Worked examples — Complex eigenvalues — rotation-scaling interpretation

This page is a drill through every kind of complex-eigenvalue problem you can meet. We build on Complex eigenvalues — rotation-scaling interpretation: the idea that a real $2\times 2$ matrix with complex eigenvalues $\lambda =\alpha \pm i\beta$ acts like rotate by $\theta$, scale by $r$ in the right basis.

Before touching numbers, let's pin down every symbol and convention we use, then list what kind of thing can go wrong or vary — so no case ambushes you.

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The scenario matrix

Every complex-eigenvalue problem is decided by two numbers read off the eigenvalue $\lambda =\alpha -i\beta$ (drawn as a point in the plane):

• $\alpha =\mathrm{Re}\lambda$ — the left/right part.
• $\beta$ — the imaginary size, which sets the spin.

From these you get $r=|\lambda |=\sqrt{{\alpha }^2+{\beta }^2}$ (scale) and $\theta =\arg \lambda$ (angle). Here is the full grid of cases, and which worked example nails each cell. (Below, "$r>1$" is the same statement as "$|\lambda |>1$".)

Case class	Trigger	What happens geometrically	Example
$r=1$, pure rotation	$\det A=1$, $(\mathrm{tr}{)}^2<4$	dots ride a circle, no growth	Ex 1
$r>1$, outward spiral (discrete)	$r=	\lambda	>1$
$r<1$, inward spiral (discrete)	$r=	\lambda	<1$
$\theta$ in each quadrant	sign of $\alpha$ and $\beta$	angle via $atan2$, not $\arctan$	Ex 4
Degenerate: distinct reals ($\beta =0$, disc $>0$)	discriminant $>0$	real eigenvalues, no rotation	Ex 5
Degenerate: repeated real root (disc $=0$)	discriminant $=0$	may fail to diagonalize (shear)	Ex 5b
Continuous flow $\dot{v}=Av$	sign of $\alpha =\mathrm{Re}\lambda$	growth ruled by $\alpha$, not $r$	Ex 6
Real-world word problem	given a spinning process	model + read off spin & scale	Ex 7
Exam twist: recover $A$	given $r,\theta ,P$	rebuild $A=PC{P}^{-1}$	Ex 8

Figure s01 — the eigenvalue as a point: its length is the scale $r$, its angle is the spin $\theta$.

Prerequisites to keep open: Trace and Determinant, Characteristic Polynomial, Complex Numbers - Polar Form, Rotation Matrices.

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Ex 1 — $r=1$: pure rotation (circle)

Forecast: guess the scale and angle before reading on. (Hint: it's already a rotation matrix.)

1. Trace and determinant. $\mathrm{tr}A=\sqrt{3}\approx 1.732$, $\det A=\frac{3}{4}+\frac{1}{4}=1$. Why this step? The characteristic quadratic is ${\lambda }^2-(\mathrm{tr}A)\lambda +\det A=0$; both numbers come straight from Trace and Determinant.
2. Discriminant. $(\mathrm{tr}A{)}^2-4\det A=3-4=-1<0$. Why this step? A negative discriminant is the signal that roots are a complex conjugate pair — a rotation is hiding here.
3. Eigenvalues. $\lambda =\frac{\sqrt{3}\pm \sqrt{-1}}{2}=\frac{\sqrt{3}}{2}\pm \frac{1}{2}i$. Why this step? Quadratic formula with $\sqrt{-1}=i$.
4. Pick $\lambda =\alpha -i\beta$: $\alpha =\frac{\sqrt{3}}{2},\beta =\frac{1}{2}$. Why this step? Our sign convention (defined at the top) fixes the choice so the block $C=\left(\begin{array}{cc}\alpha & -\beta \\ \beta & \alpha \end{array}\right)$ has $+\beta$ in its lower-left.
5. Scale and angle. $r=\sqrt{{\alpha }^2+{\beta }^2}=\sqrt{\frac{3}{4}+\frac{1}{4}}=1$; $\theta =atan2(\frac{1}{2},\frac{\sqrt{3}}{2})=30°$. Why this step? $r$ is the modulus (scale), $\theta$ the argument (spin) — the MAS rule from the top of the page.

Verify: $r=1$ means no growth, so points ride a circle; angle $30°$ matches the matrix we were handed. Twelve applications = $360°$ = back to start. ✓

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Ex 2 — $r>1$: outward spiral

Forecast: will points spiral in or out?

1. Trace/det. $\mathrm{tr}A=2$, $\det A=1-(-4)=5$.
2. Discriminant. $4-20=-16<0$ → complex pair.
3. Eigenvalues. $\lambda =\frac{2\pm \sqrt{-16}}{2}=1\pm 2i$. Take $\lambda =1-2i$, so $\alpha =1,\beta =2$. Why this step? $\sqrt{-16}=4i$, divide by 2.
4. Scale/angle. $r=\sqrt{1+4}=\sqrt{5}\approx 2.236$; $\theta =atan2(2,1)\approx 63.43°$. Why this step? Modulus = scale, argument = spin.
5. Direction. $r>1$ → each step magnifies → outward spiral (discrete map). Figure s02 traces the first steps.

Verify: $|\lambda |=\sqrt{5}$; check via $\det A=\lambda \bar{\lambda }=(1{)}^2+(2{)}^2=5$. ✓ (product of conjugate eigenvalues equals determinant.)

Figure s02 — iterating $A$ with $r>1$: each teal arrow is one step, turning $\approx 63.4°$ and growing outward.

Links: Linear Dynamical Systems, Phase Portraits.

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Ex 3 — $r<1$: inward spiral

Forecast: the numbers are small — bet on the direction.

1. Trace/det. $\mathrm{tr}A=1.2$, $\det A=0.36+0.09=0.45$.
2. Discriminant. $1.44-1.8=-0.36<0$ → complex.
3. Eigenvalues. $\lambda =\frac{1.2\pm \sqrt{-0.36}}{2}=0.6\pm 0.3i$. Take $\alpha =0.6,\beta =0.3$.
4. Scale/angle. $r=\sqrt{0.36+0.09}=\sqrt{0.45}\approx 0.6708$; $\theta =atan2(0.3,0.6)\approx 26.57°$.
5. Direction. $r<1$ → each step shrinks → inward spiral toward the origin.

Verify: $r^2=0.45=\det A$ ✓ (again $\det =\lambda \bar{\lambda }=r^2$). Since $r<1$, iterates converge to $0$. ✓

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Ex 4 — quadrant of $\theta$: why $atan2$, not $\arctan$

Forecast: naive $\arctan (\beta /\alpha )$ gives the same number for both. Why is that wrong?

1. The trap. Both have $\beta /\alpha =-1$, so $\arctan (-1)=-45°$ for both. Why this step? $\tan$ repeats every $180°$, so $\arctan$ can only ever return an angle in $(-90°,90°)$ — it cannot tell which side of the plane you're on.
2. Use signs of $\alpha$ and $\beta$ separately. That is exactly what $atan2(\beta ,\alpha )$ does — it looks at both coordinates and picks the correct quadrant.
3. Case (a): $\alpha =-1<0,\beta =1>0$ → point in the upper-left (second quadrant). $\theta =atan2(1,-1)=135°$.
4. Case (b): $\alpha =1>0,\beta =-1<0$ → lower-right (fourth quadrant). $\theta =atan2(-1,1)=-45°$.

Verify: $135°$ and $-45°$ differ by $180°$ — exactly the ambiguity $\arctan$ collapsed. The magnitudes: $r=\sqrt{2}$ for both. Figure s03 shows both points. ✓

Figure s03 — two eigenvalues with the same slope $-1$ but in opposite quadrants; only $atan2$ separates them.

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Ex 5 — degenerate: two distinct real roots ($\beta =0$, discriminant $>0$)

Forecast: upper-triangular — where do eigenvalues live for those?

1. Trace/det. $\mathrm{tr}A=5$, $\det A=6$.
2. Discriminant. $25-24=1>0$. Why this step? Positive discriminant → two distinct real roots → no rotation, $\beta =0$.
3. Eigenvalues. $\lambda =\frac{5\pm 1}{2}=3$ or $2$.
4. Interpretation. $\beta =0\Rightarrow \theta =0$; the block $C=\left(\begin{array}{cc}\alpha & -\beta \\ \beta & \alpha \end{array}\right)$ collapses to plain scaling along real eigen-lines. This is the boundary case of the whole topic: no spin.

Verify: product $3\cdot 2=6=\det A$; sum $3+2=5=\mathrm{tr}A$. ✓ Two distinct real eigenvalues, so Diagonalization over $\mathbb{R}$ works directly — no rotation-scaling block needed.

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Ex 5b — degenerate edge: a repeated real root (discriminant $=0$) that will not diagonalize

Forecast: with one eigenvalue "$2$" twice, do we get two eigenvectors or only one?

1. Trace/det. $\mathrm{tr}A=4$, $\det A=4$.
2. Discriminant. $16-16=0$. Why this step? Discriminant $=0$ is the razor's edge between two real roots and a complex pair — a repeated real root $\lambda =2$.
3. Eigenvalues. $\lambda =\frac{4\pm 0}{2}=2$ (twice).
4. Count eigenvectors. Solve $(A-2I)v=0$: $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)v=0\Rightarrow v_2=0$. Only one independent eigenvector, $\left(\begin{array}{c}1\\ 0\end{array}\right)$. Why this step? Diagonalizing needs two independent eigenvectors; here we get only one, so $A$ is not diagonalizable — it is a shear (a real eigen-line plus a slide), not a rotation and not a pure stretch.

Verify: $\det (A-2I)=0$ and rank$(A-2I)=1$, so the eigenspace is 1-dimensional → defective. No complex eigenvalues, no rotation, and no diagonalization: this is the case the discriminant-$=0$ boundary hides. ✓

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Ex 6 — continuous flow: growth ruled by $\alpha$, not $r$

Forecast: $|\lambda |$ here is bigger than 1. Does that mean it grows?

1. Eigenvalues. $\mathrm{tr}A=-0.4,\det A=0.04+9=9.04$; discriminant $0.16-36.16=-36<0$. $\lambda =-0.2\pm 3i$, so $\alpha =-0.2,\beta =3$.
2. Modulus. $r=\sqrt{0.04+9}=\sqrt{9.04}\approx 3.007$ — much bigger than 1.
3. Which rule? For the discrete map $v\mapsto Av$, $r>1$ would mean grow. For the continuous ODE $\dot{v}=Av$, the solution carries $e^{\alpha t}$, so growth is decided by $\mathrm{Re}\lambda =\alpha$. Why this step? Solutions of $\dot{v}=Av$ behave like $e^{\lambda t}$; the real part sits in the exponent, the imaginary part only spins.
4. Read $\alpha$. $\alpha =-0.2<0\Rightarrow e^{-0.2t}\to 0$: trajectory spirals inward and decays, despite $r>3$.

Verify: $\alpha =\mathrm{Re}\lambda =\frac{1}{2}\mathrm{tr}A=-0.2$ ✓ (real part = half the trace for a conjugate pair). Decay confirmed by $\alpha <0$.

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Ex 7 — word problem: a spinning, cooling turntable

Forecast: guess the spin (deg/sec) and the shrink rate.

1. Recognise the form. $A=0.9\cdot (\text{rotation by }40°)$, so $r=0.9,\theta =40°$ per second directly. Why this step? $A=rR(\theta )$ is already the rotation-scaling block $C$; the eigenvalue is $\lambda =r{e}^{-i\theta }$ (our $\lambda =\alpha -i\beta$ convention), with $|\lambda |=0.9$.
2. Spin rate. $\theta =40°$ per second.
3. Radius after $n$ seconds. Radius $\times r^n=0.9^n$. Set $0.9^n=0.5$.
4. Solve. $n=\frac{\ln 0.5}{\ln 0.9}=\frac{-0.6931}{-0.10536}\approx 6.579$ seconds. Why this step? Discrete iteration → scale is $r^n$; take logs to solve.

Verify: $0.9^{6.579}\approx 0.5$ ✓. Units: $n$ is dimensionless step count = seconds here. Spin $40°/\text{s}$ so half-life happens after about $6.58\times 40°\approx 263°$ of turning.

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Ex 8 — exam twist: rebuild $A$ from $r$, $\theta$, $P$

Forecast: would $A$ still look like a clean rotation matrix, or get distorted by $P$?

1. Build $C$. $C=r\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)=2\left(\begin{array}{cc}\cos 60°& -\sin 60°\\ \sin 60°& \cos 60°\end{array}\right)=\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)$. Why this step? $C$ is the rotation-scaling block defined at the top; $2\cos 60°=1,2\sin 60°=\sqrt{3}$.
2. Invert $P$. $\det P=2$, so $P^{-1}=\frac{1}{2}\left(\begin{array}{cc}2& -1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& -0.5\\ 0& 0.5\end{array}\right)$.
3. Assemble $A=PC{P}^{-1}$. $PC=\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)=\left(\begin{array}{cc}1+\sqrt{3}& 1-\sqrt{3}\\ 2\sqrt{3}& 2\end{array}\right).$ $A=(PC)P^{-1}=\left(\begin{array}{cc}1+\sqrt{3}& 1-\sqrt{3}\\ 2\sqrt{3}& 2\end{array}\right)\left(\begin{array}{cc}1& -0.5\\ 0& 0.5\end{array}\right)=\left(\begin{array}{cc}1+\sqrt{3}& -\sqrt{3}\\ 2\sqrt{3}& 1-\sqrt{3}\end{array}\right).$ Why this step? Similarity $A=PC{P}^{-1}$ carries the clean block $C$ into the standard basis; the skew of $P$ smears it into a non-rotation-looking matrix.

Verify: eigenvalues of $A$ must be $2e^{\pm i60°}=1\pm i\sqrt{3}$. Check via invariants: $\mathrm{tr}A=(1+\sqrt{3})+(1-\sqrt{3})=2=2\mathrm{Re}\lambda$ ✓; $\det A=r^2=4$ ✓ (see VERIFY). Numerically $\mathrm{tr}=2,\det =4$ give $\lambda =1\pm i\sqrt{3}$. ✓

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Active Recall

Below, each line uses the vault reveal format Prompt ::: Answer — everything before the ::: is the question, everything after is the hidden answer.