부산대학교 도서관

If $n \geq 2$, $k$ is a positive integer, $\Bbb{F}$ is a field, and $A \in M_n(\Bbb{F})$ is nonzero, then $A = U + N$, where $U$ is nonsingular and $N^k = 0$, if and only if ${\rm rank}\,A \geq \frac{n}{k}$ [P.~V. Danchev, E. García and M. Gómez~Lozano, Electron. J. Linear Algebra {\bf 39} (2023), 460--471; MR4657500]. In the current work, the same authors specialize to the case when $U$ is a root of $I_n$, i.e., $U^s = I_n$ for some positive integer $s$. The major results are the following: \par (a) If ${\rm char}\,\Bbb{F} = p$ with prime field $\Bbb{F}_p$, ${\rm rank}\,A \geq \frac{n}{k}$, and all the coefficients of the characteristic polynomial of $A$ are algebraic over $\Bbb{F}_p$, then $A = R +N$, where $R$ is a root of $I_n$ and $N^k =0$. \par (b) Suppose $\Delta$ is a division ring and $A \in M_n(\Delta)$ is {\it nilpotent}. (i) If the Jordan canonical form of $A$ is $J_k(0) \oplus 0_s$, where $k>1$ and $0\leq s \leq k-2$, then $A = R + N$, where $R$ is a root of $I_n$ and $N^2 = 0$. This is shown using weighted directed graphs, where the vertices of the graph are the elements of a basis for $\Delta^n$. (ii) $A = R + N$, where $R$ is a root of $I_n$ and $N^2 = 0$, if and only if ${\rm rank}\,A \geq \frac{n}{2}$. The proof of (ii) makes use of~(i).