Let λ be an arbitrary eigenvalue of a Hermitian matrix A and let x be an eigenvector corresponding to the eigenvalue λ.
Then we have
Ax=λx.(*)
Multiplying by x¯T from the left, we obtain
x¯T(Ax)=x¯T(λx)=λx¯Tx=λ||x||.
We also have
x¯T(Ax)=(Ax)Tx¯=xTATx¯.
The first equality follows because the dot product u⋅v of vectors u,v is commutative.
That is, we have
u⋅v=uTv=vTu=v⋅u.
We applied this fact with u=x¯ and v=Ax.
Thus we obtain
xTATx¯=λ||x||.
Taking the complex conjugate of this equality, we have
x¯TA¯Tx=λ¯||x||.(**)
(Note that x¯¯=x. Also ||x||¯¯¯¯¯¯¯¯¯=||x|| because ||x|| is a real number.)
Since the matrix A is Hermitian, we have A¯T=A.
This yields that
λ¯||x||=(∗∗)x¯TAx=(∗)x¯Tλx=λ||x||.
Recall that x is an eigenvector, hence x is not the zero vector and the length ||x||≠0.
Therefore, we divide by the length ||x|| and get
λ=λ¯.
It follows from this that the eigenvalue λ is a real number.
Since λ is an arbitrary eigenvalue of A, we conclude that all the eigenvalues of the Hermitian matrix A are real numbers.