THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

We shall prove that 1) implies 2), the other implication being obvious. As $\det (A_1^2+A_2^2+\cdots +A_k^2)=0$ there is a matrix $X\in {\mathcal{M}}_{n,1}(\mathbb{R})$, $X\ne O_{n,1}$, such that $(A_1^2+A_2^2+\cdots +A_k^2)X=O_{n,1}$ (the homogeneous linear system has a non-zero solution). This implies $X^t(A_1^2+A_2^2+\cdots +A_k^2)X=O_n$, or $X^t{A}_1^t{A}_{1}X+X^t{A}_2^t{A}_{2}X+\cdots +X^t{A}_k^t{A}_{k}X=O_n$. Putting $A_{i}X=Y_i$, $i=1,2,\dots ,k$, we write $Y_1^t{Y}_1+Y_2^t{Y}_2+\cdots +Y_k^t{Y}_k=O_n$. Thus $A_{i}X=O_{n,1}$, or $A_i^{t}X=O_{n,1}$, $i=1,2,\dots ,k$.

For any matrices $B_1,B_2,\dots ,B_k\in {\mathcal{M}}_n(\mathbb{R})$ we have $(B_1^t{A}_1^t+B_2^t{A}_2^t+\cdots +B_k^t{A}_k^t)X=O_{n,1}$, which is equivalent to the fact that the system $(B_1^t{A}_1+B_2^t{A}_2+\cdots +B_k^t{A}_k)X=O_{n,1}$ has a non-trivial solution, so $\det (B_1^t{A}_1+B_2^t{A}_2+\cdots +B_k^t{A}_k)=\det (A_1{B}_1+A_2{B}_2+\cdots +A_k{B}_k)=0$.