Question 376126
Consider the matrix A = [a(j,k)], where a(j,k) is the entry in the jth row and kth column. It must follow that A must be a square matrix.  The diagonal entries of A are a(1,1), a(2,2), a(3,3), ..., a(n,n).  If we take the transpose of A, the diagonal entries remain fixed. Now {{{A^T = -A}}}, and the diagonal entries of -A are -a(1,1), -a(2,2), -a(3,3), ..., -a(n,n). 
 Thus a(1,1) = -a(1,1), a(2,2) = -a(2,2), a(3,3) = -a(3,3), ..., 
a(n,n) = -a(n,n).  These imply that  a(1,1) = 0, a(2,2) = 0, a(3,3) = 0, ...,
 a(n,n) = 0, and the proof is complete.