# Numerical range in subspaces

Let $A \in \mathbb{R}^{n \times n}$ be nonsymmetric positive definite, if $A$ can be decomposed as $A = A_1 \oplus A_2$, where $A_1 \in \mathbb{R}^{p \times p}$ and $A_2 \in \mathbb{R}^{q \times q}$, $p+q=n$, it is known that \begin{align} W(A) = Co(W(A_1) \cup W(A_2)) \end{align} where $W(A) = \left\{\frac{(Av,v)}{(v,v)}, 0 \ne v \in \mathbb{C}^n \right\}$ is the numerical range anc $Co()$ is the convex hull.

Is there a similar property for when $A = R_1^T A_1 R_1 + R_2^T A_2 R_2$ but $p + q > n$?

$R_1:\mathbb{R}^n \rightarrow \mathbb{R}^p$ and $R_2:\mathbb{R}^n \rightarrow \mathbb{R}^q$ are orthogonal projections such that their ranges have a non null intersection and the union of their ranges is $\mathbb{R}^n$, i.e. $R_1 R_2 \ne \varnothing$, $R_2 R_1 \ne \varnothing$ and $R_1^TR_1 + R_2^T R_2$ has full rank.

I know of $W(A) \subseteq W(R_1^T A_1 R_1) + W(R_2^T A_2 R_2)$ where the sum is element-wise. But I know that if $A_1$ has no kernel in $\mathbb{R}^p$ and $A_2$ has no kernel in $\mathbb{R}^q$, then $A$ has no kernel in $\mathbb{R}^n$, and this doesn't show up in the formula since $R_1^T A_1 R_1$ and $R_2^T A_2 R_2$ obviously have a kernel in $\mathbb{R}^n$.