+1 vote
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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

1. $A$ has to be the $0$ matrix
2. Trace$(A)$ could be non-zero
3. $A$ is diagonalizable
4. $0$ is the only eigenvalue of $A$.
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Is it B) Trace(A) could be non-zero?

Because is not bound to affect the result of exponentiation of a matrix.

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.
by Junior

+1 vote