Let A elementof Rm times m be a symmetric matrix As you alre

Let A elementof R^m times m be a symmetric matrix. As you already learned in MAT 22A or MAT 67, an eigenvector of A is a nonzero vector x elementof R^m such that lambda x = lambda x for some lambda elementof C, the corresponding eigenvalue. (a) Prove that all eigenvalues of A are real. (b) Prove that if x and y are eigenvectors corresponding to distinct eigenvalues, then x and y

Solution

(a) Let C be an eigenvalue of a symmetric A R^nxn and let u Cⁿ be a corresponding eigenvector so that Au = u… (1)

On taking complex conjugates of both sides of (1) above, we get A* u*= * u* , i.e., Au* = ^*u* …(2)( as A is symmetric). Now, on pre-multiplying both the sides of (1) by (u*)^T,we get (u*)^T u = (u*)^TAu) =               ((u*)^T A)u = (A^Tu*)^T u ( as (AB)^T = B^T A^T).

= (Au*)^Tu (as A^T =A)= (^*u*)^T u ( as per (2) above) = (^*)(u*)^T u. Thus, ( – *)(u^*)^Tu = 0.However, u, being an eigenvector is non-zero and (u*)^T u =_i=1ⁿ u_i*u_i > 0 as at least one of the components of u is non-zero. Now, for any complex number z = a + ib we have z*z = a² + b² 0. Hence = *, i.e., and hence u are both real. Thus, all the eigenvalues of a mxm symmetric matrix are real.

(b) For any real matrix A and any arbitrary vectors x and y, we have Ax,y=x, A^T y.Now, if A is symmetric, and x and y are eigenvectors of A corresponding to distinct eigenvalues and , then x,y=x,y=Ax,y=x,A^Ty= x,Ay=x,y = x,y.Therefore, ()x,y=0. Further,since 0, hence x,y = 0 , i.e.,   x is orthogonal to y.