The Fibonacci numbers are the well-known sequence defined by the recursion relation \(F_{n + 1} = F_n + F_{n - 1}\) for \(n \geq 2\) with \(F_0 = 0\) and \(F_1 = 1\). This recursion relation has the beautiful closed form solution

\[F_n = \frac{1}{\sqrt{5}} \left(\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n\right).\]

In a previous post I showed how to derive this solution using generating functions. One aspect of this solution that fascinates me is the many ways in which it can be derived, using tools from a variety of fields of mathematics. This post will derive this solution using matrix diagonalization.

The first step in this derivation is to express the Fibonacci recurrence as a matrix product. For \(n \geq 1\),

\[ \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} F_n \\ F_{n - 1} \end{pmatrix}. \]

For conveience, let \(X = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\) be the coefficient matrix of the recurrence. We see that

\[ \begin{pmatrix} F_2 \\ F_1 \end{pmatrix} = X \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \]

so

\[ \begin{pmatrix} F_3 \\ F_2 \end{pmatrix} = X \begin{pmatrix} F_2 \\ F_1 \end{pmatrix} = X^2 \begin{pmatrix} 1 \\ 0 \end{pmatrix}. \]

Similarly,

\[ \begin{pmatrix} F_4 \\ F_3 \end{pmatrix} = X \begin{pmatrix} F_3 \\ F_2 \end{pmatrix} = X^3 \begin{pmatrix} 1 \\ 0 \end{pmatrix}. \]

We have uncovered the fact that

\[ \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix} = X^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \]

and therefore

\[ F_n = \begin{pmatrix} 0 & 1 \end{pmatrix} X^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}. \]

This formula for \(F_n\), involving powers of \(X\), provides the first clue that we should diagonalize \(X\). To diagonalize \(X\) is to write it as \(X = P D P^{-1}\), where \(D\) is a diagonal matrix. In this situation,

\[ X^2 = P D P^{-1} P D P^{-1} = P D D P^{-1} = P D^2 P^{-1}. \]

Similarly, for any natural number \(n\), \(X^n = P D^n P^{-1}\). It is simple to compute the matrix \(D^n\), since the \(n\)-th power of a diagonal matrix just involves raising the diagonal entries to the \(n\)-th power.

To diagonalize \(X\), we first calculate its eigenvalues, as these will be the diagonal entries of \(D\). To do so, we find the roots of the characteristic polynomial, \(\operatorname{det}(X - \lambda I)\), of \(X\). The characteristic polynomial of \(X\) is

\[ \operatorname{det}(X - \lambda I) = \operatorname{det} \begin{pmatrix} 1 - \lambda & 1 \\ 1 & - \lambda \end{pmatrix} = (1 - \lambda) (-\lambda) - 1 = \lambda^2 - \lambda - 1. \]

Using the quadratic formula, the roots of this equation are

\[\varphi = \frac{1 + \sqrt{5}}{2}\]

and

\[\psi = \frac{1 - \sqrt{5}}{2}.\]

The matrix \(D\) is therefore

\[ D = \begin{pmatrix} \varphi & 0 \\ 0 & \psi \end{pmatrix}. \]

These roots both appear in the closed form expression for \(F_n\) we are trying to derive, which is a good sign that we are on the right track. In fact, the expression can be written succinctly as \(F_n = \frac{1}{\sqrt{5}} (\varphi^n - \psi^n)\).

Before going any further, it will be helpful to note two relations between \(\phi\) and \(\psi\), both of which are easy to derive. First, \(\varphi + \psi = 1\). Second, \(\varphi \psi = -1\). We will make liberal use of these relations and their direct consequences without mention throughout the remainder of the post.

Now that we have determined \(D\), it remains to determine \(P\). The columns of \(P\) are the eigenvectors corresponding to the eigenvalues that form the diagonal entries of \(D\). To find the eigenvector corresponding to an eigenvalue \(\lambda\), we calculate a basis for the null space of \(X - \lambda I\).

For the eigevnalue \(\varphi\),

\[ X - \varphi I = \begin{pmatrix} \psi & 1 \\ 1 & -\varphi \end{pmatrix} \sim \begin{pmatrix} 1 & -\varphi \\ 1 & -\varphi \end{pmatrix} \sim \begin{pmatrix} 1 & -\varphi \\ 0 & 0 \end{pmatrix}. \]

Here \(\sim\) stands for row equivalence. The null space of \(X - \varphi I\) is seen to be spanned by the vector \(\begin{pmatrix}\varphi \\ 1\end{pmatrix}\). In order to simplify the calculation of \(P^{-1}\), it will be beneifical to normalize this vector to have length one. The length of this vector is

\[ \sqrt{\varphi^2 + 1^2} = \sqrt{\frac{1 + 2 \sqrt{5} + 5}{4} + 1} = \sqrt{\frac{5 + \sqrt{5}}{2}} = \sqrt[4]{5} \sqrt{\varphi}. \]

The normalized eigenvector correpsonding to \(\varphi\) is therefore

\[ \vec{v}_\varphi = \frac{1}{\sqrt[4]{5} \sqrt{\varphi}} \begin{pmatrix} \varphi \\ 1 \end{pmatrix} = \frac{1}{\sqrt[4]{5}} \begin{pmatrix} \sqrt{\varphi} \\ \sqrt{-\psi} \end{pmatrix}. \]

At first, the entry \(\sqrt{-\psi}\) of this vector may seem a bit odd, but since \(\psi < 0\), it is perfectly reasonable.

Similar calculations show that the null space of \(X - \psi I\) is spanned by the vector \(\begin{pmatrix}\psi \\ 1\end{pmatrix}\) and that the normalized eigenvector corresponding to \(\psi\) is

\[ \vec{v}_\psi = \frac{1}{\sqrt[4]{5}} \begin{pmatrix} - \sqrt{-\psi} \\ \sqrt{\varphi} \end{pmatrix}. \]

The matrix \(P\) in the diagonalization of \(X\) is therefore

\[ P = \begin{pmatrix} \vec{v}_\varphi & \vec{v}_\psi \end{pmatrix} = \frac{1}{\sqrt[4]{5}} \begin{pmatrix} \sqrt{\varphi} & -\sqrt{-\psi} \\ \sqrt{-\psi} & \sqrt{\varphi} \end{pmatrix}. \]

The advantage of normalizing the eigenvectors is that \(P\) is now an orthogonal matrix, so

\[ P^{-1} = P^\top = \frac{1}{\sqrt[4]{5}} \begin{pmatrix} \sqrt{\varphi} & \sqrt{-\psi} \\ -\sqrt{-\psi} & \sqrt{\varphi} \end{pmatrix}. \]

Finally, we are ready to derive the closed form expression for \(F_n\). We have that

\[ \begin{align*} F_n & = \begin{pmatrix} 0 & 1 \end{pmatrix} X^n \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ & = \begin{pmatrix} 0 & 1 \end{pmatrix} P D^n P^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ & = \frac{1}{\sqrt{5}} \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} \sqrt{\varphi} & -\sqrt{-\psi} \\ \sqrt{-\psi} & \sqrt{\varphi} \end{pmatrix} \begin{pmatrix} \varphi^n & 0 \\ 0 & \psi^n \end{pmatrix} \begin{pmatrix} \sqrt{\varphi} & \sqrt{-\psi} \\ -\sqrt{-\psi} & \sqrt{\varphi} \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ & = \frac{1}{\sqrt{5}} \begin{pmatrix} \sqrt{-\psi} & \sqrt{\varphi} \end{pmatrix} \begin{pmatrix} \varphi^n & 0 \\ 0 & \psi^n \end{pmatrix} \begin{pmatrix} \sqrt{\varphi} \\ -\sqrt{-\psi} \end{pmatrix} \\ & = \frac{1}{\sqrt{5}} \begin{pmatrix} \sqrt{-\psi} & \sqrt{\varphi} \end{pmatrix} \begin{pmatrix} \sqrt{\varphi}\ \varphi^n \\ -\sqrt{-\psi}\ \psi^n \end{pmatrix} \\ & = \frac{1}{\sqrt{5}} \left(\sqrt{-\psi \varphi} \varphi^n - \sqrt{-\varphi \psi} \psi^n\right) \\ & = \frac{1}{\sqrt{5}} \left(\phi^n - \psi^n\right). \end{align*} \]