# Interesting Matrix Identity

• A
Hi all,

I've come across an interesting matrix identity in my work. I'll define the NxN matrix as $$S_{ij} = 2^{-(2N - i - j + 1)} \frac{(2N - i - j)!}{(N-i)!(N-j)!}.$$ I find numerically that $$\sum_{i,j=1}^N S^{-1}_{ij} = 2N,$$ (the sum is over the elements of the matrix inverse). In fact, I expected to get 2N based on the problem I'm studying, but I don't know what this complicated matrix expression is doing or why it equals 2N. Does any of this look familiar to anyone here?

P.S. If this is in the wrong subforum, please move it.

Mentor
Interesting, how did you come across this? using some numerical computing software like matlab?

@fresh_42 or @Mark44 might be interested in how you discovered this.

Gold Member
I haven't run through the math, but keep in mind that the inverse matrix element can be expressed as:
$$(S^{-1})_{ij} = \frac{1}{\det{S}}C_{ji}$$
where ##C_{ji}## is the element of the transposed cofactor matrix. Also remember that the determinant can be expressed as a cofactor expansion:
$$\det{S} = \sum_{i=1}^{N} S_{ij} C_{ij}$$
Also keep in mind that the cofactor expansion works for any row or any column, so that
$$(S^{-1})_{ij} = \frac{C_{ji}}{ \sum_{i=1}^{N} S_{ji} C_{ji}}$$
I dunno, maybe that helps. It might not hurt, too, to see if you can pull out a general formula for the cofactor.

$$\frac{\left[ \int_0^\infty f(t) dt \right]^2}{\int_0^\infty f^2(t) dt }$$ where $$f(t) = \sum_{i=1}^N w_i \frac{(ct)^{N-i}}{(N-i)!} e^{\lambda t}$$ and $$w_i$$ are weights which I want to maximise with respect to. I can show that the maximum is $$\frac{1}{-\lambda} \sum_{ij} \left(S^{-1}\right)_{ij}$$ and using Matlab this turns out to be $$\frac{2N}{-\lambda}$$ for N=1...15 (I stopped here as it became numerically unstable).