Johnson's determinantal identity for contiguous minors of Toeplitz matrices, with an accretive extension

Abstract

Let $A$ be an $n\times n$ real Toeplitz matrix satisfying $A+A^{\top}=2\mathbb J_n$, where $\mathbb J_n$ is the all-ones matrix.If $A_r(i,j)$ denotes the $r\times r$ contiguous submatrix of $A$ consisting of rows $i,i+1,\dots,i+r-1$ and columns $j,j+1,\dots,j+r-1$, then for every $n\ge 2$ one has $$\det A_{n-1}(1,2)+\det A_{n-1}(2,1)=2\det A_{n-1}(1,1).$$ This confirms a conjecture of Charles R.~Johnson (2003). The proof combines a rank-one determinant expansion with Dodgson's condensation formula, and then invokes a polynomial-identity argument in the Toeplitz parameters: after obtaining an equality of squares in the integral domain $\mathbb{Z}[b_1,\dots,b_{n-1}]$, we factor it to deduce an identity up to sign and determine the sign by a suitable specialization.We also give an extension of the Bayat--Teimoori arithmetic--geometric mean identity: for every real accretive matrix $A$, one has the sharp inequality $$ \sqrt{\det A_{n-1}(1,1)\ \det A_{n-1}(2,2)} \ \ge\ \left|\frac{\det A_{n-1}(1,2)+\det A_{n-1}(2,1)}{2}\right|,$$ with equality whenever the symmetric part has rank one, i.e.\ $A+A^{\top}=\alpha\,ww^{\top}$ for some $\alpha\in\mathbb R$ and $w\in\mathbb R^n\setminus\{0\}$,recovering the Bayat--Teimoori equality as a special case.