Romanian Mathematical Olympiad

Observe that $\text{tr}(A)=0$ implies $A^2=O_{n\times n}$ and $$ABA\cdot ACA=O_{n\times n}=ACA\cdot ABA,$$

If $\text{tr}(A)\ne 0$, we show that $\text{rank}(A)=1$. Let $r=\text{rank}(A)$. Then we can find matrices $X\in {\mathcal{M}}_{n\times r}(\mathbb{C})$ and $Y\in {\mathcal{M}}_{r\times n}(\mathbb{C})$ with $\text{rank}(X)=\text{rank}(Y)=r$ such that $A=XY$. Then $YX\in {\mathcal{M}}_r(\mathbb{C})$ and, as $\text{tr}(A)\ne 0$, we have $$r\ge \text{rank}(YX)\ge \text{rank}(X(YX)Y)=\text{rank}(A^2)=\text{rank}(A)=r,$$ That is $\text{rank}(YX)=r$ and the matrix $YX$ is non-singular. Equality $A^2=\text{tr}(A)\cdot A$, can be written $(X{)}^2=\text{tr}(A)\cdot XY$, obtaining $(YX{)}^3=\text{tr}(A)\cdot (YX{)}^2$. As $YX$ is non-singular, we get $YX=\text{tr}(A)\cdot I_r$. In conclusion, $$\text{tr}(A)=\text{tr}(XY)=\text{tr}(YX)=\text{tr}(A)\cdot r,$$ giving $r=1$. To conclude, let $B,C\in {\mathcal{M}}_n(\mathbb{C})$. We have $ABA=XYBXY=pXY=pA$, where $p=YBX\in {\mathcal{M}}_1(\mathbb{C})$. In the same way $ACA=qA$ with $q\in \mathbb{C}$, concluding the result of the problem.