Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Digital systems, linear algebra, introduction to formal proof resit / 6 / c

If $B^k=0$ with $k\ge 1$, put $S=I+B+\cdots +B^{k-1}$. The matrix geometric series telescopes on both sides:Thus $I-B$ is nonsingular, andThis is a finite polynomial identity for a nilpotent matrix; no convergence argument or bound $\Vert B\Vert <1$ is required.

Write $B=QD{Q}^{-1}$ with $D$ a diagonal matrix. Then $0=B^k=Q{D}^k{Q}^{-1}$, so each diagonal entry ${\lambda }_i$ satisfies ${\lambda }_i^k=0$ and therefore ${\lambda }_i=0$. Hence $D=0$ and $B=0$. Conversely the zero matrix is diagonalizable and nilpotent. This proves the existing result that a diagonalizable nilpotent matrix is zero; nilpotence alone does not imply a zero matrix, because nonzero nilpotent Jordan blocks exist.