In this post we solve the Fibonacci sequence using linear algebra. The Fibonacci equation is a *second-order difference equation* which is a particular
type of sequence.

## Sequences

A sequence is a (possibly infinite) set of numbers with a defined order.

\[ a_n = \frac{1}{n}, \textit{ for } n = 0, 1, 2, ..., \]

### Difference Equations

A difference equation is a type of sequence. A difference equation expresses the value of the sequence relative to the value of another point in the sequence.

\[ x_{n+1} = 1.5x_n, \textit{ for } n = 0, 1, 2, ..., \]

To evaluate this equation at \(n=3\) we need to know the initial value \(x_0\), and evaluate \(n=1, 2, 3\) in succession. For example, if \(x_0=100\):

\[\begin{align} x_0 & && = 100 \\ x_1 & = 1.5 \times 100 && = 150\\ x_2 & = 1.5 \times 150 && = 225\\ x_3 & = 1.5 \times 225 && = 337.5 \end{align}\]

In the example above, we evaluate the equation recursively. It would be better to be able to evaluate the equation at \(n = 100\) without performing \(100\) calculations. Note that we can also write the equation above in this form:

\[ x_3 = 100 \times 1.5 \times 1.5 \times 1.5 = 337.5 \]

Solving a difference equation means writing the same equation such that it can be evaluated at a given \(n\) without first evaluating \(n=1,2,etc\). The solution to the equation above is:

\[ x_{n} = 1.5^{n} \times x_0, \textit{ for } n = 0, 1, 2, ..., \]

### Second Order Difference Equations

The previous example is a **first order difference equation** because it expresses the value of \(x_n\) as a function
of one other sample, \(x_{n-1}\). A **second order difference equation** is a function of two other samples, for
example

\[ x_{n+2} = 1.5x_n + 1.4x_{n+1}, \textit{ for } n = 0, 1, 2, ..., \]

## Fibonacci Numbers

The easiest way to describe the Fibonacci sequence is with a second order difference equation. Assume \(F_0 = 0\) and \(F_1 = 1\)

\[ F_{k+2} = F_{k+1} + F_{k} \]

The first eight Fibonacci numbers are:

\[ F_k=0,1,1,2,3,5,8,13,... \]

We can solve the sequence with linear algebra. Let us start our sequence with a vector, \(u_0=\begin{bmatrix}1\\0\end{bmatrix}\). Find the next vector in our sequence, \(u_1\) like this:

\[ u_{1} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

More generally:

\[ u_{k+1} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} u_k \]

Note that \(u_0\) contains \(\begin{bmatrix}F_1\\F_0\end{bmatrix}\).

The key to solving the sequence is in the eigenvalues and eigenvectors of \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\). We can write \(u_3\) as

\[ u_3 = \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}3 \\ 2\end{bmatrix} = \begin{bmatrix}F_4 \\ F_3\end{bmatrix} \]

```
import numpy as np
= np.matrix([[1,1],[1,0]])
fib = np.matrix([[1],[0]])
u_0
* fib * fib * u_0 # The * operator is dot product for np.matrix (but not np.array) fib
```

```
matrix([[3],
[2]])
```

### Eigenvector Solution

The general case of the previous example is:

\[ \begin{align} u_k &= \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}^k\begin{bmatrix}1 \\ 0\end{bmatrix}\\ u_k &= A^ku_0 \end{align} \]

We can use eigenvectors and eigenvalues to solve the matrix power (see Eigenvalues and Eigenvectors). First, find the eigenvalues of \(A\):

\[ \begin{align} det\begin{vmatrix} 1 - \lambda & 1\\ 1 & - \lambda \end{vmatrix} & = 0\\ -\lambda(1-\lambda) - 1 & = 0\\ \lambda{}^2 - \lambda - 1 & = 0 \end{align} \]

Solve using the quadratic formula \(\frac{-b \pm \sqrt{b^2-4ac}}{2a} \implies \frac{1 \pm \sqrt{5}}{2} \approx \frac{1 \pm 2.236}{2}\\\), which gives the eigenvalues for our matrix: \[ \begin{align} \lambda_1 & \approx &1.618 \\ \lambda_2 & \approx &-0.618 \end{align} \]

Now we plug our constant eigenvalues (\(\lambda_1\) and \(\lambda_2\)) into \((A-\lambda I)\vec{x}=0\) to find the eigen vectors.

\[ \begin{bmatrix}1-\lambda_1 & 1 \\ 1 & -\lambda_1 \end{bmatrix}\vec{x}_1 = 0 \implies \vec{x}_1 = \begin{bmatrix}\lambda_1 \\ 1\end{bmatrix} \]

Note that this solution is not necessarily intuitive. This is **not** a valid solution for *any* arbitrary value of \(\lambda\):
our values, \(\lambda_1\) and \(\lambda_2\) are valid solutions. We can use the same process
to solve for \(\lambda_2\), revealing eigenvectors \(\begin{bmatrix}\lambda_1 \\1\end{bmatrix}\) and \(\begin{bmatrix}\lambda_2 \\1\end{bmatrix}\). Now create the eigenvector matrix, \(S\) and eigenvalue matrix, \(\Lambda\).

\[ \begin{align} S &= \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix}\\ \Lambda &= \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix} \end{align} \]

Remeber that \(A^k = S^{-1}\Lambda^kS\). To proceed, invert \(S\). Remember this shortcut for inverting a \(2 \times 2\) matrix.

\[ \begin{align} \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} =&& \frac{1}{ad-cb} &\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ S^{-1} =&& &\begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix}^{-1} \\ S^{-1} =&& \frac{1}{\lambda_1-\lambda_2} &\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix}\\ S^{-1} =&& \tfrac{1}{\sqrt{5}} &\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} \end{align} \]

Using the components \((A, \Lambda, S, S^{-1},u_0)\):

\[ \begin{align} u_k &= A^ku_0 \\ u_k &= S\Lambda^kS^{-1}u_o \\ u_k &= S\Lambda^k\frac{1}{\sqrt{5}}\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix}u_0 \\ u_k &= \frac{1}{\sqrt{5}}S\Lambda^k\begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix}u_0 \\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}^k \begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} u_0 \\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1^k & 0 \\ 0 & \lambda_2^k\end{bmatrix} \begin{bmatrix}1 & -\lambda_2\\ -1 & \lambda_1 \end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} \\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1^k & 0 \\ 0 & \lambda_2^k\end{bmatrix} \begin{bmatrix}1 \\ -1 \end{bmatrix}\\ u_k &= \tfrac{1}{\sqrt{5}} \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 1\end{bmatrix} \begin{bmatrix}\lambda_1^k \\ -\lambda_2^k\end{bmatrix} \end{align} \]

## Verify Solution in Python

```
import numpy as np
= (1+np.sqrt(5))/2
lmbd_1 = (1-np.sqrt(5))/2
lmbd_2
= np.matrix([[1, 1], [1, 0]])
a = np.matrix([[lmbd_1, lmbd_2], [1, 1]])
s = np.matrix([[1, -lmbd_2], [-1, lmbd_1]]) * (1/np.sqrt(5))
s_inv = np.matrix([[lmbd_1, 0], [0, lmbd_2]])
eig_val = np.matrix([[1], [0]])
u_0
def eig_k(k):
return eig_val ** k
def u_k(k):
return s * eig_k(k) * s_inv * u_0
def fib(n):
= u_k(n)[1,0]
f_n return int(round(f_n))
for n in range(15)] [fib(n)
```

`[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]`