Eigenvalues of a Hermitian Matrix are Real

In this post, we prove the following:

Statement: All eigenvalues of a Hermitian matrix are real.

Proof:

Since 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}\,.

Then,

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)

The sum in the right hand side is real. It follows that \lambda will be real if and only if q is real.  Let us examine: Using A = A^*,

q^*=\left(\mathbf{x}^*A\mathbf{x}\right)^*=\mathbf{x}^*A^*\mathbf{x}=\mathbf{x}^*A\mathbf{x}=q,

thus q must be real. Consequently, \lambda must be real.

\square

In another post, we have shown that for a Hermitian matrix, the eigenvectors pertaining to different eigenvalues are orthogonal.

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