Determinant of an infinite matrix

The Endeavour 2024-02-26

What does the infinite determinant

$D = \left| \begin{array}{lllll} 1 & a_1 & 0 & 0 & \cdots \\ b_1 & 1 & a_2 & 0 & \cdots \\ 0 & b_2 & 1 & a_3 & \\ 0 & 0 & b_3 & 1 & \ddots \\ \vdots & \vdots & & \ddots & \ddots \\ \end{array} \right|$

mean and when does it converge?

The determinant D above is the limit of the determinants D_n defined by

$D_n = \left| \begin{array}{lllll} 1 & a_1 & & & \\ b_1 & 1 & a_2 & & \\ & b_2 & 1 & \ddots & \\ & & \ddots & \ddots & a_{n-1} \\ & & & b_{n-1} & 1 \\ \end{array} \right|$

If all the a‘s are 1 and all the b‘s are −1 then this post shows that D_n = F_n, the nth Fibonacci number. The Fibonacci numbers obviously don’t converge, so in this case the determinant of the infinite matrix does not converge.

In 1895, Helge von Koch said in a letter to Poincaré that the infinite determinant is absolutely convergent if and only if the sum

$\sum_{i=1}^\infty a_ib_i$

is absolutely convergent. A proof is given in [1].

The proof shows that the D_n are bounded by

$\prod_{i=1}^n\left(1 + |a_ib_i| \right)$

and so the infinite determinant converges if the corresponding infinite product converges. And a theorem on infinite products says

$\prod_{i=1}^\infty\left(1 + |a_ib_i| \right)$

converges absolute if the sum in Koch’s theorem converges. In fact,

$\prod_{i=1}^\infty\left(1 + |a_ib_i| \right) \leq \exp\left(\sum_{i=1}^\infty |a_ib_i| \right )$

and so we have an upper bound on the infinite determinant.

[1] A. A. Shaw. H. von Koch’s First Lemma and Its Generalization. The American Mathematical Monthly, April 1931, Vol. 38, No. 4, pp. 188–194

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