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The Eigenvalues of a Hermitian Matrix Are Always Real 📂Matrix Algebra

The Eigenvalues of a Hermitian Matrix Are Always Real

Theorem

Let $A$ be an $n \times n$ Hermitian matrix. Then all eigenvalues of $A$ are real.

Explanation

In a general matrix there is no guarantee that the eigenvalues are real, but for Hermitian matrices we can confirm through a proof that they are real.

It is not easy to grasp intuitively, but the proof itself is fairly simple, and it is quite useful as a fact. Combined with concepts such as positive definiteness that follow later, it yields various good results, so be sure to remember it.

Proof

Let the eigenvalue of $A$ be $\lambda$, and let the eigenvector corresponding to $\lambda$ be $\mathbf{x}$. Then the eigenvalue equation is as follows.

$$ A \mathbf{x} = \lambda \mathbf{x} $$

Multiplying both sides on the left by $\mathbf{x}^{\ast}$ gives the following.

$$ \begin{equation} \mathbf{x}^{\ast} A \mathbf{x} = \lambda \mathbf{x}^{\ast} \mathbf{x} \end{equation} $$

Taking the conjugate transpose $^{ \ast }$ of both sides, by the properties of the conjugate transpose we obtain the following.

$$ \begin{align*} \left( \mathbf{x}^{\ast} A \mathbf{x} \right) ^{\ast} = & \left( \lambda \mathbf{x}^{\ast} \mathbf{x} \right) ^{\ast} \\ \implies \mathbf{x}^{\ast} A^{ \ast } \mathbf{x} = & \overline{\lambda} \mathbf{x}^{ \ast } \mathbf{x} \end{align*} $$

Since $A$ is a Hermitian matrix, $A=A^{\ast}$, and the equation above becomes the following.

$$ \begin{equation} \mathbf{x}^{\ast} A \mathbf{x} = \overline{\lambda} \mathbf{x}^{ \ast } \mathbf{x} \end{equation} $$

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 $$ ( \lambda - \overline{\lambda} ) \mathbf{x}^{ \ast } \mathbf{x} = 0 $$

But since $\mathbf{x}$ is an eigenvector, it is not $\mathbf{0}$. Therefore $\mathbf{x}^{\ast} \mathbf{x} \ne \mathbf{0}$. Hence

$$ \lambda = \overline{\lambda} $$

holds, and this means that $\lambda$ is real.

See Also