IMC 2009 · Day 2 · P8

Let us fix the matrix $A \in M_n(\mathbb{C})$. For every matrix $X \in M_n(\mathbb{C})$, let $\Delta X := AX - XA$. We need to prove that the matrix $\Delta B$ is nilpotent.

Observe that the condition $A^2 B + B A^2 = 2 A B A$ is equivalent to $$\tag{1} \Delta^2 B = \Delta(\Delta B) = 0.$$ $\Delta$ is linear; moreover, it is a derivation, i.e. it satisfies the Leibniz rule: $$\Delta(XY) = (\Delta X) Y + X (\Delta Y), \quad \forall X, Y \in M_n(\mathbb{C}).$$ Using induction, one can easily generalize the above formula to $k$ factors: $$\tag{2} \Delta(X_1 \cdots X_k) = (\Delta X_1) X_2 \cdots X_k + \dots + X_1 \cdots X_{j-1} (\Delta X_j) X_{j+1} \cdots X_k + \dots + X_1 \cdots X_{k-1} \Delta X_k,$$ for any matrices $X_1, X_2, \dots, X_k \in M_n(\mathbb{C})$. Using the identities (1) and (2) we obtain the equation for $\Delta^k(B^k)$: $$\tag{3} \Delta^k(B^k) = k! (\Delta B)^k, \quad \forall k \in \mathbb{N}.$$ By the last equation it is enough to show that $\Delta^n(B^n) = 0$.

To prove this, first we observe that equation (3) together with the fact that $\Delta^2 B = 0$ implies that $\Delta^{k+1} B^k = 0$, for every $k \in \mathbb{N}$. Hence, we have $$\tag{4} \Delta^k(B^j) = 0, \quad \forall k, j \in \mathbb{N},\ j < k.$$ By the Cayley–Hamilton Theorem, there are scalars $\alpha_0, \alpha_1, \dots, \alpha_{n-1} \in \mathbb{C}$ such that $$B^n = \alpha_0 I + \alpha_1 B + \dots + \alpha_{n-1} B^{n-1},$$ which together with (4) implies that $\Delta^n B^n = 0$.

Set $X = AB - BA$. The matrix $X$ commutes with $A$ because $$AX - XA = (A^2 B - ABA) - (ABA - BA^2) = A^2 B + B A^2 - 2 ABA = 0.$$ Hence for any $m \ge 0$ we have $$X^{m+1} = X^m (AB - BA) = A X^m B - X^m B A.$$ Take the trace of both sides: $$\operatorname{tr} X^{m+1} = \operatorname{tr} A (X^m B) - \operatorname{tr} (X^m B) A = 0$$ (since for any matrices $U$ and $V$, we have $\operatorname{tr} UV = \operatorname{tr} VU$). As $\operatorname{tr} X^{m+1}$ is the sum of the $m+1$-st powers of the eigenvalues of $X$, the values of $\operatorname{tr} X$, …, $\operatorname{tr} X^n$ determine the eigenvalues of $X$ uniquely, therefore all of these eigenvalues have to be 0. This implies that $X$ is nilpotent.