The eigenvalues of a Hermitian matrix are real.

Explanation:

A Hermitian matrix is a square matrix that is equal to its conjugate transpose. In other words, if A is a Hermitian matrix, then A = A* (where A* denotes the conjugate transpose of A).

Eigenvalues are the values λ for which there exist nonzero vectors v such that Av = λv. In other words, λ is an eigenvalue of A if and only if there exists a nonzero vector v such that Av = λv.

Proof:

Let A be a Hermitian matrix, and let λ be an eigenvalue of A. Then, there exists a nonzero vector v such that Av = λv.

Take the conjugate transpose of both sides of the equation: (Av)* = (λv)*. Since A = A*, we have (Av)* = (A*)*v = Av.

Therefore, Av = λv implies Av = (Av)* = (λv)*.

Now, let's multiply both sides of the equation by v*: v*(Av) = v*((λv)*).

Using the properties of complex conjugates, we have v*(Av) = (v*A)*v = (Av)*v = (λv)*v.

Since Av = (λv)*, we can substitute Av in the equation above: v*(Av) = (Av)*v = (λv)*v.

Expanding the equation, we have v*(Av) = (λv)*v = λ*(v*v).

Since v is nonzero, v*v is a positive real number. Therefore, λ*(v*v) is a real number.

So, v*(Av) = λ*(v*v) is a real number.

Since v*(Av) is a complex number and λ*(v*v) is a real number, λ must be a real number.

Hence, the eigenvalues of a Hermitian matrix are real.

Therefore, the correct answer is option 'C': Real.