LINEAR-ALGEBRA

Why Fibonacci Numbers Hide the Golden Ratio

IKey Formulas

Formula	Name	Description
$u_{k+1} = A u_k \Rightarrow u_k = A^k u_0$	Difference equation	One-step update and $k$ -step solution
$u_k = c_1 \lambda_1^k x_1 + c_2 \lambda_2^k x_2 + \cdots$	Eigenvector expansion	Each mode evolves independently
$\det(A - \lambda I) = 0$	Characteristic equation	Solve for eigenvalues $\lambda$
$F_k = \dfrac{\lambda_1^k - \lambda_2^k}{\sqrt{5}}$	Binet's formula	Closed form for Fibonacci, with $\lambda_{1,2} = \dfrac{1 \pm \sqrt{5}}{2}$

IIVisualization 1 — Two Modes of Fibonacci

Fibonacci splits into two geometric modes — one grows like $\varphi^k$ , the other decays — so the golden mode dominates.

step k 10

IIIVisualization 2 — Consecutive Ratios Converge to φ

The ratios $F_{k+1}/F_k$ zigzag toward the golden ratio $\varphi$ , alternating sides while the error collapses.

up to k 9

IVVisualization 3 — Stability Zoo of 2×2 Systems

Every system $u_{k+1} = Au_k$ lives or dies by its eigenvalues; the trace–determinant plane charts its fate.

💡 Inside the green triangle every $|\lambda| < 1$ , so the state decays to the origin; the Fibonacci point $(T, D) = (1, -1)$ sits below it — one eigenvalue exceeds $1$ , so it grows.

VQuiz Questions

Question 1

The Fibonacci recurrence $F_{k+2} = F_{k+1} + F_k$ is encoded by the matrix

$A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}.$

What is the characteristic equation $\det(A - \lambda I) = 0$ ?

✅ Correct! The characteristic equation is the Fibonacci recurrence in disguise.

❌ Not quite. Expand $(1-\lambda)(-\lambda) - 1$ carefully and watch the signs.

Show solution

Solution:

Compute $\det(A - \lambda I)$ for $A - \lambda I = \begin{bmatrix} 1-\lambda & 1 \\ 1 & -\lambda \end{bmatrix}$ :

$\det(A - \lambda I) = (1 - \lambda)(-\lambda) - (1)(1) = -\lambda + \lambda^2 - 1 = \lambda^2 - \lambda - 1.$

Setting this to zero gives $\lambda^2 - \lambda - 1 = 0$ .

Notice this is exactly the Fibonacci recurrence $F_{k+2} - F_{k+1} - F_k = 0$ rearranged — the characteristic polynomial re-exposes the recurrence we started with.

Question 2

Solving $\lambda^2 - \lambda - 1 = 0$ gives two eigenvalues. One is the golden ratio $\lambda_1 = \dfrac{1+\sqrt{5}}{2} \approx 1.618$ .

What happens to $\lambda_2^{100}$ , where $\lambda_2 = \dfrac{1-\sqrt{5}}{2} \approx -0.618$ , in Binet's formula for $F_{100}$ ?

✅ Correct! Since $|\lambda_2| < 1$ , its powers decay exponentially — the dominant eigenvalue $\lambda_1$ wins.

❌ Not quite. Think about what happens to $r^{100}$ when $|r| < 1$ .

Show solution

Solution:

Since $|\lambda_2| \approx 0.618 < 1$ , raising it to a large power shrinks it:

$|\lambda_2|^{100} = (0.618)^{100} \approx 1.3 \times 10^{-21}.$

This is utterly negligible compared with $\lambda_1^{100} \approx 3.54 \times 10^{20}$ . So Binet's formula

$F_{100} = \frac{\lambda_1^{100} - \lambda_2^{100}}{\sqrt{5}} \approx \frac{\lambda_1^{100}}{\sqrt{5}}$

is dominated entirely by $\lambda_1$ . This is the general principle: in any linear dynamic system, eigenvalues with $|\lambda| < 1$ die away, and the largest $|\lambda|$ wins.

Question 3

True or False: For a dynamic system $u_{k+1} = A u_k$ , if every eigenvalue of $A$ satisfies $|\lambda| < 1$ , then $u_k \to 0$ as $k \to \infty$ , regardless of the starting vector $u_0$ .

✅ Correct! Every component $c_i \lambda_i^k x_i$ decays to zero when $|\lambda_i| < 1$ , so the whole state vector goes to zero.

❌ Not quite. Decompose $u_0$ into eigenvectors and track what happens to each piece.

Show solution

Solution:

True. Expand $u_0$ in the eigenvector basis:

$u_0 = c_1 x_1 + c_2 x_2 + \cdots + c_n x_n.$

After $k$ steps, each eigenvector scales by its own $\lambda^k$ :

$u_k = c_1 \lambda_1^k x_1 + c_2 \lambda_2^k x_2 + \cdots + c_n \lambda_n^k x_n.$

If every $|\lambda_i| < 1$ , then every $\lambda_i^k \to 0$ , so every term vanishes and $u_k \to 0$ regardless of the constants $c_i$ (which are determined by $u_0$ ).

This is stability: the origin is an attractor. Conversely, if any $|\lambda_i| > 1$ with $c_i \neq 0$ , that component blows up and the system is unstable.

Question 4

Consider the matrix $B = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}$ with a starting vector $u_0 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ .

Using the eigenvector expansion, what is $u_{10}$ ?

✅ Correct! Each eigenvector evolves independently, so component $i$ just gets scaled by $\lambda_i^k$ .

❌ Not quite. The matrix is diagonal, so each entry evolves by its own eigenvalue. Match the eigenvalue to the correct component.

Show solution

Solution:

The matrix $B$ is already diagonal, so its eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = 2$ , with eigenvectors $x_1 = (1, 0)^T$ and $x_2 = (0, 1)^T$ .

Decompose $u_0$ : $u_0 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} = 1 \cdot \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 1 \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix},$

so $c_1 = 1$ and $c_2 = 1$ . Applying the general formula:

$u_k = c_1 \lambda_1^k x_1 + c_2 \lambda_2^k x_2 = \begin{bmatrix} 3^k \\ 2^k \end{bmatrix}.$

At $k = 10$ : $u_{10} = \begin{bmatrix} 3^{10} \\ 2^{10} \end{bmatrix} = \begin{bmatrix} 59049 \\ 1024 \end{bmatrix}.$

The dominant eigenvalue $\lambda_1 = 3$ causes the top entry to overwhelm the bottom entry by a factor of $(3/2)^{10} \approx 57.7$ .

Solved: 0 / 4