Track / Stage 6 / 20 of 400 #1020 of 1964

Problem 1020

National olympiad, first round

Algebra Difficulty 6.0 Prove it

4. Let $A, B \in M_{2}(\mathbb{C})$ with the property that $A B - B A = A$ . Prove that $A B A = A B^{2} A = O_{2}$ .

Marian Cucoanes, Gazeta Matematică 11/2012 Problem elaborated by prof. Cristian Lazăr

## 11th Grade

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

4. Since the matrices $A B$ and $B A$ have the same trace, it follows that $\operatorname{tr} A=\operatorname{tr}(A B-B A)=0$ . The characteristic equation of matrix $A$ leads to $A^{2}=-(\operatorname{det} A) I_{2}$ , hence the matrix $A^{2}$ commutes with any other matrix; thus, $A^{2} B=B A^{2}$ .

Multiplying the relation from the statement by $A$ , on the left, and then on the right, and taking into account the previous results, we deduce that $A^{2}=-A^{2}$ , so $A^{2}=O_{2}$ . From here,

$A B A=(A+B A) A=\left(I_{2}+B\right) A^{2}=O_{2}$

Now we square both sides of the relation from the statement; we obtain that

$(A B A) B-A B^{2} A-B A^{2} B+B(A B A)=A^{2}$

Since $A B A=A^{2}=O_{2}$ , it follows that $A B^{2} A=O_{2}$ , which completes the solution.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.