Let A be a real n times n matrix Suppose A is skewsymmetric

Let A be a real n times n matrix. Suppose A is skew-symmetric, that is A^T = -A. Prove that each nonzero eigenvalue lambda of A is a pure imaginary number, i.e. lambda = ib for some real number b NotEqual 0.

Solution

Let A be a skew symmetric matrix. Then that for all x, y Cⁿ we have<Ax, y>=<x, A*y>,where A* is the adjoint or conjugate transpose of A. Here, since A is a real matrix, the adjoint of A is equal to its transpose, so that for every x, y Cⁿ , we have <Ax, y> =<x, A^Ty> = <x,-Ay> = - <x, Ay>. Now assuming that x is an eigenvector of A with corresponding eigenvalue , when y = x, we have Ax, x = x, x=x². Also x, Ax=x, x = ^*x,x = *x², where ^* is complex conjugate of . Now, since x is an eigenvector of A, hence x0. Further, Ax, x =x, Ax. Therefore, x² =*x² . Hence, =* , and hence is either 0 or a purely imaginary number.