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I am trying to find the simplest way to get an expression of the determinant of the following infinite matrix as m tends to infinity.

$$ \left[\begin{array}{cccccc} 1 & a_{1} & 0 & \cdots & 0 & 0 \\ \beta_{1} & 1 & a_{2} & \cdots & 0 & 0\\ 0 & \beta_{2} & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 & a_{m}\\ 0 & 0 & 0 & \cdots & \beta_{m} & 1 \end{array} \right]$$

I have considered using both the Leibniz formula or the Laplace formula. Leibniz formula required considering sums over permutations which I was hoping to avoid and Laplace formula seems somewhat recursive even though I have only 2-3 elements in each row it. Is there a simpler solution to this problem which I am overlooking ?

Edit: I am just after an algebraic expansion of the determinant into and infinite series of some kind ?

Any help would be much appreciated.

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The criteria for absolute convergence of this determinant is given by Whittaker & Watson section 2.82 p37 as an example due to von Koch. It is probably a good idea to confirm your case satisfies this necessary and sufficient condition. –  John Spitzer Sep 5 at 7:45

1 Answer 1

up vote 1 down vote accepted

Try for $\lambda=-1$

$$ \det(I-\lambda K) = \left[ \sum_{n=0}^\infty (-\lambda)^n \operatorname{Tr } K^n \right]= \exp{(\sum_{n=0}^\infty(-1)^{n+1}\frac{\operatorname{Tr} A^n}{n}z^n})$$

copied from


Note that $I-K$ can be put as $D^+ + D^-$, where $D^+$ are strict upper triangular and $D^-$ are strict lower diagonal. Then use the binomial theorem for $(D^+ + D^-)^n$.

Note that $(D^-)^k$ and $(D^+)^k$ are fairly simple to compute;)

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I am not sure the method you are suggesting is much simpler, if anything it seems quite the opposite. –  Hardy Mar 6 '12 at 19:00
I provided you with a further hint;) The binomial theorem is now doing the combinatorics for you;) –  plusepsilon.de Mar 7 '12 at 8:25
thanks i 'll try it out and would post back if i run into trouble. –  Hardy Mar 7 '12 at 9:56

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