Overlapping positive definite block matrices Let $A \in \mathbb{R}^{p \times p}$ and $B \in \mathbb{R}^{q \times q}$ be nonsymmetric positive definite matrices in the sense that $v^T A v > 0, \forall v \in \mathbb{R}^p$ and $v^T B v > 0, \forall v \in \mathbb{R}^q$.
I know that $A \oplus B$, i.e. a matrix with a block structure
\begin{align}
\pmatrix{A & 0 \\
         0 & B}
\end{align}
is positive definite and I also know that if $p=q$, then $A+B$ is positive definite as well.
Can I say that as well if the blocks overlap partially? When? 
For instance a matrix as follows:
\begin{align}
\pmatrix{a_{11} & a_{12} & a_{13} & a_{14} & 0& 0\\
         a_{21} & a_{22} & a_{23} & a_{24}& 0& 0\\
         a_{31} & a_{32} & a_{33} + b_{11} & a_{34} + b_{12} &b_{13}&b_{14}\\
         a_{41} & a_{42} & a_{43} + b_{21} & a_{44} + b_{22} &b_{23}&b_{24}\\
         0 & 0 & b_{31} & b_{32} &b_{33}&b_{34} \\
         0 & 0 & b_{41} & b_{42} &b_{43}&b_{44}}
\end{align}
 A: Let us first formalize what could be meant in general with the overlapping matrices you describe. 
Let $n$ be a natural number such that $p,q \leq n \leq p + q$ and $A' \in \Bbb R^{n \times n}$ denote the matrix given by $A'_{i,j} = A_{i,j}$ for $1 \leq i,j \leq p$  and $A'_{i,j} = 0$ if $i > p$ or $j > p$. Similarly, let $B' \in \Bbb R^{n \times n}$ be such that $B'_{i + (n-q),j + (n-q)} = B_{i,j}$ and $B_{i,j} = 0$ if $i < n-q$ or $j < n-q$.
So $A' = \begin{pmatrix} A &0 \\ 0 & 0 \end{pmatrix}$ and $B' = \begin{pmatrix} 0 &0 \\ 0 & B \end{pmatrix}$ with zero matrices of appropriate sizes.
Now the overlapping of $A$ and $B$ in $\Bbb R^{n \times n}$ is the matrix $A' + B'$.
To show that this is positive definite, let $\pi_1 : \Bbb R^n \to \Bbb R^p$ denote the projection to the first $p$ entries and $\pi_2: \Bbb R^n \to \Bbb R^q$ the projection to the last $q$ entries. Note that if $v \in \Bbb R^n$ is not zero, then $\pi_1(v) \neq 0$ or $\pi_2(v) \neq 0$ (because $n \leq p + q$).
For any nonzero $v \in \Bbb R^n$ we thus get 
$$ v^T (A' + B') v = \pi_1(v)^T A \pi_1(v) + \pi_2(v)^T B \pi_2(v) > 0$$
and so $A ' + B'$ is positive definite.
