Let $A$ consist of the characteristic vectors of the union of two nested families. Show that $A$ is totally unimodular. A family of sets is nested if for any two members $A$ and $B$ we have that 
$A\subseteq B$, $B\subseteq A$ or $A\cap B = \emptyset$.
Exercise: Let $\mathcal{F}_1$ and $\mathcal{F}_2$ be two nested families of subsets of $\{1,\ldots,n\}$ and let $A$ be the matrix whose rows are the characteristic vectors of the sets in $\mathcal{F}_1 \cup \mathcal{F}_2$. Show that $A$ is totally unimodular.
What I've tried: $A$ is totally unimodular if for every square submatrix $B$ of $A$, we have that $\det B\in\{-1,0,1\}$. The rows of $A$ are the characteristic vectors of the sets in $\mathcal{F}_1 \cup \mathcal{F}_2$, meaning that a square submatrix $B$ will look something like this:
$$B = \begin{bmatrix}a_{11} & a_{12} & \ldots & a_{1k}\\
a_{21} & a_{22} & \ldots & a_{2k}\\
\vdots & \vdots & \vdots & \vdots\\
a_{m1} & a_{m2} & \dots & a_{mk} \end{bmatrix},$$ where $k<n$ and where $a_{ij} \in\{0,1\}$.  
Showing that $\det B \in\{-1,0,1\}$ would be a lot easier if  $\mathcal{F}_1\cup \mathcal{F}_2$ was a nested family. In that case there'd exist a minimal vector, and we could use row operations to clear all the $1's$ in the columns corresponding to a $1$ in the minimal vector. After that we could use induction to show that for any square submatrix $B$, $\det B\in\{-1,0,1\}$. 
However, those row operations wouldn't give the same results in this case, because we are not sure that a minimal vector exists. I think that I have to use the fact that $\mathcal{F}_1$ and $\mathcal{F}_2$ are nested families and split the matrix up, but I don't know how!
Question: How do I show that $A$ is totally unimodular?
Thanks in advance!
Edit: Consider the matrices $A_{\mathcal{F}_1}$ and $A_{\mathcal{F}_2}$ whose rows are the characteristic vectors of $\mathcal{F}_1$ and $\mathcal{F}_2$ respectively. That means that $A_{\mathcal{F}_1\cup \mathcal{F}_2}$ looks like this:
$$A_{\mathcal{F}_1 \cup \mathcal{F}_2} = \begin{bmatrix}\begin{bmatrix}A_{\mathcal{F}_1}\end{bmatrix}\\\begin{bmatrix}A_{\mathcal{F}_2}\end{bmatrix}\end{bmatrix}.$$
It can be shown that $A_{\mathcal{F}_1}$ and $A_{\mathcal{F}_2}$ are both totally unimodular, so I could rephrase my question like this: 
Suppose that $A$ is a block matrix whose two blocks are totally unimodular matrices. Show that $A$ is totally unimodular as well. 
 A: Oh, huh, I finally figured this out! Looking at any submatrix $B$, its rows also correspond to two nested families. Using row operations, you can make it so all the sets in family 1 are disjoint from each other, and the same for family 2. Now there are just two cases:


*

*The union of the sets in family 1 is not equal to the whole set. In this case, column expand along any element which is not in the union, and proceed by induction. Same applies if the union of family 2 is not everything.

*The union of both families is equal to the whole set. In this case, $\det B=0$ because the rows of $B$ are linearly dependent; namely, the sum of the rows in family 1 is equal to the all ones vector, and same for family 2!
A: We can write $A_{\mathcal{F}_1\cup\mathcal{F}_2}$ as a block matrix, consisting of $A_{\mathcal{F}_1}$ and $A_{\mathcal{F}_2}$. 
$$
A_{\mathcal{F}_1\cup\mathcal{F}_2}=\begin{bmatrix}
A_{\mathcal{F}_1}\\
A_{\mathcal{F}_2}
\end{bmatrix}
$$
We know that $A_{\mathcal{F}_1}$ and $A_{\mathcal{F}_1}$ both have a minimal row vector, and that we can use row permutations in such a way that all the elements in columns corresponding to a $1$ in the minimal vector are zero. Following the separate row permutations on $A_{\mathcal{F}_1}$ and $A_{\mathcal{F}_2}$,  $A_{\mathcal{F}_1\cup \mathcal{F}_2}$ will look like this:
$$
    A_{\mathcal{F}_1\cup \mathcal{F}_2} = \begin{bmatrix}\begin{bmatrix}a_{11} & a_{12} &\dots & 0 & \ldots & a_{1n}\\a_{21} & a_{22} & \ldots & 0 & \ldots & a_{2n}\\a_{i1}&a_{i2}&\ldots & 1 & \ldots & a_{in}\\\vdots &\vdots&\vdots&\vdots&\vdots&\vdots\\a_{m1}&a_{m2}&\ldots & 0 & \ldots & a_{mn}\end{bmatrix}\\\begin{bmatrix}a_{11} & a_{12} &0&\dots & \ldots & a_{1n}\\a_{21} & a_{22} & 0 & \ldots & \ldots & a_{2n}\\a_{j1}&a_{j2}&1&\ldots & \ldots & a_{jn}\\\vdots &\vdots&\vdots&\vdots&\vdots&\vdots\\a_{k1}&a_{k2}& 0 &\ldots  & \ldots & a_{kn}\end{bmatrix}\end{bmatrix}    
$$
where $m$ is the number of sets in $A_{\mathcal{F}_1}$ and $k$ is the number of sets in $A_{\mathcal{F}_2}$ and where $i$ and $j$ are the minimal rows of $A_{\mathcal{F}_1}$ and $A_{\mathcal{F}_2}$ respectively (the minimal row can very well be the first, the second or the last row in both matrices, other than in this representation). Now compare the $i$-th row of $A_{\mathcal{F}_1}$ and the $j$-th row of $A_{\mathcal{F}_2}$. Suppose that the set corresponding to $i$ is smaller that the set corresponding to $j$. Using column permutations we can rearrange the columns in $A_{\mathcal{F}_1}$ in such a way that for every $1$ in $i$ there's a $1$ in $j$. We can now use induction to prove that for any submatrix $B$, $\det B\in\{-1,0,1\}$. This works similar if we suppose that the set corresponding to $j$ is smaller than the set corresponding to $i$.
