Let \(A\) be an \(nxn\) matrix with real entries such that \(A^k=0\) (0-matrix) for some \(k\in\mathbb{N}\).


A. A has to be the 0-matrix.

B. trace(A) could be non-zero.

C. A is diagonalizable.

D. 0 is the only eigenvalue of A


Let \(v\) be an eigenvector of \(A\) with eigenvalue \(\lambda\).

Then \(v \neq 0\) and \(Av=\lambda v\).

Again, \(A^2 v=A(Av)=A(\lambda v)=\lambda Av= (\lambda)^2v\).

We continue to apply A, applying it k times gives: \(A^k v=(\lambda)^k v\).

By given information, the left hand side of the above equality is 0.

So \(\lambda^k v=0\) and remember \(v \neq 0\).

So \(\lambda =0\).

Therefore \(0\) is the only eigenvalue for \(A\).

So D is true.

We analyse the question a little bit further, to check it satisfies no other options above.

We know \(trace(A)=\) sum of eigenvalues of A= \(\sum 0 =0\)

So option B is false.

Take \(A=\begin{bmatrix} 0 & 1 // 0 & 0 \end{bmatrix} \).

Then \(A^2 =0\). But \(A\) is not the zero matrix.

Also, if \(A\) were diagonalizable then the corresponding diagonal matrix would be the zero matrix. Which would then imply that \(A\) is the zero matrix, which in this case it is not. (See diagonalizable-nilpotent-matrix-tifr-2013-problem-8 ) So this disproves options A and C.