Discussion
Suppose A is a non-zero real nilpotent matrix. Then :
Answer: C
The true statement is that the transpose of A is nilpotent.
If \(A^{k} = 0\), then \((A^{T})^{k} = (A^{k})^{T} = 0\), so \(A^{T}\) is nilpotent too.
The other statements fail. Every nilpotent matrix has only the eigenvalue 0, so its characteristic polynomial \(t^{n}\) splits and A is triangulable (it is similar to a strictly upper triangular matrix); for the same reason it has exactly one eigenvalue, not two. The minimal polynomial is \(t^{k}\) where k is the index of nilpotency, which can be smaller than n: for the 3 × 3 matrix with a single 1 in the corner, \(A^{2} = 0\), so the minimal polynomial is \(t^{2}\) while the characteristic polynomial is \(t^{3}\).
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