3
Eigenvalues
Eigenstuff ยท Long-term behavior ยท RS coords ยท Complex ฮป
๐ Eigenvalue & Eigenvector Basics
The core idea
An eigenvector $\mathbf{v}$ of matrix $M$ is a special direction: when you apply $M$ to $\mathbf{v}$, the vector only gets scaled (not rotated):
$$M\mathbf{v} = \lambda\mathbf{v}$$
The scalar $\lambda$ is the eigenvalue โ it says by how much $\mathbf{v}$ gets stretched or flipped.Graphically: eigenvectors are the directions that the matrix leaves unchanged (up to scaling). Every other direction gets rotated. This is why they determine long-term behavior โ a system moving along an eigenvector stays on that line.
If you're given the eigenvector, finding the eigenvalue is easy
Compute $M\mathbf{v}$ and see what scalar gives $M\mathbf{v} = \lambda\mathbf{v}$.Example: $M=\begin{pmatrix}3&1\\0&2\end{pmatrix}$, $\mathbf{v}=\begin{pmatrix}1\\0\end{pmatrix}$.
$M\mathbf{v}=\begin{pmatrix}3\\0\end{pmatrix}=3\begin{pmatrix}1\\0\end{pmatrix}$ โ $\lambda=3$.
โก Diagonal & Triangular Shortcut
๐ Just read the diagonal โ no computation needed
For diagonal and triangular (upper or lower) matrices, the eigenvalues ARE the diagonal entries. Period.| Matrix type | Example | Eigenvalues |
|---|---|---|
| Diagonal | $\begin{pmatrix}3&0\\0&-2\end{pmatrix}$ | $\lambda_1=3,\lambda_2=-2$ |
| Upper triangular | $\begin{pmatrix}3&7\\0&-2\end{pmatrix}$ | $\lambda_1=3,\lambda_2=-2$ (ignore the 7) |
| Lower triangular | $\begin{pmatrix}3&0\\5&-2\end{pmatrix}$ | $\lambda_1=3,\lambda_2=-2$ (ignore the 5) |
๐ข Computing Eigenvalues โ Step by Step
The characteristic polynomial approach
For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, eigenvalues satisfy:
$$\lambda^2 - \underbrace{(a+d)}_{\text{sum of diag.}}\lambda + (ad-bc) = 0$$
Then use the quadratic formula (provided on exam):
$$\lambda = \frac{(a+d) \pm \sqrt{(a+d)^2 - 4(ad-bc)}}{2}$$
Full worked example
Find eigenvalues of $A=\begin{pmatrix}3&2\\3&-2\end{pmatrix}$Step 1: $(a+d)=3+(-2)=1$, $(ad-bc)=(3)(-2)-(2)(3)=-6-6=-12$
Step 2: Characteristic polynomial: $\lambda^2-\lambda-12=0$
Factor: $(\lambda-4)(\lambda+3)=0$ โ $\boxed{\lambda_1=4,\ \lambda_2=-3}$
Check: $(a+d)=\lambda_1+\lambda_2=4+(-3)=1$ โ $(ad-bc)=\lambda_1\lambda_2=4(-3)=-12$ โ
๐ Finding Eigenvectors
Procedure once you have an eigenvalue $\lambda$
Solve $(A-\lambda I)\mathbf{v}=\mathbf{0}$. For 2ร2 this means:Plug in the eigenvalue: write out $A-\lambda I=\begin{pmatrix}a-\lambda&b\\c&d-\lambda\end{pmatrix}$
Take one row (usually row 1): $(a-\lambda)v_1 + b\cdot v_2 = 0$
Set $v_2=1$ (or any nonzero value) and solve for $v_1$, OR rearrange to get $v_1/v_2$
Eigenvectors for $A=\begin{pmatrix}3&2\\3&-2\end{pmatrix}$, $\lambda_1=4$, $\lambda_2=-3$
For $\lambda_1=4$:$A-4I=\begin{pmatrix}-1&2\\3&-6\end{pmatrix}$ โ Row 1: $-v_1+2v_2=0$ โ $v_1=2v_2$
Set $v_2=1$: $\mathbf{v}_1=\begin{pmatrix}2\\1\end{pmatrix}$ โ
For $\lambda_2=-3$:
$A+3I=\begin{pmatrix}6&2\\3&1\end{pmatrix}$ โ Row 1: $6v_1+2v_2=0$ โ $v_2=-3v_1$
Set $v_1=1$: $\mathbf{v}_2=\begin{pmatrix}1\\-3\end{pmatrix}$ โ
๐ RS/UV Coordinates (Eigenbasis)
What are RS coordinates?
Instead of the standard $(x,y)$ axes, we can use the eigenvectors as our new axes (basis). The coordinates in this new system are called RS or UV coordinates.If $\{\mathbf{v}_1, \mathbf{v}_2\}$ are eigenvectors, and a vector $\mathbf{x} = \alpha\mathbf{v}_1 + \beta\mathbf{v}_2$, then $(\alpha, \beta)$ are the RS coordinates of $\mathbf{x}$.
Why this is powerful: In the eigenbasis, the matrix becomes diagonal (uncoupled). Each axis evolves independently: $u_N = \lambda_1^N u_0$ and $w_N = \lambda_2^N w_0$.
How to find RS coordinates โ solve a 2ร2 system
Given eigenvectors $\mathbf{v}_1$ and $\mathbf{v}_2$, and a vector $\mathbf{x}$, find $\alpha,\beta$ such that:
$$\alpha\mathbf{v}_1 + \beta\mathbf{v}_2 = \mathbf{x}$$
This is just a system of 2 equations in 2 unknowns. Solve by substitution or elimination.
Full worked example โ RS coordinates
Find RS coords of $\mathbf{x}=\begin{pmatrix}-8\\12\end{pmatrix}$ with respect to basis $\mathbf{r}=\begin{pmatrix}2\\3\end{pmatrix}$, $\mathbf{s}=\begin{pmatrix}-2\\1\end{pmatrix}$.Set up: $\alpha\begin{pmatrix}2\\3\end{pmatrix}+\beta\begin{pmatrix}-2\\1\end{pmatrix}=\begin{pmatrix}-8\\12\end{pmatrix}$
Two equations: $2\alpha-2\beta=-8$ โ $\alpha-\beta=-4$ (divide by 2)
$\phantom{Two equations:}$ $3\alpha+\beta=12$
Add: $4\alpha=8$ โ $\alpha=2$. Then $\beta=12-3(2)=6$.
RS coordinates: $(\alpha,\beta)=(2,6)$. Verify: $2(2,3)+6(-2,1)=(4,6)+(-12,6)=(-8,12)$ โ
Using RS coords for long-term prediction
If $\mathbf{x}_0 = \alpha\mathbf{v}_1+\beta\mathbf{v}_2$, then after $N$ steps (discrete):$$\mathbf{x}_N = \alpha\lambda_1^N\mathbf{v}_1 + \beta\lambda_2^N\mathbf{v}_2$$ This works because $M\mathbf{v}_i=\lambda_i\mathbf{v}_i$, so $M^N\mathbf{v}_i=\lambda_i^N\mathbf{v}_i$.
Example from class: $f(\mathbf{U})=1.15\mathbf{U}$, $f(\mathbf{V})=0.4\mathbf{V}$. If $\mathbf{x}=5\mathbf{U}+\mathbf{V}$, then after 1 application: $f(\mathbf{x})=5(1.15)\mathbf{U}+0.4\mathbf{V}=5.75\mathbf{U}+0.4\mathbf{V}$.
Discrete-time: stability by |ฮป|
โ Schematic illustration only โ not drawn to scale or from simulation.
๐ Long-Term Behavior โ Discrete-Time
The complete classification table (from review session)
| Magnitude | Sign of $\lambda$ | Type | Equilibrium |
|---|---|---|---|
| $|\lambda_1|,|\lambda_2|>1$ | $\lambda>0$ | Real | Unstable node (grows) |
| $|\lambda_1|>1,|\lambda_2|<1$ | $\lambda>0$ | Real | Saddle point |
| $|\lambda_1|,|\lambda_2|<1$ | $\lambda>0$ | Real | Stable node (decays to 0) |
| $|\lambda_1|,|\lambda_2|>1$ | $\lambda<0$ | Real | Unstable node with oscillation |
| $|\lambda_1|>1,|\lambda_2|<1$ | $\lambda<0$ | Real | Saddle with oscillation |
| $|\lambda_1|,|\lambda_2|<1$ | $\lambda<0$ | Real | Stable node with oscillation |
| $|\lambda|>1$ | N/A | Complex | Unstable spiral |
| $|\lambda|<1$ | N/A | Complex | Stable spiral |
| $|\lambda|=1$ | N/A | Complex | Neutral rotation (center) |
Dominant eigenvalue = long-term direction
For large $N$, the term with largest $|\lambda|$ dominates. The system grows/decays along the corresponding eigenvector.Example (from HW and exam): System with $\lambda_1=-1.4$ and $\lambda_2=0.9$ (discrete).
$|\lambda_1|=1.4>1$ dominates. System grows without bound. Since $\lambda_1<0$, it oscillates (sign flips each step). Long-term growth rate: 1.4 per step, with oscillation.
๐ฅ Negative eigenvalue = oscillation along that axis
If $\lambda<0$: each step multiplies by a negative number, so the component along that eigenvector alternates sign (+, โ, +, โ, ...). This is not a spiral โ it's a flip back and forth along the eigenvector direction.
๐ Long-Term Behavior โ Continuous-Time
Solution and stability
$$\mathbf{x}(t)=c_1 e^{\lambda_1 t}\mathbf{v}_1+c_2 e^{\lambda_2 t}\mathbf{v}_2$$
Stability depends on the real part of each eigenvalue:• Re$(\lambda)<0$: $e^{\lambda t}\to 0$ โ stable (decays)
• Re$(\lambda)=0$: $e^{\lambda t}=1$ โ neutral (constant amplitude)
• Re$(\lambda)>0$: $e^{\lambda t}\to\infty$ โ unstable (grows)
| Eigenvalues | Equilibrium type |
|---|---|
| Both real, both negative | Stable node |
| Both real, both positive | Unstable node |
| Real, opposite signs | Saddle point |
| Complex $a\pm bi$, $a<0$ | Stable spiral |
| Complex $\pm\beta i$ (pure imaginary) | Center / Neutral EP |
| Complex $a\pm bi$, $a>0$ | Unstable spiral |
Dominant eigenvalue โ continuous vs discrete
Continuous: dominant = largest real part (not absolute value). Example: $\lambda_1=-1.4$, $\lambda_2=0.9$ โ dominant is $\lambda_2=0.9$ (larger real value). System grows since $e^{0.9t}\to\infty$.Discrete: dominant = largest absolute value $|\lambda|$. Example: $\lambda_1=-1.4$, $\lambda_2=0.9$ โ dominant is $\lambda_1=-1.4$ since $|-1.4|=1.4>0.9$.
๐ Complex Eigenvalues
What complex eigenvalues mean geometrically
Complex eigenvalues $\lambda=a\pm bi$ produce rotation + scaling:• The rotation angle is $\theta=\arctan(b/a)$
• The scaling factor per step/time is $|\lambda|=\sqrt{a^2+b^2}$
This gives spiral behavior. Whether it spirals inward or outward depends on $|\lambda|$ (discrete) or $\alpha=\text{Re}(\lambda)$ (continuous).
The Locust model โ neutral rotation
Three life stages: Eggs (E), Hoppers (H), Adults (A). The matrix has complex eigenvalues with $|\lambda|=1$ โ neutral rotation โ locust population oscillates forever at constant amplitude that depends on initial conditions.
PLF Final Review โ Identify eigenvectors from given Mv products
Given: $M\begin{pmatrix}3\\1\end{pmatrix}=\begin{pmatrix}9\\3\end{pmatrix}$, $M\begin{pmatrix}-10\\1\end{pmatrix}=\begin{pmatrix}-25\\-17\end{pmatrix}$, $M\begin{pmatrix}1\\9\end{pmatrix}=\begin{pmatrix}13\\-13\end{pmatrix}$, $M\begin{pmatrix}-1\\4\end{pmatrix}=\begin{pmatrix}2\\-8\end{pmatrix}$For each candidate, check if $M\vec{v} = \lambda\vec{v}$ (output = scalar ร input):
• $M\begin{pmatrix}3\\1\end{pmatrix}=\begin{pmatrix}9\\3\end{pmatrix}= 3\begin{pmatrix}3\\1\end{pmatrix}$ โ โ eigenvector $\vec{v}_1=\begin{pmatrix}3\\1\end{pmatrix}$, $\lambda_1=3$
• $M\begin{pmatrix}-1\\4\end{pmatrix}=\begin{pmatrix}2\\-8\end{pmatrix}= -2\begin{pmatrix}-1\\4\end{pmatrix}$ โ โ eigenvector $\vec{v}_2=\begin{pmatrix}-1\\4\end{pmatrix}$, $\lambda_2=-2$
• $\begin{pmatrix}-10\\1\end{pmatrix}$: would need $\lambda(-10)=-25$ โ $\lambda=2.5$ and $\lambda(1)=-17$ โ $\lambda=-17$. Contradiction โ NOT an eigenvector.
• $\begin{pmatrix}1\\9\end{pmatrix}$: would need $\lambda(1)=13$ โ $\lambda=13$ and $\lambda(9)=-13$ โ $\lambda=-1.4$. Contradiction โ NOT an eigenvector.
Key method: Divide output by input component-by-component. If all ratios are equal, that ratio is $\lambda$.
PLF Final Review โ RS coordinates (second example)
Find $(R, S)$ of $\vec{w}=\begin{pmatrix}5\\9\end{pmatrix}$ with basis $\vec{u}=\begin{pmatrix}2\\3\end{pmatrix}$, $\vec{v}=\begin{pmatrix}3\\5\end{pmatrix}$Set up: $R\begin{pmatrix}2\\3\end{pmatrix}+S\begin{pmatrix}3\\5\end{pmatrix}=\begin{pmatrix}5\\9\end{pmatrix}$
Two equations: $2R+3S=5$ and $3R+5S=9$
Multiply eq1 by 3, eq2 by 2: $6R+9S=15$ and $6R+10S=18$. Subtract: $S=3$.
Then $2R+9=5$ โ $R=-2$.
$(R,S) = (-2, 3)$ โ Check: $-2(2,3)+3(3,5)=(-4,-6)+(9,15)=(5,9)$ โ
Note: negative R or S is completely fine โ the eigenbasis spans all directions.
PLF Final Review โ Finding matrix M from eigenvector knowledge
Given eigenvector $\vec{v}_1=\begin{pmatrix}1\\2\end{pmatrix}$ with $\lambda=2$ (i.e. $M\vec{v}_1=\begin{pmatrix}2\\4\end{pmatrix}$) and eigenvector $\vec{v}_2=\begin{pmatrix}3\\5\end{pmatrix}$ with $\lambda=3$ (i.e. $M\vec{v}_2=\begin{pmatrix}9\\15\end{pmatrix}$)Strategy: find $f(\vec{e}_1)$ and $f(\vec{e}_2)$ using linearity
Note: $\vec{e}_2=\begin{pmatrix}0\\1\end{pmatrix}=3\vec{v}_1-\vec{v}_2$, so $f(\vec{e}_2)=3f(\vec{v}_1)-f(\vec{v}_2)=3\begin{pmatrix}2\\4\end{pmatrix}-\begin{pmatrix}9\\15\end{pmatrix}=\begin{pmatrix}-3\\-3\end{pmatrix}$
And: $\vec{e}_1=\begin{pmatrix}1\\0\end{pmatrix}=\vec{v}_1-2\vec{e}_2$, so $f(\vec{e}_1)=f(\vec{v}_1)-2f(\vec{e}_2)=\begin{pmatrix}2\\4\end{pmatrix}-2\begin{pmatrix}-3\\-3\end{pmatrix}=\begin{pmatrix}8\\10\end{pmatrix}$
Therefore $M=\begin{pmatrix}8&-3\\10&-3\end{pmatrix}$
โฆ Unit 3 Practice Quiz
โฆ Quiz โ Eigenvalues & Long-Term Behavior
Q1. Matrix $A=\begin{pmatrix}4&0\\0&-3\end{pmatrix}$ (diagonal). The eigenvalues are:
Q2. A discrete-time system has eigenvalues $\lambda_1=-1.4$ and $\lambda_2=0.9$. In the long run, what happens?
Q3. Continuous-time system has eigenvalues $\lambda_1=-1.4$ and $\lambda_2=0.9$. Which is dominant and what happens?
Q4. Find RS coords of $\begin{pmatrix}7\\7\end{pmatrix}$ in basis $\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}$, $\mathbf{v}_2=\begin{pmatrix}-2\\1\end{pmatrix}$. What is $(\alpha,\beta)$?
Q5. After many time steps, a discrete system has $W_N=200$ bears. The dominant eigenvector is $\mathbf{v}_1=\begin{pmatrix}3/4\\1\end{pmatrix}$. How many $L_N$ (the other stage) are there?