Toeplitz Matrices are Hermitian Matrices
Proof
A positive-definite matrix $A \in \mathbb{C}^{n \times n}$ is a Hermitian matrix. Naturally, a positive semi-definite matrix is also a Hermitian matrix.
Proof 1
$$ \mathbf{x}^{\ast} A \mathbf{x} = \lambda $$ If $A$ is a positive-definite matrix, for all $\mathbf{x} \in \mathbb{C}^{n}$, the quadratic form $\mathbf{x}^{\ast} A \mathbf{x}$ is expressed as some real number $\lambda \in \mathbb{R}$ as above. Taking the conjugate transpose on both sides, the complex conjugate of $\lambda \in \mathbb{R}$ is $\overline{\lambda} = \lambda$, which is itself, hence we obtain the following. $$ \begin{align*} & \mathbf{x}^{\ast} A \mathbf{x} = \lambda \\ \implies & \mathbf{x}^{\ast} A^{\ast} \mathbf{x} = \overline{\lambda} = \lambda \\ \implies & \mathbf{x}^{\ast} \left( A - A^{\ast} \right) \mathbf{x} = 0 \end{align*} $$
A necessary and sufficient condition for a quadratic form to be 0: A necessary and sufficient condition for the quadratic form $\mathbf{x}^{\ast} A \mathbf{x}$ to become $0$ for all $\mathbf{x} \in \mathbb{C}^{n}$ is that $A$ is a zero matrix: $$ \mathbf{x}^{*} A \mathbf{x} = 0 , \forall \mathbf{x} \in \mathbb{C}^{n} \iff A = O $$
Since $\left( A - A^{\ast} \right) = O$, $A$ is a Hermitian matrix.
The proof process shows that it does not particularly matter whether it is positive-definite or positive semi-definite, as seen in taking the conjugate of $\lambda$.
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