_{k+1} = A\mathbf{x}_k in general, with real and complex eigenvalues side by side; a conjugate pair with |\lambda| < 1 is one more way for a system to settle. Diagonalization is the real-eigenvalue counterpart of the factorization A = PCP^{-1}. The rule C^k = r^kR_{k\theta} is the matrix form of De Moivre’s theorem.

Exercises

Day 1: complex eigenvalues, eigenvectors and the rotation-scaling form

Find the eigenvalues of \begin{bmatrix} 2 & -3 \\ 3 & 2 \end{bmatrix}, its scale r and its angle \theta.
Find the eigenvalues of \begin{bmatrix} -1 & -1 \\ 1 & -1 \end{bmatrix}, its scale and its angle.
Write the rotation-scaling matrix that turns by 90^\circ and doubles every length, and give its eigenvalues.
Find the eigenvalues of A = \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix} and an eigenvector for 2 + i.
For A in Exercise 4, use \lambda = 2 - i to find P and C with A = PCP^{-1}, and give r and \theta.
For B = \begin{bmatrix} 4 & -5 \\ 2 & -2 \end{bmatrix}, find the eigenvalues, an eigenvector for 1 + i, and P and C built from 1 - i.
Is (1, i) an eigenvector of \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}? If so, for which eigenvalue?
Find all eigenvalues of Q = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} and the modulus and argument of each non-real one.
For which real k does \begin{bmatrix} 3 & k \\ 2 & 3 \end{bmatrix} have non-real eigenvalues, and what are they? For which k is their modulus 5?

Day 2: powers and spirals

For \displaystyle A = \tfrac12\begin{bmatrix} 1 & -\sqrt3 \\ \sqrt3 & 1 \end{bmatrix}, find r, \theta, A^3 and A^6.
For C = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}, find C^2, C^4 and C^{10}.
Say whether repeated application spirals in, stays on a closed path, or spirals out: (a) \begin{bmatrix} 0.6 & -0.8 \\ 0.8 & 0.6 \end{bmatrix}; (b) \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix}; (c) \tfrac15\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}.
Netsanet’s whirlpool has N = \begin{bmatrix} \tfrac12 & -\tfrac12 \\ \tfrac12 & 0 \end{bmatrix}. Find its eigenvalues, r and \theta, then N^3, and where a leaf starting at (4, 0) is after six seconds.
For which real t does \begin{bmatrix} t & -1 \\ 1 & t \end{bmatrix} make points spiral in?
A real 2 \times 2 matrix A has trace 0 and determinant 9. Find its eigenvalues, r and \theta, and show that A^2 = -9I whatever its entries.
For W_k in the last worked example, find the k for which the leaf turns 30^\circ per step in the new coordinates, and say whether it then spirals in.

Reasoning

Explain why a real 2 \times 2 matrix with non-real eigenvalues has no real eigenvector.
Prove that a real 2 \times 2 matrix with eigenvalues a \pm bi has \operatorname{tr} A = 2a and \det A = a^2 + b^2.
Explain why no real 3 \times 3 matrix has eigenvalues i, 2i and 3i.
Show that R_\theta has eigenvalues \cos\theta \pm i\sin\theta, and that it has a real eigenvalue only when \theta is a multiple of 180^\circ.
In the Uncovering result, use the eigenvalue a + bi and its eigenvector \bar{\mathbf{v}} = \mathbf{x} - i\mathbf{y} instead. Show that P' = \begin{bmatrix} \mathbf{x} & -\mathbf{y} \end{bmatrix} gives A = P'C'P'^{-1} with C' = \begin{bmatrix} a & b \\ -b & a \end{bmatrix}, and explain why the turn reverses.

Answers

2 \pm 3i; \displaystyle r = \sqrt{13}, \theta = \arctan\tfrac32 \approx 56.31^\circ. It is in rotation-scaling form with a = 2, b = 3.
-1 \pm i; \displaystyle r = \sqrt2, \theta = 135^\circ. Here a = -1, b = 1, and -1 + i lies in the second quadrant.
\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}, with eigenvalues \pm 2i. It is 2R_{90^\circ}, so a = 0, b = 2.
2 \pm i; \mathbf{v} = (-1 + i, 1). Trace 4, determinant 5, \lambda^2 - 4\lambda + 5 = 0. The second row of A - (2 + i)I is (1,\ 1 - i), so x = -(1 - i)y.
P = \begin{bmatrix} -1 & -1 \\ 1 & 0 \end{bmatrix}, C = \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}, \displaystyle r = \sqrt5, \theta \approx 26.57^\circ. The eigenvector for 2 - i is the conjugate (-1 - i, 1) = (-1, 1) + i(-1, 0), and AP = PC = \begin{bmatrix} -3 & -1 \\ 2 & -1 \end{bmatrix}.
1 \pm i; for 1 + i, (3 + i, 2); P = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}, C = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}. Trace 2, determinant 2. The second row of B - (1 + i)I gives 2x = (3 + i)y. For 1 - i use (3 - i, 2) = (3, 2) + i(-1, 0); both BP and PC equal \begin{bmatrix} 2 & -4 \\ 2 & -2 \end{bmatrix}.
Yes, for -i: the product is (-i, 1), and -i \cdot (1, i) = (-i, 1).
1 and \displaystyle -\tfrac12 \pm \tfrac{\sqrt3}{2}i; the pair has modulus 1 and arguments \pm 120^\circ. Expanding along the first row, \det(Q - \lambda I) = -\lambda^3 + 1, so the eigenvalues are the cube roots of 1. (Q sends (x, y, z) to (z, x, y), a turn of space about the line through (1, 1, 1).)
k < 0, with eigenvalues \displaystyle 3 \pm i\sqrt{-2k}; modulus 5 when k = -8. The equation (3 - \lambda)^2 = 2k gives \displaystyle \lambda = 3 \pm \sqrt{2k}, non-real exactly when k < 0. Then |\lambda|^2 = 9 - 2k, the determinant, and 9 - 2k = 25 gives k = -8, with eigenvalues 3 \pm 4i.
r = 1, \theta = 60^\circ, A^3 = -I, A^6 = I. A = R_{60^\circ}, so A^3 = R_{180^\circ} and A^6 = R_{360^\circ}.
C^2 = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}, C^4 = -4I, C^{10} = \begin{bmatrix} 0 & -32 \\ 32 & 0 \end{bmatrix}. \displaystyle C = \sqrt2 R_{45^\circ}, so C^2 = 2R_{90^\circ}, C^4 = 4R_{180^\circ}, and C^{10} = C^8C^2 = 16C^2.
(a) Closed path, r = 1: it is a rotation. (b) Out: the eigenvalues are 2 \pm i and r^2 = \det = 5. (c) In: trace \tfrac35, determinant \tfrac3{25}, discriminant \tfrac9{25} - \tfrac{12}{25} < 0, and \displaystyle r = \tfrac{\sqrt3}{5} < 1.
\displaystyle \tfrac14 \pm \tfrac{\sqrt3}{4}i, r = \tfrac12, \theta = 60^\circ, N^3 = -\tfrac18 I, and the leaf is at \left(\tfrac1{16}, 0\right). Trace \tfrac12 and determinant \tfrac14 give \lambda^2 - \tfrac12\lambda + \tfrac14 = 0. Three steps turn 180^\circ and scale by \tfrac18; six steps give N^6 = \tfrac1{64}I.
No t. The eigenvalues are t \pm i, with \displaystyle r = \sqrt{t^2 + 1} \ge 1: a closed path when t = 0, outward otherwise.
\pm 3i, r = 3, \theta = 90^\circ (using 3i). Here C = \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix} and C^2 = -9I, so A^2 = PC^2P^{-1} = -9I.
k = \tfrac13, and it spirals in. For k > \tfrac14, \displaystyle \lambda = \tfrac12 \pm \tfrac12\sqrt{4k - 1}\,i, so \displaystyle \tan\theta = \sqrt{4k - 1}. Setting this to \displaystyle \tan 30^\circ = \tfrac{1}{\sqrt3} gives 4k - 1 = \tfrac13, and \displaystyle r = \sqrt{k} = \tfrac{\sqrt3}{3} < 1.
A real eigenvector \mathbf{w} would give A\mathbf{w} = \lambda\mathbf{w} with the left side real and \mathbf{w} \neq \mathbf{0} real, forcing \lambda real.
p(\lambda) = \lambda^2 - (\operatorname{tr} A)\lambda + \det A, and also p(\lambda) = (\lambda - (a + bi))(\lambda - (a - bi)), which expands to \lambda^2 - 2a\lambda + (a^2 + b^2); compare coefficients.
The non-real eigenvalues of a real matrix come in conjugate pairs, so -i, -2i and -3i would be eigenvalues too, six roots for a polynomial of degree 3.
\det(R_\theta - \lambda I) = (\cos\theta - \lambda)^2 + \sin^2\theta, so \lambda = \cos\theta \pm i\sin\theta, real only when \sin\theta = 0.
A\bar{\mathbf{v}} = (a + bi)(\mathbf{x} - i\mathbf{y}) has real part a\mathbf{x} + b\mathbf{y} and imaginary part b\mathbf{x} - a\mathbf{y}, so A\mathbf{x} = a\mathbf{x} - b(-\mathbf{y}) and A(-\mathbf{y}) = b\mathbf{x} + a(-\mathbf{y}): the columns of AP' = P'C'. C' = rR_{-\theta} turns the other way because \det P' = -\det P: the new basis has the opposite orientation, which swaps clockwise and counterclockwise.