# Kostka Matrix Identity

How does one prove $K^{-1}K = I$ using a more general sign reversing involution, where $K$ is the Kostka matrix and $I$ is the identity matrix?

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What is J? What is this sign reversing involution supposed to be more general than? – Qiaochu Yuan Jul 26 '10 at 7:20
Sorry it should be K. – Alex Jul 26 '10 at 7:25
I know, I know... I was merely trying to show by example that the question could use some rewriting! – Mariano Suárez-Alvarez Jul 26 '10 at 7:38
@Qiaochu: Here is the paper by Sagan and Lee journals1.scholarsportal.info/details.xqy?uri=/01966774/… – Alex Jul 26 '10 at 7:44
Alex, please edit the relevant information in the text of the question itself. That way, it is mch more easy to see what is being asked. – Mariano Suárez-Alvarez Jul 26 '10 at 8:10

As I imagine you already know, Egecioglu and Remmel gave an elegant bijective proof that $K K^{-1}=I$. And, as you say, Sagan and Lee gave a bijective proof that $K^{-1} K = I$, but it is much more complicated. I imagine this is the proof you want to improve.
One of the surprising facts about bijective proofs is that sometimes there is an elegant proof for $AB=I$ but not for $BA=I$. I use "elegant" in a vague sense, because Loehr and Mendes have an algorithm which takes a bijective proof of $AB=I$ and outputs a bijective proof of $BA=I$, so there are no examples where there is a bijective proof of one equality and not the other. Loehr and Mendes work out the bijection proving $K^{-1} K=I$ as one of their examples.
It occurs to me that it might be interesting to try to show, in the sense of computational complexity, that there are situations where there is a short bijective proof of $AB=I$ but only long bijective proofs for $BA=I$. Speaking vaguely, the obstacle should be that the implication $(AB=I) \implies (BA=I)$, although true in the $n \times n$ matrices for any $n$, is not true in an arbitrary ring, so you somehow need to know that you are dealing with matrices. I haven't thought about this question in any depth.