Properties of the Cone of Positive Semidefinite Matrices

The set of positive semidefinite symmetric real matrices form a cone. We can define an order over the set of matrices by saying $X\geq Y$ if and only if $X-Y$ is positive semidefinite. I suspect that this order does not have the lattice property, but I would still like to know which matrices are candidates for the meet and join of two matrices.

In other words, let $P$ be the cone of positive semidefinite matrices. Is there a nice characterization of the set $(X+P)\cap (Y+P)$, for two given matrices? What are the minimal points in this intersection?

Solutions Collecting From Web of "Properties of the Cone of Positive Semidefinite Matrices"

Geometric Interpretation. Consider a positive definite matrix $A$. It defines the ellipsoid
$${\cal E}_A = \{u: u^T A u \leq 1\}.$$
Note that the correspondence between $A$ and ${\cal E}_A$ is one-to-one. Moreover, $A\succeq B$ if and only if ${\cal E}_B$ contains ${\cal E}_A$. Using this fact, we can give the following characterization.

$C \succeq A$ and $C \succeq B$ if and only if ${\cal E}_C \subset {\cal E}_A \cap {\cal E}_B$.

Similarly,

$C \preceq A$ and $C \preceq B$ if and only if ${\cal E}_C \supset {\cal E}_A \cup {\cal E}_B$.

Now a matrix $C$ is a minimal matrix s.t. $C \succeq A$ and $C \succeq B$ if and only if ${\cal E}_C \subset {\cal E}_A \cap {\cal E}_B$ and there is no ellipsoid “sandwiched” between ${\cal E}_C$ and ${\cal E}_A \cap {\cal E}_B$. It is easy to see that that happens iff $\partial{\cal E}_C$ intersects $\partial{\cal E}_A$ by a $k$-dimensional ellipsoid and $\partial{\cal E}_C$ intersects $\partial{\cal E}_B$ by an $n-k$ dimensional ellipsoid. This can be stated in terms of matrices $A$, $B$, and $C$.

Consider a minimal $C$ s.t. $C \succeq A$ and $C \succeq B$. Then there are two subspaces $U$ and $V$ with ${\mathbb R}^n = U \oplus V$ such that $u^T Cu = u^T Au$ for $u\in U$ and $u^T Cu = u^T Bu$ for $u\in V$; in particular, $\operatorname{rank}(C-A) + \operatorname{rank}(C-B) \leq n$.

Visualization. The following animation shows ellipses ${\cal E}_C$ inscribed in ${\cal E}_A \cap {\cal E}_B$ in two dimensions. Every ellipse ${\cal E}_C$ corresponds to a minimal matrix $C$ ($C \succeq A$ and $C \succeq B$)

Ellipse ${\cal E}_C$ inscribed in ${\cal E}_A \cap {\cal E}_B$

And this animation shows minimum ellipses ${\cal E}_C$ containing ${\cal E}_A \cup {\cal E}_B$.

minimum ellipses ${\cal E}_C$ containing ${\cal E}_A \cup {\cal E}_B$

I’m afraid that there is no more explicit characterization of sets $\{C: C \succeq A \text{ and } C \succeq B\}$ and $\{C: C \preceq A \text{ and } C \preceq B\}$.

Correspondence between “meet” and “join” matrices. Note that if $C_1$ is a “join” then $C_2 = A+B-C_1$ is a “meet” and vice versa. That follows from the fact that $C_2 \preceq A$ iff $A+B-C_1 \preceq A$ iff $B \preceq C_1$; similarly, $C_2 \preceq B$ iff $A \preceq C_1$.

Characterization for $2\times 2$ matrices. If $A$ and $B$ are $2\times 2$ matrices (s.t. $A\not\preceq B$ and $A\not\succeq B$) then meet and join matrices $C$ must satisfy the following equations $\det(C-A) = 0$ and $\det(C-A) =0$. The set of symmetric matrices that satisfy this system of equations forms a one dimensional curve in the space of matrices. Let us write
$$C_{xyz} = \begin{pmatrix}x&y\\y&z\end{pmatrix}.$$ Then the set
$$\{(x,y,z): \det(C_{xyz} – A) = \det(C_{xyz} – B) = 0\}$$
is a hyperbola (that lies in a plane in ${\mathbb R}^3$). Points on one branch of the hyperbola correspond to join matrices; points on the other branch correspond to meet matrices.

Notes. Note that by changing the basis we can always assume that $A=I$ and $B$ is a diagonal matrix, but I don’t think that this observation leads to a very explicit characterization of $\{C: C \succeq A \text{ and } C \succeq B\}$. In particular, $C$ does not have to be a diagonal matrix. For example, let
$$A=\begin{pmatrix} 1& 0\\0& 1\end{pmatrix}\quad B =\begin{pmatrix} 2& 0\\0& 1/2\end{pmatrix}.$$ Then the following matrices $C$ are minimal matrices greater ($\succeq$) than $A$ and $B$:
$$C=\begin{pmatrix} 2& 0\\0& 1\end{pmatrix}\quad C=\begin{pmatrix} 3& 1\\1& 3/2\end{pmatrix} \quad C=\begin{pmatrix} 3& -1\\-1& 3/2\end{pmatrix}.$$
(The set of all such matrices $C$ forms a one dimensional curve in the space of all $2\times 2$ matrices.)