Given a zonotope $Z$ generated by vectors $\{\vec v_i\}$, where a zonotope is a $convex$ set of points $\{z\in Z|z=\sum_ic_i\vec v_i;0\le c_i\le1~\forall_i\}$. Replace one of the generating vectors by two new generators that sum to the first. That is, wlog, replace $\vec v_1$ by $\vec w_1=p\vec v_1+\vec\Delta$ and $\vec w_2=(1-p)\vec v_1-\vec\Delta$, where $0\le p\le 1$, and we can take $\vec\Delta$ to be orthogonal to $\vec v_1$.
Then my question is this: is it possible to prove that the volume of the new zonotope is always strictly greater than that of the original. By volume, I mean the $d$-dimensional volume (e.g., ordinary area in $d$=2), where $d$ is the dimension of the space spanned by $\vec w_1,\vec w_2$, and $\{\vec v_i\}_{i\ne1}$. The conjecture is trivially true when $\vec\Delta$ does not lie in the span of $\{\vec v_i\}$, since then the original volume vanishes in the new, higher-dimensional space, and the final volume does not. Intuitively, one might think that it ought to be true in general, at least from the simplistic view of considering a $d=2$ example where the original generating set has only two vectors. Then, one starts with the area of a parallelogram, and the new zonotope just adds two extra triangular pieces, each being of area equal to $v_1\Delta/2$ ($v_1=|\vec v_1|$, etc.), which is just the usual formula for the area of a triangle as one-half the base times the height.
One finds that the volume of a zonotope can be calculated as the sum of absolute values of determinants; see High-dimensional shapes with known volume formulas for a precise definition. For each of these determinants that includes $\vec v_1$, call it $D_n(\vec v_1)$, one obtains new contributions by replacing that original determinant by two new determinants, each of these two replacing the column $\vec v_1$ by $\vec w_1$ or $\vec w_2$, respectively. [In addition to these new determinants, one will also have new contributions from determinants that include both $\vec w_1$ and $\vec w_2$ as columns; see below.] Defining $D_n(\vec X)$ as this particular determinant that has $\vec X$ in the column where $\vec v_1$ used to be, then $D_n(\vec w_1)=pD_n(\vec v_1)+D_n(\vec\Delta)$ and $D_n(\vec w_2)=(1-p)D_n(\vec v_1)-D_n(\vec\Delta)$. However, since to obtain the new volume, we are adding ${absolute~values}$ of these determinants, the result depends on the signs and relative magnitudes of $D_n(\vec v_1)$ and $D_n(\vec\Delta)$, which does not seem to lead to an obvious conclusion. In some cases, one finds that $|D_n(\vec w_1)|+|D_n(\vec w_2)|=|D_n(\vec v_1)|$, but there are other cases where it appears possible that $|D_n(\vec w_1)|+|D_n(\vec w_2)|<|D_n(\vec v_1)|$.
As mentioned above, there are the contributions from new determinants that have both $\vec w_1,\vec w_2$ appearing as columns. Since these are entirely new and not just replacing determinants that were previously present, taken by themselves they would increase the volume. However, there is still the issue noted at the end of the preceding paragraph, which appears to allow a possibility that the volume decreases.
Finally, let me mention that if we can't prove the desired result for general zonotopes, consider this: I am actually working with (complex) positive semidefinite (PSD, and thus, Hermitian) matrices. That is, the generating "vectors" of the zonotopes are PSD matrices, and in particular, the generating set always sums to the identity matrix on the overall (Hilbert) space. Then, in calculating determinants, one just reshapes each of these matrices into a column vector. So, if the conjecture isn't provable for general zonotopes, can it be proved for this special case involving PSD matrices?
If $\vec{\Delta}=0$ then the volume does not grow because the new zonotope $Z'$ is equal to the original zonotope $Z$. Suppose then that $\vec{\Delta} \ne 0$. We'll first show there is some new point added to $Z'$. Consider the maximal point of $Z$ in the $\vec{\Delta}$ direction:
$$ \begin{align} m &= \max\left\{ \vec{\Delta} \cdot \vec{z} \middle| \vec{z} \in Z \right\} \\ &= \max\left\{ \sum_{i \ne 1} c_i \vec{\Delta} \cdot \vec{v_i} \right\} \end{align}. $$
Note we need only sum over $i \ne 1$ because $\vec{\Delta} \cdot \vec{v}_1 = 0$. Maximization over $Z'$ gives a strictly greater value. The maximal value will require the coefficient of $\vec{w}_1$ to be $1$ and that of $\vec{w}_2$ to be zero.
$$ \begin{align} \max\left\{ \vec{\Delta} \cdot \vec{z} \middle| \vec{z} \in Z' \right\} &= \max\left\{ \vec{\Delta} \cdot \vec{\Delta} + \sum_{i \ne 1} c_i \vec{\Delta} \cdot \vec{v_i} \right\} \\ &> m \end{align}. $$
Therefore $Z' \ne Z$. Since $Z' \supseteq Z$, there is some point $\vec{x} \in Z'$ that is not in $Z$. It should be intuitively clear that the volume has increased, if $Z$ was of non-zero volume, because we have extended a closed convex set. Since $Z$ is closed and convex, the Hyperplane Separation Theorem guarantees a hyperplane separating $\vec{x}$ from $Z$. Let $B$ be a ball in the interior of $Z$. For some $\alpha$ sufficiently small, $\alpha B + (1-\alpha)\vec{x}$ will be on the same side of the hyperplane as $\vec{x}$, and in particular will be disjoint from $Z$. Because $Z'$ is convex, $\alpha B + (1-\alpha)\vec{x} \in Z'$. This is a subset of non-zero area in $Z' - Z$. Thus the volume has increased.