74th Romanian Mathematical Olympiad

For $Z\in M_n(\mathbb{C})$, there are $X,Y\in M_n(\mathbb{R})$ such that $Z=X+iY$, and $\bar{Z}=X-iY$. Thus $f(Z)=(A+B)X+i(A-B)Y$.

(1) $\Rightarrow$ (3). Suppose that $A+B$ or $A-B$ are singular. In case $A+B$ is singular, $\det (A+B)=0$, so there is $C\in M_{n,1}(\mathbb{R})\setminus \{O_{n,1}\}$ such that $(A+B)C=O_{n,1}$. Define $X\in M_n(\mathbb{R})$, $X\ne O_n$, with the $n$ columns the same as $C$. Then $f(X)=(A+B)X=O_n=f(O_n)$, in contradiction with (1). The case when $A-B$ is singular can be treated in the same manner.

(3) $\Rightarrow$ (1). Let $Z_1=X_1+i{Y}_1$, with $X_1,Y_1\in {\mathcal{M}}_n(\mathbb{R})$, and $Z_2=X_2+i{Y}_2$, with $X_2,Y_2\in {\mathcal{M}}_n(\mathbb{R})$, such that $f(Z_1)=f(Z_2)$. Then $(A+B)X_1+i(A-B)Y_1=(A+B)X_2+i(A-B)Y_2$. It follows that $(A+B)(X_1-X_2)=O_2$ and $(A-B)(Y_1-Y_2)=O_2$, and by (3) we get $X_1-X_2=O_2$ and $Y_1-Y_2=O_2$. So $X_1=X_2$ and $Y_1=Y_2$, that is $Z_1=Z_2$.

(2) $\Rightarrow$ (3). Suppose $A+B$ or $A-B$ is singular. In the case that $A+B$ is singular, $\det (A+B)=0$. For $Z=X+iY\in {\mathcal{M}}_n(\mathbb{C})$, we have $\det ((A+B)X)=0$, implying $f(Z)\ne I_n$ for all $Z\in {\mathcal{M}}_n(\mathbb{C})$, in contradiction with (2). The same proof works in case $A-B$ is singular.

(3) $\Rightarrow$ (2). Let $Z=X+iY\in {\mathcal{M}}_n(\mathbb{R})$. Define $U=(A+B{)}^{-1}X$ and $V=(A-B{)}^{-1}Y$. Then $f(U+iV)=X+iY=Z$, so $f$ is surjective.