1 votes 1 votes Let $A$ be an $n \times n$ matrix with real entries such that $A^{k}=0$ (0-matrix), for some $k \in \mathbb{N}$. Then $A$ has to be the $0$ matrix Trace$(A)$ could be non-zero $A$ is diagonalizable $0$ is the only eigenvalue of $A$ Linear Algebra tifrmaths2014 linear-algebra matrix + – makhdoom ghaya asked Dec 17, 2015 edited Aug 18, 2020 by soujanyareddy13 makhdoom ghaya 431 views answer comment Share Follow See 1 comment See all 1 1 comment reply Debanjan Dey commented Aug 25, 2018 reply Follow Share Is it B) Trace(A) could be non-zero? Because is not bound to affect the result of exponentiation of a matrix. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Matrix A is basically nilpotent matrix in case of nilpotent matrix eigen values are 0 and as it has same eigen values it is not diagonalizable . So answer is option D. Yash4444 answered May 31, 2019 Yash4444 comment Share Follow See all 0 reply Please log in or register to add a comment.