Determinant of an infinite matrix - SoatDev IT Consulting

February 26, 2024
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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

$prod_{i=1}^nleft(1 + |a_ib_i| right)$

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

$prod_{i=1}^inftyleft(1 + |a_ib_i| right)$

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

$prod_{i=1}^inftyleft(1 + |a_ib_i| right) leq expleft(sum_{i=1}^infty |a_ib_i| right )$

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

The post Determinant of an infinite matrix first appeared on John D. Cook.

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