Matrices

We’ll approach continued fractions from a completely different viewpoint.

Let $A_1, ...$ be a sequence of $2\times 2$ matrices. Let

\[ \begin{pmatrix} p_n & r_n \\ q_n & s_n \end{pmatrix} = A_1 ... A_n \]

If $\frac{p_n}{q_n}$ and $\frac{r_n}{s_n}$ tend to the same limit $\alpha$ as $n \rightarrow \infty$ then $\alpha$ is the limit of the infinite product of matrices, and we write

\[ \alpha ~ \prod_{n=1}^\infty A_n \]

Example: Let $a_1, ...$ be a sequence of positive integers, except we allow $a_1 = 0$. Let

\[ A_k = \begin{pmatrix} a_k & 1 \\ 1 & 0 \end{pmatrix} \]

Then $[a_1; a_2, ...] ~ \prod_{n=0}^\infty A_n$. The convergents are $p_n, q_n$, and we have $r_n = p_{n-1}, s_n = q_{n-1}$. Since $\det A_n = -1$, we see the difference between successive convergents can be determined by studying

\[ \det \begin{vmatrix} p_n & r_n \\ q_n & s_n \end{vmatrix} = (-1)^n \]

Example: Suppose $\alpha ~ \prod_{k=1}^\infty A_k$.

Let $c_k$ be nonzero numbers. Then $\alpha ~ \prod_{k=1}^\infty c_k A_k$.

Let $ A = \left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) $. Then

\[ \frac{a\alpha +b}{c\alpha +d} ~ A \prod_{k=1}^\infty A_k \]

provided $c\alpha +d \ne 0$. For some proofs, we exploit this and ignore a finite number of matrices.

Example: Given that $\left( \begin{smallmatrix} 1 & 2 \\ 1 & 1 \end{smallmatrix} \right) \left( \begin{smallmatrix} 1 & 2 \\ 1 & 1 \end{smallmatrix} \right) ... $ converges, find its limit.

Let

\[ \alpha ~ \prod_{n=1}^\infty \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \]

Then

\[ \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \prod_{n=2}^\infty \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} ~ \frac{\alpha + 2}{\alpha+1} \]

Hence $\alpha = \frac{\alpha + 2}{\alpha + 1}$, and therefore $\alpha = \sqrt{2}$.

Theorem: Consider a sequence of matrices of the form $ A_k = \left( \begin{smallmatrix} a_k & c_k \\ b_k & 0 \end{smallmatrix} \right) $. where $a_k, b_k, c_k$ are positive integers satisfying

\[ a_k \ge (b_k c_k)^{1+\delta} \]

for some fixed $\delta \gt 0$. Then $\prod_{n=1}^\infty ~ \alpha$ for some irrational $\alpha \gt 1$.

Proof: Let $ \prod_{k=1}^n A_k = \left( \begin{smallmatrix} p_n & r_n \\ q_n & s_n \end{smallmatrix} \right) $. We have $r_n = c_n p_{n-1}, s_n = c_n q_{n-1}$, $q\rightarrow\infty$ as $n \rightarrow \infty$ and $p_n\ge q_n$. The last fact implies $\alpha \gt 1$.

Also,

\[ \frac{p_{n+1}}{q_{n+1}} = \frac{ a_{n+1}p_n + b_{n+1}c_n p_{n-1} }{ a_{n+1}q_n + b_{n+1}c_n q_{n-1} } \]

so $\frac{p_{n+1}}{q_{n+1}}$ lies strictly between $\frac{p_n}{q_n}$ and $\frac{p_{n-1}}{q_{n-1}}$.

Then:

\[ \left| \frac{p_n}{q_n} - \frac{p_{n-1}}{q_{n-1}} \right| = \frac{ \prod_{k=1}^n \left| \det A_k \right| }{c_n q_n q_{n-1}} = \frac {b_1 c_1 ... b_n c_n}{c_n q_n q_{n-1} } \lt \frac{b_1 c_1 ... b_{n-1} c_{n-1}}{(q_{n-1})^2} \]

since $q_n \gt b_n q_{n-1}$

Now $q_{n-1}\gt(k b_1 c_1 ... b_{n-1}c_{n-1})^{1+\delta}$ for some $\kappa \gt 0$ because

\[ \begin{aligned} q_{n-1} &=& a_{n-1} q_{n-2} + b_{n-1} c_{n-2} q_{n-3} \\ &\ge& a_{n-1}q_{n-2} \ge (b_{n-1}c_{n-1})^{1+\delta}q_{n-2} \end{aligned} \]

and if we asssume inductively that $q_{n-2} \gt (\kappa b_1 c_1...b_{n-2}c_{n-2})^{1+\delta}$ it follows that

\[ q_{n-1} \gt (\kappa b_1 c_1 ... b_{n-1} c_{n-1})^{1+\delta} \]

So $\left|\frac{p_n}{q_n} - \frac{p_{n-1}}{q_{n-1}}\right|\rightarrow 0$ as $n \rightarrow \infty$ and

\[ \left|\alpha - \frac{p_{n-1}}{q_{n-1}}\right| \le \left|\frac{p_n}{q_n} - \frac{p_{n-1}}{q_{n-1}}\right| \lt \frac{b_1 c_1 ... b_{n-1}c_{n-1}}{q_{n-1}^2} \lt \frac{1}{\kappa q_{n-1}^{1+\sigma}} \]

where $\sigma \gt 0$ is defined by $\frac{1}{1+\delta} = 1-\sigma$.

Hence $\alpha$ is irrational, because if $\alpha = k/l$ then:

\[ |k q_n - l p_n| = l q_n \left|\alpha - \frac{p_n}{q_n} \right| \lt \frac{l q_n}{\kappa q_n^{1+\sigma}} \rightarrow 0 \]

as $n\rightarrow\infty$, a contradiction because $|k q_n - l p_n|$ is always a nonzero integer (since $\frac{p_{n+1}}{q_{n+1}}$ lies strictly between the previous two convergents).

Corollary: Let $a_k$ be positive integers. Then $[a_1; a_2, ....]$ converges to an irrational $\alpha \gt 1$.

Proof: Apply theorem to continued fraction matrices.

Example: $\prod_{k=1}^\infty \left(\begin{smallmatrix} k&1\\2&0 \end{smallmatrix}\right)$ converges to an irrational number because apart from the first two matrices, we have

\[ a_k = k \gt 2^{1.5} = (b_k c_k)^{1.5} \]

We can develop this theory further to derive continued fraction expansions of hyperbolic trignometric functions.