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 $\mathbb{R}^n$.



Numerical range in subspaces
The question is equivalent to search for the numerical range of overlapping block matrices.


The question is answered by studying overlapping matrices [here](https://math.stackexchange.com/questions/2416236/overlapping-positive-definite-block-matrices).


