Is it B) Trace(A) could be non-zero?

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

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

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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

- $A$ has to be the $0$ matrix
- Trace$(A)$ could be non-zero
- $A$ is diagonalizable
- $0$ is the only eigenvalue of $A$.

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