Hermitian Matrices have Real Eigenvalues

Theorem

Every Hermitian matrix has real eigenvalues.

Corollary

Every real symmetric matrix has real eigenvalues.

Proof 1

Let $\mathbf A$ be a Hermitian matrix.

Then, by definition, $\mathbf A = \mathbf{A}^*$, where $^*$ designates the conjugate transpose.

Let $\lambda$ be an eigenvalue of $\mathbf A$.

Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$, of the matrix $\mathbf A$.

Now, by definition of eigenvector, $\mathbf{A v} = \lambda \mathbf v$.

Left-multiplying both sides by $\mathbf{v}^*$, we obtain:

$(1):\quad \mathbf{v}^* \mathbf{A v} = \mathbf{v}^* \lambda \mathbf v = \lambda \mathbf{v}^* \mathbf v$

Firstly, note that both $\mathbf{v}^* \mathbf{A v}$ and $\mathbf{v}^* \mathbf{v}$ are $1 \times 1$-matrices.

Now observe that, using Conjugate Transpose of Matrix Product: General Case:

$\left({\mathbf{v}^* \mathbf{A v}}\right)^* = \mathbf{v}^* \mathbf{A}^* \left({\mathbf{v}^*}\right)^*$

As $\mathbf A$ is Hermitian, and $\left({\mathbf{v}^*}\right)^* = \mathbf v$ by Double Conjugate Transpose is Itself, it follows that:

$\mathbf{v}^* \mathbf{A}^* \left({\mathbf{v}^*}\right)^* = \mathbf{v}^* \mathbf{A v}$

That is, $\mathbf{v}^* \mathbf{A v}$ is also Hermitian.

By Product with Conjugate Transpose Matrix is Hermitian, $\mathbf{v}^* \mathbf v$ is Hermitian.

So both $\mathbf{v}^* \mathbf{A v}$ and $\mathbf{v}^* \mathbf v$ are Hermitian $1 \times 1$ matrices.

Now suppose that we have for some $a,b \in \C$:

$\mathbf{v}^* \mathbf{A v} = \begin{bmatrix}a\end{bmatrix}$

$\mathbf{v}^* \mathbf v = \begin{bmatrix}b\end{bmatrix}$

Note that $b \ne 0$ as an eigenvector is non-zero.

By definition of a Hermitian matrix, $a = \bar a$ and $b = \bar b$.

By Complex Number Equals Conjugate iff Wholly Real, it follows that $a,b \in \R$ are real.

From equation $(1)$, it follows that $\begin{bmatrix}a\end{bmatrix} = \lambda \begin{bmatrix}b\end{bmatrix}$.

Thus, $a = \lambda b$, i.e. $\lambda = \dfrac a b$ (recall that $b \ne 0$).

Hence $\lambda$, being a quotient of real numbers, is real.

$\blacksquare$

Proof 2

Let $\mathbf A$ be a Hermitian matrix.

Then, by definition, $\mathbf A = \mathbf{A}^*$, where $^*$ designates the conjugate transpose.

Let $\lambda$ be an eigenvalue of $\mathbf A$.

Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$, of the matrix $\mathbf A$.

Denote with $\left\langle{\cdot, \cdot}\right\rangle$ the inner product on $\C$.

Because of the properties of a complex innerproduct, $\lambda * \left\langle v, v\right\rangle = \left\langle \lambda*v, v\right\rangle$. Since $\lambda*v = A*v$, we know $\left\langle\lambda*v, v\right\rangle = \left\langle A*v, v\right\rangle$. Using the properties of the adjugate, we know $\left\langle A*v, v\right\rangle = \left\langle v, A^**v\right\rangle$. Because $A^* = A$ and $A*v = \lambda * v$, it must be that $\left\langle v, A^**v\right\rangle = \left\langle v, A*v\right\rangle = \left\langle v, \lambda*v\right\rangle$. And again we use the properties of the innerproduct to conclude that $\left\langle v, \lambda*v\right\rangle = \overline{\lambda}*\left\langle v, v\right\rangle$.

Now we know that $\lambda*\left\langle v, v\right\rangle = \overline{\lambda}*\left\langle v, v\right\rangle$, and since $\left\langle v, v\right\rangle \neq 0$ ($v \neq 0$, and because of the positive definiteness, it must be that $\left\langle v, v\right\rangle \neq 0$), we can divide both sides by $\left\langle v, v\right\rangle$. And thus $\lambda = \overline{\lambda}$, and therefore every eigenvalue is a real number.

Therefore, Hermitian matrices have real eigenvalues.

$\blacksquare$