On the span invariant for cubic similarity (Q2724124)

Let ``\(*\)'' denote the entrywise product on \(\mathbb{R}^n\) or on the space \(n\times n\) matrices with real entries (also called ``Hadamard product''). The authors propose the following conjecture: for any real \(n\times n\) matrix \(A\) and any integer \(k\geq 1\), the span of the image of the algebraic map: \(\mathbb{R}^n\to \mathbb{R}^n\), \(x\mapsto (Ax)^{*k}\) equals the image of the linear map: \(\mathbb{R}^n\to\mathbb{R}^n\) defined by the matrix \((AA^t)^{* k}\) (also called the range of this last matrix). Since \(\text{Im\,}A^t\cap \text{Ker\,}A= (0)\), it follows that \(\text{Im\,}A= \text{Im}(AA^t)\) hence the conjecture is equivalent to the following one: for any symmetric positive semidefinite real \(n\times n\) matrix \(B\) and any \(k\geq 1\), \(\text{span}(x\mapsto (Bx)^{*k})= \text{Im}(B^{*k})\). The inclusion ``\(\supseteq\)'' is obvious.NEWLINENEWLINE Using Schur's product theorem which asserts that if \(B\), \(B'\) are symmetric positive definite matrices then \(B*B'\) is also positive definite, the authors remark that, for any symmetric positive semidefinite matrix \(B\), \(\text{rank}(B^{*k})\geq \text{rank\,}B\), with equality if \(B\) is invertible or if \(\text{rank\,}B= 1\). One deduces that the second conjecture is true for \(n= 2\). The main result of the paper is the proof of the second conjecture for \(n= 3\).NEWLINENEWLINE Concerning the title of the paper: two real \(n\times n\) matrices \(A\) and \(B\) are cubic-similar if there exists an invertible matrix \(T\) such that \((Bx)^{*3}= T^{-1}(ATx)^{*3}\) for every \(x\in \mathbb{R}^n\). In this case, \(\text{span}(x\mapsto (Ax)^{*3})\) and \(\text{span}*x\mapsto (Bx)^{*3})\) have the same dimension. Cubic-similarity is related to the Jacobian conjecture via the work of \textit{L. M. Drużkowski} [Math. Ann. 264, 303--313 (1983; Zbl 0504.13006)].