Romanian Mathematical Olympiad - Final Round

As $A$ is non-singular, we get $\text{rank}(A) \le n - 1$. Distinguish two cases:

i) $\text{rank}(A) \le n - 2$. Then $A^* = O_n$ and the conclusion follows immediately.

ii) $\text{rank}(A) = n-1$. Then $AA^* = O_n$ and by Sylvester's inequality $0 \ge \text{rank}(A) + \text{rank}(A^*) - n$, that is $\text{rank}(A^*) \le 1$.
It follows that $A = CL$ where $L \in \mathcal{M}_{n,1}(\mathbb{R})$ and $C \in \mathcal{M}_{1,n}(\mathbb{R})$. It follows that $LC = (a) \in \mathcal{M}_1(\mathbb{R})$, with $a = \text{tr}(A^*)$ and $(A^*)^2 = CLCL = C(a)L = aA^*$. Denote by $B = I_n + A^*$. Then $(B - I_n)^2 = a(B - I_n)$ or, equivalently, $B((a+2)I_n - B) = (a+1)I_n$, which, in turn, implies $a \ne -1$, that is $B$ is invertible.

Moreover, if $B$ is invertible but $a = -1$, then from $B(I_n - B) = O_n$ we deduce $B = I_n$. Thus $A^* = O_n$ implying $-1 = \text{tr}(A^*) = 0$, a contradiction.