Let A and B be \(n \times n\) matrices over a field of characteristic zero. Prove that the condition \(B A - A B = A\) implies that A is nilpotent. (Hint: what does A do to eigenvectors of B?)
Solution
Let \(K\) be the ground field.
1. It suffices to prove the result after extending scalars from \(K\) to analgebraic closure \(\overline K\).
The relation \[ BA-AB=A \] is unchanged by scalar extension. If the resulting matrix over \(\overline K\) satisfies \(A^N=0\), then the same matrix power, whose entries already lie in \(K\), is zero over \(K\). Thus nilpotence after scalar extension implies nilpotence over the original field.
2. Suppose the ground field is algebraically closed. If \(v\) is aneigenvector of \(B\) with eigenvalue \(\lambda\) and \(Av\neq0\), then \(Av\) isan eigenvector of \(B\) with eigenvalue \(\lambda+1\).
The given relation is equivalent to \[ BA=A(B+I). \] Hence \[ B(Av) = A(B+I)v = (\lambda+1)Av. \] If \(Av\neq0\), this says exactly that \(Av\) is an eigenvector with the stated eigenvalue.
3. Under the hypotheses of step 2, the kernel of \(A\) is nonzero.
Because the field is algebraically closed, \(B\) has an eigenvector \(v\neq0\), say \(Bv=\lambda v\). Repeated application of step 2 shows that, as long as \(A^rv\neq0\), the vector \(A^rv\) is an eigenvector of \(B\) with eigenvalue \[ \lambda+r. \] In characteristic zero the scalars \[ \lambda,\lambda+1,\lambda+2,\ldots \] are all distinct. Since an \(n\times n\) matrix has only finitely many eigenvalues, \(A^rv\) must vanish for some \(r\geq1\). For the least such \(r\), the vector \[ w\coloneqq A^{r-1}v \] is nonzero and satisfies \(Aw=0\). Hence \(\ker A\neq0\).
4. The subspace \(W\coloneqq\ker A\) is invariant under \(B\).
Rearranging the relation gives \[ AB=BA-A. \] If \(w\in W\), then \(Aw=0\), so \[ A(Bw) = B(Aw)-Aw = 0. \] Thus \(Bw\in W\).
5. Over an algebraically closed field, \(A\) is nilpotent.
Proceed by induction on \(n\). The assertion is immediate for \(n=0\). Assume \(n>0\). By step 3, the subspace \[ W=\ker A \] is nonzero, and by step 4 it is invariant under both \(A\) and \(B\). Therefore \(A\) and \(B\) induce endomorphisms \(\overline A,\overline B\) on the quotient \(V/W\), where \(V=\overline K^n\). Passing the relation to the quotient gives \[ \overline B\,\overline A - \overline A\,\overline B = \overline A. \] Since \(\dim(V/W)<n\), the induction hypothesis gives \[ \overline A^N=0 \] for some \(N\). Thus \[ A^N(V)\subseteq W=\ker A, \] and consequently \[ A^{N+1}=0. \] So \(A\) is nilpotent.
6. The original matrix \(A\) over \(K\) is nilpotent.
Step 5 proves nilpotence after extending scalars to \(\overline K\). Step 1 then descends the same matrix identity \(A^N=0\) to \(K\).
7. Q.E.D.
Step 6 is the required conclusion.