[Original question] Given symmetric positive semi-definite (SPSD) matrices $P=\begin{pmatrix}A&B\\B^T&C\\\end{pmatrix}$ and $Q=\begin{pmatrix}D&E\\E^T&F\\\end{pmatrix}$ with SPSD diagonal blocks $A, C, D$, and $F$.
None of these blocks is necessarily diagonally-dominant. How can we identify the condition for $r$ or otherwise test that the matrix $X=\begin{pmatrix}A&(B+rE)/2\\(B+rE)^T/2&rF\\\end{pmatrix}$ is also SPSD? For practical considerations (yes, I'm an engineer), my application typically has a large size of $A$, with hundreds or thousands of rows and columns, while $F$ typically has only a few to a few dozen.
Oops! My edit of earlier today giving a potential solution was wrong. Removed it.
This problem arises from a continuous unconstrained game (each player optimizing a different objective of a set of variables, of which each player controls only a subset of the variables), and Rosen's (1965) analysis that the Nash equilibrium is unique if the assembled Hessian is N.D. for some scaled combination (and my source matrices are actually N.D.) My problem is two-player.
With the help of the Schur complement, we can say that $X$ is positive semidefinite if and only if the matrix $$ X/(rF) = A - \frac r4 (B + rE)(rF)^+ (B + rE)^T \\ = A - \frac 14 (B + rE)F^+ (B + rE)^T $$ is positive semidefinite (where $F^+$ denotes the Moore-Penrose pseudoinverse of $F$).
With that said, it is not clear to me if there is a nice way to take full advantage of the positivity of $P$ and $Q$. The fact that $P$ and $Q$ tell us that $A - BC^+B^T$ and $D - EF^+E^T$ are positive definite, but it is not clear if there is any way to take advantage of that here.