This is elementary, yet useful and interesting.

Statement: All eigenvalues of a Hermitian matrix are real.


If the matrix A is Hermitian, by definition, A^*=A\,.\ \ \ \ (1)

Let \lambda be an eignevalue of A, and \mathbf{x} be the corresponding eigenvector. Let

\mathbf{x}=\begin{bmatrix}a_1+ib_1\\a_2+ib_2\\\vdots\\a_n+ib_n\end{bmatrix}\,, and

\mathbf{x}^*=\begin{bmatrix}a_1-ib_1\ \ a_2-ib_2\ \ \cdots\ \ a_n-ib_n\end{bmatrix}\,.


q=\mathbf{x}^*A\mathbf{x}=\mathbf{x}^*\lambda \mathbf{x}=\lambda \mathbf{x}^*\mathbf{x}=\lambda \sum_{i=i}^n{a_i^2+b_i^2}\,.\ \ \ \ (2)

Since \mathbf{x} is an n\times 1 vector, q is just a number (real or complex).

Now, it follows from (1) that q^*=\left(\mathbf{x}^*A\mathbf{x}\right)^*=\mathbf{x}^*A^*\mathbf{x}=\mathbf{x}^*A\mathbf{x}=q.

Any complex number which is equal to its conjugate must have the imaginary part equal to zero. Therefore, q must be real, which implies \displaystyle \lambda \sum_{i=i}^n{a_i^2+b_i^2} must also be real. Since the sum \sum_{i=i}^n{a_i^2+b_i^2} is also real, it follows the \lambda must be real.