Question

Given a symmetric nxn matrixA, how do you prove that an eigenvalue of A is a real number?

Answer 1

The only way i know how is to use the complex properties of the inner product:\[\langle cx,y\rangle=c\langle x,y\rangle\]\[\langle x,cy \rangle=\bar{c}\langle x,y\rangle\] where c bar is the complex conjugate of c. We also use the property:\[\langle Ax,y\rangle =\langle x,A^T y\rangle\] Let x be a non zero eigenvector of a symmetric n x n matrix with lambda as its eigenvalue. then we have:\[\langle Ax,x\rangle=\langle \lambda x,x\rangle =\lambda \langle x,x\rangle\]On the other hand:\[\langle Ax,x\rangle = \langle x,A^T x\rangle = \langle x, Ax\rangle=\langle x,\lambda x\rangle = \bar{\lambda} \langle x,x\rangle\]So we have:\[\lambda \langle x,x\rangle = \bar{\lambda}\langle x,x\rangle\]since x is not zero, the inner product of x with itself isnt zero, so we divde by the inner product of x with itself to obtain:\[\lambda = \bar{\lambda}\]But when does a complex number equal its conjugate? only when the imaginary part is 0. Hence the eigenvalue must be real.