74th Romanian Mathematical Olympiad

a) Assume $A=(a_{ij}{)}_{1\le i,j\le n}$ and $A^2=(m_{ij}{)}_{1\le i,j\le n}$. The property $A^T=-A$ leads to the relations $a_{ji}=-a_{ij}$, for $i,j=1,\dots ,n$, that is $A$ is antisymmetric. Then $$m_{ii}=\sum _{j=1}^n{a}_{ij}{a}_{ji}=-\sum _{j=1}^n{a}_{ij}^2,i=1,\dots ,n.$$ If $A^2=O_n$, then $m_{ii}=0$, $i=1,\dots ,n$. Since $A\in {\mathcal{M}}_n(\mathbb{R})$, we obtain $a_{ij}=0$, for $i,j=1,\dots ,n$. So $A=O_n$.

b) From the assumption, $A=B^{\ast }$, where $B^{\ast }$ is the adjoint of $B$. Since $n$ is an odd integer number, we obtain $\det (A)=\det (A^T)=\det (-A)=(-1{)}^n\det (A)=-\det (A)$. Then we obtain $\det (A)=0$. Therefore $\det (B{B}^{\ast })=\det (B)\cdot \det (B^{\ast })=\det (B)\cdot \det (A)=0$. From the relation $B{B}^{\ast }=\det (B)I_n$, we get $\det (B)=0$. Hence $\text{rank}(B)\le n-1$ and $B{B}^{\ast }=O_n$.

Case 1. If $\text{rank}(B)\le n-2$, then $B^{\ast }=O_n$. Thus, $A^2=(B^{\ast }{)}^2=O_n$.

Case 2. If $\text{rank}(B)=n-1$, then $B^{\ast }\ne O_n$. From the Sylvester rank inequality, we have $\text{rank}(B^{\ast })\le \text{rank}(B{B}^{\ast })+n-\text{rank}(B)=1$. We conclude $\text{rank}(B^{\ast })=1$ and $(B^{\ast }{)}^2=\text{tr}(B^{\ast })B^{\ast }$. But $\text{tr}(B^{\ast })=\text{tr}(A)=0$ because $a_{ii}=-a_{ii}$, $i=1,2,\dots ,n$. Finally, we get $A^2=(B^{\ast }{)}^2=O_n$.