LINEAR-ALGEBRA · Interactive Practice | Unit 23 · Video 3

Where Will Everyone Live? Predicting Population Equilibrium with Eigenvalues

IKey Formulas

Formula	Name	Description
$u_k = A^k u_0$	Population evolution	State vector after $k$ steps
columns sum to $1$	Markov / stochastic matrix	Conservation: no one is created or destroyed
$\lambda_1 + \lambda_2 = \text{trace}(A)$	Trace identity (2×2)	Sum of eigenvalues = sum of diagonal
$u_k = c_1 \lambda_1^k x_1 + c_2 \lambda_2^k x_2$	Eigenvector decomposition	General solution to $u_k = A^k u_0$
$\lambda = 1$ always exists	Stochastic matrix property	Steady-state eigenvalue

IIVisualization 1 — Watching a Population Reach Equilibrium

Two states trade residents every year — where does the split settle, and does the starting point matter?

p: leaving CA/yr 0.10 q: leaving MA/yr 0.20 start CA 0 year k 0

💡 The yearly update is $u_{k+1} = A\,u_k$ with $A = \begin{bmatrix} 1-p & q \\ p & 1-q \end{bmatrix}$ . Challenge: change only the starting split — the dashed steady-state levels never budge.

IIIVisualization 2 — How Fast the Transient Dies

The steady part stays pinned at $1$ forever, while the transient shrinks like $\lambda_2^k$ — how fast?

λ₂ 0.70

💡 Diagonalization trades one hard matrix power $A^k$ for cheap scalar powers $\lambda^k$ — that is the whole reason eigenvalues make this tractable.

IVVisualization 3 — The Eigenvector Sets the Destination

Start anywhere on the conservation line; every trajectory bends onto the eigenvector's ray.

start CA 0

💡 The eigenvector $(2, 1)$ fixes the long-run ratio; the conserved total $1000$ fixes the size — together they pin the single steady state.

VQuiz Questions

Question 1

Which of the following matrices is a Markov (stochastic) matrix?

$M_1 = \begin{bmatrix} 0.6 & 0.5 \\ 0.5 & 0.7 \end{bmatrix}, \quad M_2 = \begin{bmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{bmatrix}, \quad M_3 = \begin{bmatrix} 1.2 & 0.0 \\ -0.2 & 1.0 \end{bmatrix}, \quad M_4 = \begin{bmatrix} 0.6 & 0.4 \\ 0.4 & 0.6 \end{bmatrix}$

Recall: every column must sum to 1, and every entry must be non-negative.

✅ Correct! Both $M_2$ and $M_4$ have columns summing to $1$ with non-negative entries — while $M_1$ 's first column sums to $1.1$ and $M_3$ has a negative entry.

❌ Not quite. Remember: a stochastic matrix needs (1) every column to sum to exactly 1, AND (2) every entry to be non-negative. Check $M_3$ carefully — does it have any negative entries?

Show solution

Solution:

Check each column sum, and check for non-negative entries.

$M_1$ : column 1 = $0.6 + 0.5 = 1.1$ . Column sum is not 1. Not stochastic.
$M_2$ : column 1 = $0.5 + 0.5 = 1.0$ , column 2 = $0.5 + 0.5 = 1.0$ . All entries non-negative. Stochastic.
$M_3$ : column 1 = $1.2 + (-0.2) = 1.0$ . Column sum is fine, but it has a negative entry ( $-0.2$ ). Not stochastic.
$M_4$ : column 1 = $0.6 + 0.4 = 1.0$ , column 2 = $0.4 + 0.6 = 1.0$ . All entries non-negative. Stochastic.

The answer is "M2 and M4". Both conditions — non-negative entries and columns summing to exactly 1 — must hold.

Question 2

Suppose $A$ is a $2 \times 2$ Markov matrix with diagonal entries $0.6$ and $0.85$ . What is the second eigenvalue $\lambda_2$ ?

Hint: Use the trace identity and the fact that every Markov matrix has $\lambda_1 = 1$ .

✅ Correct! Trace = 1.45, and since $\lambda_1 = 1$ , we get $\lambda_2 = 0.45$ .

❌ Not quite. That's the trace itself, which equals $\lambda_1 + \lambda_2$ , not $\lambda_2$ alone. Subtract $\lambda_1 = 1$ .

❌ Not quite. Compute the trace (sum of diagonal), then subtract $\lambda_1 = 1$ to get $\lambda_2$ .

Show solution

Solution:

Every Markov matrix has $\lambda_1 = 1$ .

The trace of $A$ equals the sum of eigenvalues: $\text{trace}(A) = 0.6 + 0.85 = 1.45 = \lambda_1 + \lambda_2$

Since $\lambda_1 = 1$ : $\lambda_2 = 1.45 - 1 = 0.45$

The answer is λ₂ = 0.45.

Question 3

Using the migration matrix $A = \begin{bmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{bmatrix}$ from the video, suppose the starting population is $u_0 = (500, 500)$ — split evenly. What is $u_1 = A u_0$ ?

✅ Correct! Row 1: $0.9(500) + 0.2(500) = 550$ . Row 2: $0.1(500) + 0.8(500) = 450$ . Total still 1000.

❌ Not quite. You may have swapped the rows. Row 1 of $A$ produces the new CA population.

❌ Not quite. $(500, 500)$ is not the steady state — the steady state is $(2000/3, 1000/3) \approx (667, 333)$ . Multiply $A \cdot u_0$ carefully.

❌ Not quite. Compute each row separately: row $i$ of $A$ dotted with $u_0$ .

Show solution

Solution:

Multiply $A \cdot u_0$ row by row:

Row 1 (CA next year): $0.9 \cdot 500 + 0.2 \cdot 500 = 450 + 100 = 550$

Row 2 (MA next year): $0.1 \cdot 500 + 0.8 \cdot 500 = 50 + 400 = 450$

So $u_1 = (550, 450)$ . Total: $550 + 450 = 1000$ — conservation holds.

The answer is u₁ = (550, 450).

Question 4

True or False: For the migration matrix $A = \begin{bmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{bmatrix}$ , if we change the starting population from $(0, 1000)$ to $(1000, 0)$ , the long-run steady-state population $u_\infty$ will be different.

✅ Correct! The steady state is $(2000/3, 1000/3)$ in both cases. The eigenvector $(2,1)$ and the conserved total ( $1000$ ) determine $u_\infty$ — initial conditions only change the path, not the destination.

❌ Not quite. This is the key insight from the video! The transient term $c_2 (0.7)^k (-1, 1)$ decays to zero regardless of $c_2$ . Only $c_1 (2, 1)$ survives, and $c_1$ is determined by total population, which is conserved.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

False. The steady state depends only on:

The matrix $A$ (which determines the eigenvector for $\lambda = 1$ , here $(2, 1)$ ), and
The total population (which is conserved — columns sum to 1).

The general solution is $u_k = c_1 \cdot 1^k \cdot (2,1) + c_2 \cdot (0.7)^k \cdot (-1, 1).$

As $k \to \infty$ , the second term vanishes (since $|0.7| < 1$ ), leaving only $c_1 (2, 1)$ .

Changing the initial condition only changes $c_1$ and $c_2$ . But $c_1$ is fixed by total population: with total $1000$ , $c_1 = 1000/3$ in both cases. So $u_\infty = (2000/3, 1000/3)$ regardless of the starting split.

Initial conditions affect how fast you reach the steady state, not what the steady state is.

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