📂Matrix Algebra

Hermitian Matrix Spaces and Convex Cones of Positive Semidefinite Matrices

Definition

Let’s denote as $n \in \mathbb{N}$.

Hermitian Matrix Space

The set of Hermitian matrices of size $n \times n$ is denoted as follows. $$\mathbb{H}_{n} := \left\{ A \in \mathbb{C}^{n \times n} : A = A^{\ast} \right\}$$

Positive Definite Matrix Set

The set of positive definite matrices of size $n \times n$ is denoted as $\mathbb{P}_{n}$.

Theorem

$\mathbb{H}_{n}$ is a Vector Space

• [1]: With respect to the scalar field $\mathbb{R}$, $\mathbb{H}_{n}$ is a vector space.

$\mathbb{P}_{n}$ is a Convex Cone of $\mathbb{H}_{n}$

• [2]: For all $a, b > 0$ and $X, Y \in \mathbb{P}_{n}$, the following holds: $$aX + bY \in \mathbb{P}_{n}$$ In other words, $\mathbb{P}_{n} \subset \mathbb{H}_{n}$ is a convex cone of $\mathbb{H}_{n}$.

Proof

[1]

As long as the scalar field of $\mathbb{H}_{n}$ is given as the set of real numbers $\mathbb{R}^{1}$, the scalar multiplication for any $X, Y \in \mathbb{H}_{n}$ is irrelevant to both matrix transpose and complex conjugate, satisfying various conditions of the vector space trivially.

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[2] ¹

Firstly, since a positive definite matrix is a Hermitian matrix, $\mathbb{P}_{n} \subset \mathbb{H}_{n}$ holds. If any $X , Y$ is a positive definite matrix, then for all vectors $\mathbf{v} \in \mathbb{C}^{n}$, $\mathbf{v}^{\ast} X \mathbf{v} > 0$ holds and since $\mathbf{v}^{\ast} Y \mathbf{v} > 0$, for all scalars $a, b > 0$, the following holds: $$\begin{align*} & \mathbf{v}^{\ast} \left( aX + bY \right) \mathbf{v} \\ =& \mathbf{v}^{\ast} aX \mathbf{v} + \mathbf{v}^{\ast} bY \mathbf{v} \\ =& a \mathbf{v}^{\ast} X \mathbf{v} + b \mathbf{v}^{\ast} Y \mathbf{v} \\ >& 0 \end{align*}$$

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