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

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

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 \([a_1; a_2, ...] \sim \prod_{n=0}^\infty A_n\). 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 \(\alpha \sim \prod_{k=1}^\infty A_k\).

Let \(c_k\) be nonzero numbers. Then \(\alpha \sim \prod_{k=1}^\infty c_k A_k\).

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

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

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

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

Then

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

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 \(\delta > 0\). Then \(\prod_{n=1}^\infty \sim \alpha\) for some irrational \(\alpha > 1\).

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 \(\alpha > 1\).

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:

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

since \(q_n > b_n q_{n-1}\)

Now \(q_{n-1}>(k b_1 c_1 ... b_{n-1}c_{n-1})^{1+\delta}\) for some \(\kappa > 0\) 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 \(q_{n-2} > (\kappa b_1 c_1...b_{n-2}c_{n-2})^{1+\delta}\) it follows that

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

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

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

where \(\sigma > 0\) is defined by \(\frac{1}{1+\delta} = 1-\sigma\).

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

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

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 \(\alpha > 1\).

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

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

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

Ben Lynn blynn@cs.stanford.edu 💡

Matrices

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