OMOUS 2024
Let A be a real matrix such that A+A^2A^t+(A^t)^2=0. Prove that A=0. \textbf{Solution}. Multiply the given equation by A^t from the right to get AA^t+A^2(A^2)^t+(A^t)^3=0 and hence (A^t)^3=-AA^t-A^2(A^2)^t . Observe that RHS is negative semi-definite symmetric matrix. Thus (A^t)^3 and hence A^3 is symmetric negative semi-definite. Now multiply the given equation with A from the left to get A^2+A^3A^t+A(A^t)^2=0 and using A^3=(A^t)^3 we get A^2+(A^t)^4+A(A^t)^2=0. Transposing this equation we get (A^t)^2+A^4+A^2 A^t=0. Comparing with the original equation we get A^4=A. Thus A is diagonalizable and \lambda^4=\lambda for every characteristic root \lambda of A. Thus \lambda=0 or \lambda^3=1. But A^3 is negative semi-definite, so \lambda^3\le 0. Thus \lambda=0 and hence all roots of A are 0. Since A is diagonalizable, A=0.