Hermitian Matrix's Loewner Order
Definition
Loewner Order
Let’s assume two matrices $A, B \in \mathbb{C}^{n \times n}$ are Hermitian matrices. If $\left( A - B \right)$ is positive semidefinite, it is denoted as $A \ge B$, and if $\left( A - B \right)$ is positive definite, it is denoted as $A > B$. This kind of partial order $\ge$, $>$ are called the Loewner order.
Explanation
Unlike commonly considered scalars, and indeed even with complex numbers, defining a natural order among them is not an easy task. Especially in the case of matrices, which have many more components, defining a valid order for them is difficult. However, the notion of positive definiteness in matrices provides a way to assign a kind of sign to matrices, enabling the establishment of an order among them.
Theorem
$$ \begin{align*} \mathbb{H}_{n} :=& \left\{ A \in \mathbb{C}^{n \times n} : A = A^{\ast} \right\} \\ \mathbb{P}_{n} :=& \left\{ A \in \mathbb{C}^{n \times n} : \mathbf{x}^{\ast} A \mathbf{x} > 0 \right\} \\ \overline{\mathbb{P}_{n}} :=& \left\{ A \in \mathbb{C}^{n \times n} : \mathbf{x}^{\ast} A \mathbf{x} \ge 0 \right\} \end{align*} $$ For the set of positive definite matrices in the vector space of Hermitian matrices $\mathbb{H}_{n}$, denoted as $\mathbb{P}_{n}$, let’s denote the set of positive semidefinite matrices as $\overline{\mathbb{P}_{n}}$. $\overline{\mathbb{P}_{n}}$ is a partially ordered set with respect to $\ge$, but cannot be a totally ordered set.
Proof
It suffices to provide a counterexample. $$ \begin{align*} A :=& \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \\ B :=& \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \end{align*} $$ The two matrices $A, B$ given above are positive semidefinite; however, $$ \begin{align*} A - B =& \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \\ B - A =& \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \end{align*} $$ for any vector $\left( x_{1} , x_{2} \right) \in \mathbb{R}^{2}$, its quadratic form always becomes of the form $x_{1}^{2} - x_{2}^{2}$, which cannot all be positive semidefinite. In this case, since neither $A > B$ nor $B > A$ holds, $\overline{\mathbb{P}_{n}}$ cannot form a totally ordered set.
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