Permanent inequality for rank-at-most-two matrices

Let TT be a matrix of size n×nn\times n, and let per(T)\operatorname{per}(T) denote its permanent. Form the 2n×2n2n\times 2n block matrix with every block equal to TT:

(TT\TT).\begin{pmatrix}T&T\T&T\end{pmatrix}.

Rank-two permanent inequality. For any matrix TT with rank at most 22,

per(TT\TT)(2nn)per(T)2.\operatorname{per}\begin{pmatrix}T&T\T&T\end{pmatrix}\leq \binom{2n}{n}\operatorname{per}(T)^2.

The conjecture was proposed as a potentially useful tool in finite free probability. The source describes it as an unproved conjecture motivating the determinantal identity, and does not provide evidence of a resolution.

Sources & referencesView supporting material

Primary source

Adam W. Marcus, “A Determinantal Identity for the Permanent of a Rank 2 Matrix”, arXiv:2108.02528 (2021).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.