The eigenvalues of a Hermitian matrix are always real
Theorem
Let $A$ be a Hermitian matrix of size $n \times n$. Then, the eigenvalues of $A$ are all real.
Explanation
In general, there is no guarantee that the eigenvalues of a matrix are real, but for Hermitian matrices, this can be verified through proof.
It may not be intuitively obvious, but the proof itself is relatively simple, and it is quite useful as a fact. It yields various good results when combined with concepts like positive-definiteness, so it is well worth knowing.
Proof
Let’s say the eigenvalue of $A$ is $\lambda$ and the corresponding eigenvector is $\mathbf{x}$. Then, the eigenvalue equation is as follows.
$$ A \mathbf{x} = \lambda \mathbf{x} $$
Multiplying both sides by $\mathbf{x}^{\ast}$ yields:
$$ \begin{equation} \mathbf{x}^{\ast} A \mathbf{x} = \lambda \mathbf{x}^{\ast} \mathbf{x} \end{equation} $$
Taking the conjugate transpose $^{ \ast }$ on both sides, by the properties of conjugate transpose, we get:
$$ \begin{align*} \left( \mathbf{x}^{\ast} A \mathbf{x} \right) ^{\ast} = & \left( \lambda \mathbf{x}^{\ast} \mathbf{x} \right) ^{\ast} \\ \implies \mathbf{x}^{\ast} A^{ \ast } \mathbf{x} = & \overline{\lambda} \mathbf{x}^{ \ast } \mathbf{x} \end{align*} $$
Since $A$ is a Hermitian matrix, we have $A=A^{\ast}$, and the equation can be rewritten as:
$$ \begin{equation} \mathbf{x}^{\ast} A \mathbf{x} = \overline{\lambda} \mathbf{x}^{ \ast } \mathbf{x} \end{equation} $$
By $(1)$ and $(2)$, the following equation holds:
$$ \lambda \mathbf{x}^{ \ast } \mathbf{x} = \mathbf{x}^{ \ast } A \mathbf{x} = \mathbf{x}^{ \ast } A^{ \ast } \mathbf{x} = \overline{\lambda} \mathbf{x}^{ \ast } \mathbf{x} $$
Therefore, $$ ( \lambda - \overline{\lambda} ) \mathbf{x}^{ \ast } \mathbf{x} = 0 $$
However, since $\mathbf{x}$ is an eigenvector, it is not $\mathbf{0}$. Therefore, $\mathbf{x}^{\ast} \mathbf{x} \ne \mathbf{0}$ must be true. Hence,
$$ \lambda = \overline{\lambda} $$
This implies that $\lambda$ is real.
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