Exercises — Complex eigenvalues — rotation-scaling interpretation

WHAT we compute: the discriminant $D=(\mathrm{tr}A{)}^2-4\det A$. WHY: its sign alone decides real vs complex (a negative number under a square root is what manufactures the $i$).

$M_1$: $\mathrm{tr}=4,\det =(2)(2)-(-3)(1)=4+3=7.$ $D=16-28=-12<0\Rightarrow$ complex. $M_2$: $\mathrm{tr}=3,\det =(3)(0)-(1)(4)=-4.$ $D=9-4(-4)=9+16=25>0\Rightarrow$ real (in fact $\lambda =\frac{3\pm 5}{2}=4,-1$). $M_3$: $\mathrm{tr}=10,\det =25-(-2)(2)=25+4=29.$ $D=100-116=-16<0\Rightarrow$ complex.

False. It has eigenvalues $\pm i$ — they are complex, not absent. WHY the confusion: over the real line $\mathbb{R}$ there is no real eigenvector (this is the $90^{\circ }$ rotation, every arrow turns, so no nonzero real vector is merely stretched). But over $\mathbb{C}$ the characteristic equation ${\lambda }^2+1=0$ still has two roots. "No real eigenvectors" $\ne$ "no eigenvalues."

Step 1 — eigenvalues. $\mathrm{tr}=4,\det =7$, so $\lambda =\frac{4\pm \sqrt{16-28}}{2}=\frac{4\pm \sqrt{-12}}{2}=\frac{4\pm 2\sqrt{3}i}{2}=2\pm \sqrt{3}i.$ WHY $\sqrt{-12}=2\sqrt{3}i$: $-12=(-1)(4)(3)$, and $\sqrt{-1}=i$.

Step 2 — pick the convention. Take $\lambda =\alpha -i\beta =2-\sqrt{3}i$, so $\alpha =2,\beta =\sqrt{3}$.

Step 3 — scale. $r=|\lambda |=\sqrt{{\alpha }^2+{\beta }^2}=\sqrt{4+3}=\sqrt{7}\approx 2.646$.

Step 4 — angle. $\theta =atan2(\beta ,\alpha )=atan2(\sqrt{3},2)=\arctan \left(\frac{\sqrt{3}}{2}\right)\approx 40.9^{\circ }$. Reading: each application of $A$ spins by $\approx 40.9^{\circ }$ and magnifies by $\sqrt{7}$ — an outward spiral, since $r>1$.

$\mathrm{tr}=1.2,\det =(0.6)(0.6)-(-0.8)(0.8)=0.36+0.64=1.$ $\lambda =\frac{1.2\pm \sqrt{1.44-4}}{2}=\frac{1.2\pm \sqrt{-2.56}}{2}=0.6\pm 0.8i.$ Take $\lambda =0.6-0.8i$: $r=\sqrt{0.36+0.64}=\sqrt{1}=1$, and $\theta =atan2(0.8,0.6)\approx 53.13^{\circ }$. Motion: $r=1$ means no scaling — a pure rotation by $\approx 53.13^{\circ }$. Indeed this is exactly $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$ with $\cos \theta =0.6,\sin \theta =0.8$. See the figure.

Recover $\lambda$ from trace and determinant. For a conjugate pair, $\mathrm{tr}A=\lambda +\bar{\lambda }=2\alpha$ and $\det A=\lambda \bar{\lambda }=|\lambda {|}^2=r^2$.

• $2\alpha =1\Rightarrow \alpha =\frac{1}{2}.$
• $r^2=\det A=1\Rightarrow r=1.$
• $\beta$: from $r^2={\alpha }^2+{\beta }^2$, $1=\frac{1}{4}+{\beta }^2\Rightarrow \beta =\sqrt{3}/2\approx \mathrm{0.866.}$
• $\theta =atan2(\beta ,\alpha )=atan2(0.866,0.5)=60^{\circ }.$

Discrete $A^n$: the growth per step is $r=1$ → neither grows nor shrinks; it stays bounded. WHY the orbits are ellipses: in the $P$-frame (columns $x,y$) the motion is a pure rotation on circles; back in the standard basis we apply $P$, which shears each circle into a tilted ellipse. So $|A^{n}v|$ neither grows nor decays because $r^n=1^n=1$, and each orbit traces one fixed ellipse.

Continuous $\dot{v}=Av$: growth is set by $\mathrm{Re}\lambda =\alpha =\frac{1}{2}>0$ → solutions grow like $e^{\alpha t}$ — an outward spiral. WHY the difference: discrete iteration multiplies by $\lambda$ each step ($|\lambda |=r$ matters); the ODE exponentiates, $e^{\lambda t}$, whose magnitude is $e^{\alpha t}$ ($\mathrm{Re}\lambda$ matters). See Linear Dynamical Systems and Phase Portraits.

Real vs complex is decided by $D=(\mathrm{tr}A{)}^2-4\det A$. Write $T=\mathrm{tr}A,\Delta =\det A$.

• $D>0⟺\Delta <\frac{T^2}{4}$: two real eigenvalues.
• $D<0⟺\Delta >\frac{T^2}{4}$: complex conjugate pair.
• $D=0⟺\Delta =\frac{T^2}{4}$: a repeated real eigenvalue — the boundary parabola.

So in the plane with horizontal axis $T$ and vertical axis $\Delta$, the parabola $\Delta =T^2/4$ separates the two worlds; above it (large determinant relative to trace) you get rotation-scaling. See the figure.

Step 1 — eigenvalues. $\mathrm{tr}=4,\det =(1)(3)-(-1)(2)=3+2=5.$ $\lambda =\frac{4\pm \sqrt{16-20}}{2}=\frac{4\pm 2i}{2}=2\pm i.$ Convention: $\lambda =\alpha -i\beta =2-i$, so $\alpha =2,\beta =1$.

Step 2 — eigenvector. Solve $(A-\lambda I)v=0$ with $\lambda =2-i$: $A-\lambda I=\left(\begin{array}{cc}1-(2-i)& -1\\ 2& 3-(2-i)\end{array}\right)=\left(\begin{array}{cc}-1+i& -1\\ 2& 1+i\end{array}\right).$ Row 2 is cleaner: $2v_1+(1+i)v_2=0\Rightarrow v_1=-\frac{1+i}{2}{v}_2$. Take $v_2=2$: $v=\left(\begin{array}{c}-(1+i)\\ 2\end{array}\right)=\underset{x}{\underbrace{\left(\begin{array}{c}-1\\ 2\end{array}\right)}}+i\underset{y}{\underbrace{\left(\begin{array}{c}-1\\ 0\end{array}\right)}}.$ Check row 1: $(-1+i)(-(1+i))-1\cdot 2=(-1+i)(-1-i)-2=(1+i-i-i^2)-2=(1+1)-2=0.$ ✓

Step 3 — build $P$ and $C$. $P=\left(\begin{array}{cc}x& y\end{array}\right)=\left(\begin{array}{cc}-1& -1\\ 2& 0\end{array}\right),\det P=(-1)(0)-(-1)(2)=2,P^{-1}=\frac{1}{2}\left(\begin{array}{cc}0& 1\\ -2& -1\end{array}\right).$ $C=P^{-1}AP=\frac{1}{2}\left(\begin{array}{cc}0& 1\\ -2& -1\end{array}\right)\left(\begin{array}{cc}1& -1\\ 2& 3\end{array}\right)\left(\begin{array}{cc}-1& -1\\ 2& 0\end{array}\right)=\left(\begin{array}{cc}2& -1\\ 1& 2\end{array}\right).$ This equals $\left(\begin{array}{cc}\alpha & -\beta \\ \beta & \alpha \end{array}\right)$ with $\alpha =2,\beta =1$. ✓ It reads as rotate by $\theta =atan2(1,2)\approx 26.57^{\circ }$, scale $r=\sqrt{5}\approx 2.236$.

WHAT we want: rotation-scaling in the standard basis means $P=I$, so $A=C$ directly. Use $C=r\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$ with $r=0.9,\theta =30^{\circ }$. $\cos 30^{\circ }=\frac{\sqrt{3}}{2}\approx 0.8660,\sin 30^{\circ }=\frac{1}{2}=\mathrm{0.5.}$ $A=0.9\left(\begin{array}{cc}0.8660& -0.5\\ 0.5& 0.8660\end{array}\right)=\left(\begin{array}{cc}0.7794& -0.45\\ 0.45& 0.7794\end{array}\right).$ Eigenvalues: by construction $\alpha =r\cos \theta =0.9(0.8660)=0.7794$, $\beta =r\sin \theta =0.9(0.5)=0.45$, so $\lambda =0.7794\pm 0.45i,|\lambda |=r=0.9,\arg \lambda =30^{\circ }.$ Check: $\det A=r^2=0.81$, $\mathrm{tr}A=2r\cos \theta =1.5588$; discriminant $=1.5588^2-4(0.81)=2.4299-3.24=-0.8101<0$ ✓ complex. Long run: each step multiplies length by $r=0.9<1$, so $|A^{n}v|=0.9^n|v|\to 0$ — an inward spiral collapsing to the origin (a stable spiral in the discrete sense). See the figure.

Both are similarity transforms turning $A$ into simplest coordinates. Over $\mathbb{C}$, Diagonalization gives a diagonal $D=\left(\begin{array}{cc}\lambda & 0\\ 0& \bar{\lambda }\end{array}\right)$ — but the change-of-basis matrix $S=\left(\begin{array}{cc}v& \bar{v}\end{array}\right)$ has complex columns (the eigenvectors $v,\bar{v}$). The real block $C=\left(\begin{array}{cc}\alpha & -\beta \\ \beta & \alpha \end{array}\right)$ has $P=\left(\begin{array}{cc}x& y\end{array}\right)$ with real columns ($v=x+iy$).

The explicit bridge. The real and complex frames are related by the fixed complex matrix $W=\frac{1}{2}\left(\begin{array}{cc}1& -i\\ 1& i\end{array}\right),W^{-1}=\left(\begin{array}{cc}1& 1\\ i& -i\end{array}\right),$ which just repackages "real part, imaginary part" into "$\lambda$-component, $\bar{\lambda }$-component." A direct multiply gives $WD{W}^{-1}=\frac{1}{2}\left(\begin{array}{cc}1& -i\\ 1& i\end{array}\right)\left(\begin{array}{cc}\lambda & 0\\ 0& \bar{\lambda }\end{array}\right)\left(\begin{array}{cc}1& 1\\ i& -i\end{array}\right)=\left(\begin{array}{cc}\alpha & -\beta \\ \beta & \alpha \end{array}\right)=C,$ using $\lambda =\alpha -i\beta$, $\bar{\lambda }=\alpha +i\beta$. WHY this works: $\frac{\lambda +\bar{\lambda }}{2}=\alpha$ lands on the diagonal and $\frac{\lambda -\bar{\lambda }}{2i}=-\beta$ (with its conjugate) fills the off-diagonal — exactly the block entries. So $C=WD{W}^{-1}$: $C$ and $D$ share the same eigenvalues $\alpha \pm i\beta$ (its characteristic equation is $t^2-2\alpha t+({\alpha }^2+{\beta }^2)=0$), and $S=PW$. See Complex Numbers - Polar Form for why $\lambda =r{e}^{i\theta }$ makes both views interchangeable.

Advantage of the real form: $P$ and $C$ are real, so you can iterate $A^n=P{C}^n{P}^{-1}$ and read the geometry (spin $n\theta$, scale $r^n$) without ever leaving real arithmetic — no complex bookkeeping, and the phase portrait falls right out.