Real Positive Semidefinite Matrix Order
Why
Can we order the cone of positive semidefinite matrices?
Definition
The positive semidefinite matrix
order (or Loewner
order) is a partial ordering $\geq$ on
$\mathbfsf{S} ^d$ defined by
\[
A \geq B \quad \iff \quad A - B \geq 0 \quad \iff \quad
A - B \in \mathbfsf{S} _+^d.
\] \[
A > B \quad \iff \quad A - B > 0 \quad \iff \quad A -
B \in \mathbfsf{S} _{++}^d.
\]
Properties
Each of the following results from the
geometric properties of the positive semidefinite
cone:
\[
\begin{aligned}
\alpha A \geq 0 \quad & \text{ for all } \delta > 0, A
\geq 0, \\
A + B \geq 0 \quad & \text{ for all } A, B \geq 0, \\
A \geq B \text{ and } B \geq A \implies A = B \quad &
\text{ for all } A, B \in \mathbfsf{S} ^d, \\
\lim_{n \to \infty} A_n = A \implies A \geq 0 \quad &
\text{ for all sequences } (A_n)_n \text{ in }
\mathbfsf{S} _+^d.
\end{aligned}
\]
Partial Order
$A \geq B$ and $B \geq A$ giving $A = B$
means that $\geq$ is antisymmetric.
Moreover,
\[
\begin{aligned}
A \geq A \quad & \text{ for all } A \in \mathbfsf{S} ^d,
\text{ and } \\
A \geq B \text{ and } B \geq C \implies A \geq C & \text
{ for all } A, B, C \in \mathbfsf{S} ^d.
\end{aligned}
\]
For $d = 1$, $\geq$ reduces to the familiar total order of the real line (see Real Order). The converse perspective is to see the positive semidefinite order as an extension of the order on $\R $ to the space $\mathbfsf{S} ^d$. Of course, the key difference is that two matrices may not be comparable. The order is partial.
For example, the matrices $A, B \in \mathbfsf{S} ^2$ defined by
\[ A = \bmat{ 1 & 0 \\ 0 & 0 }, \quad B = \bmat{ 0 & 0 \\ 0 & 1 } \]
are not comparable. Neither $A \geq B$ nor $B \geq A$ holds.- Future editions will include more formal accounts. ↩︎