Lecture 26 - Pitch on Why Diagonalization is Cool

We can use diagonalization to solve the Fibonacci Sequence:

A Tale of Fibonacci

Consider $f_n = {0,1,1,2,3,5,8,13,...}. Namely:

f_{0} = 0, f_{1} = 1, f_{n} = f_{n - 1} + f_{n - 2}

Is it possible to get an iterative formula for $f_{n}$ ? Notice that: we get make a forumla:

F_{1} = [\begin{matrix} 0 \\ 1 \end{matrix}], F_{2} = [\begin{matrix} 1 \\ 1 \end{matrix}], F_{3} = [\begin{matrix} 1 \\ 2 \end{matrix}], . . .

Namely, we take the first two numbers and put them in $F_{1}$ , and the 2nd and 3rd are in $F_{2}$ and so on. But consider the transformation $T$ from $F_{1} \to F_{2}, F_{2} \to F_{3}, . . .$ . This is defined as:

T ([\begin{matrix} x \\ y \end{matrix}]) = [\begin{matrix} y \\ x + y \end{matrix}]

implies that:

F_{n} = T^{n - 1} (F_{1})

but we can turn $T$ into $M (T)$ , so what if I have a diagonal matrix. Consider such a matrix:

{[\begin{matrix} λ_{1} & 0 & 0 \\ 0 & ⋱ & 0 \\ 0 & 0 & λ_{n} \end{matrix}]}^{k}

if we want $M (T)^{k}$ then it's just:

{[\begin{matrix} λ_{1} & 0 & 0 \\ 0 & ⋱ & 0 \\ 0 & 0 & λ_{n} \end{matrix}]}^{k} = [\begin{matrix} λ_{1}^{k} & 0 & 0 \\ 0 & ⋱ & 0 \\ 0 & 0 & λ_{n}^{k} \end{matrix}]

Notice that we've turned something that was $O (k)$ and turned it into $O (1)$ (assuming exponentiation is easy to do). Thus, the idea is to find a basis $v_{1}, v_{2}$ such that $M (T)$ is diagonal. Because if we can then:

M (F_{n}) = M (T)^{n - 1} M (F_{1})

where we know the left matrix will be easy to compute if it's diagonalizable. Hence, we need to make the basis vectors as eigenvectors. Hence, ask:

T ([\begin{matrix} x \\ y \end{matrix}]) = [\begin{matrix} y \\ x + y \end{matrix}] = λ [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} λ x \\ λ y \end{matrix}]

Thus since $y = λ x$ then $x + λ x = λ (λ x)$ so then since $x \neq \vec{0}$ as we want nonzero eigenvectors:

\begin{aligned} (1 + λ) x & = λ^{2} x \\ λ^{2} x - λ x - 1 x & = 0 \\ λ^{2} - λ - 1 & = 0 & x is non-zero, so divide it out \end{aligned}

Thus solve the quadratic:

λ = \frac{1 \pm \sqrt{5}}{2}

this is the golden ratio. Now we need the eigenvectors for these eigenvalues. We ideally want simple eigenvectors where we have our first $x$ is simple, so use $x = 1$ , then $y = λ = \frac{1 + \sqrt{5}}{2}$ hence:

v_{1} = [\begin{matrix} 1 \\ \frac{1 + \sqrt{5}}{2} \end{matrix}]

and similarly for the other eigenvalue:

v_{2} = [\begin{matrix} 1 \\ \frac{1 - \sqrt{5}}{2} \end{matrix}]

So my basis $β = {v_{1}, v_{2}}$ . These are eigenvectors so we'll have a diagonal matrix representation, where:

M (T) = [\begin{matrix} \frac{1 + \sqrt{5}}{2} & 0 \\ 0 & \frac{1 - \sqrt{5}}{2} \end{matrix}]

so:

M (F_{n}) = {[\begin{matrix} \frac{1 + \sqrt{5}}{2} & 0 \\ 0 & \frac{1 - \sqrt{5}}{2} \end{matrix}]}^{n - 1} M (F_{1})

But what is $M (F_{1})$ ? We'll talk about this next time.