How to prove this theorem? Let S be a real symmetric matrix. Then S has only real eigenvalues.?

1 Answer
Apr 5, 2018

See below:

Explanation:

Let #A# be a real symmetric matrix with an eigenvector #x# corresponding to an eigenvalue #lambda#

#Ax = lambda x#

Consider the number (strictly speaking - #1 times 1# matrix)

#x^+Ax = lambda x^+x#

Here "+" denotes Hermitian adjoint - i,e, a transpose followed by a complex conjugation.

Taking the adjoint of both sides, we get

#(x^+Ax)^+ = lambda^**(x^+x)^+#

Using #(AB)^+ = B^+A^+#, this becomes

#x^+A^+(x^+)^+ =lambda^**x^+(x^+)^+ #

or

#x^+A^+x = lambda^**(x^+x)#

Since #A# is real symmetric, we have

#A^+ = A#

and so #x^+A^+x = x^+Ax = lambda x^+x#, leading to

#lambda (x^+x)= lambda^**(x^+x)#

Since #x# is not a null vector #x^+x ne 0#, and so

#lambda=lambda^**#