## 2 Preliminaries
### Interaction Matrix  
We propose an interaction matrix \(G\) for representing the interconnecting relations of quadrotors with leader-follower (L-F) configuration as follows.  
\[G=G^{D}-G^{A}+G^{L} \tag{1}\]  
**Remark 1**: _Let us note that the first two terms \(G^{D}-G^{A}\) are normally called the Laplacian in graph theory._  
**Proposition 1**: _Let \(\mathcal{G}\) be an undirected simple graph, then the interconnection matrix \(G\) in Equation (1), is positive-definite, if (i) \(\mathcal{G}\) is connected; (ii) \(\mathcal{V}_{L}\
eq\Phi\), where \(\Phi\) represents a null set._  
Since \(\mathcal{G}\) is connected, we have \(G^{D}-G^{A}\geq 0\). Considering the definition of \(G^{L}\), we have \(G^{L}\geq 0\). We prove this proposition by contradiction. Firstly, we suppose that there exists a nonzero vector \(x\in\mathbb{R}^{n}\), which renders  
\[x^{T}Gx=x^{T}(L+G^{L})x=x^{T}Lx+x^{T}G^{L}x=0\]  
Therefore, we must have \(x^{T}Lx=0\) and \(x^{T}G^{L}x=0\).  
Since \(x^{T}Lx=0\), we obtain that \(x=\alpha\mathbf{1}_{n}\), where \(\alpha\) is a nonzero scalar. According to the fact that \(\mathcal{V}_{L}\
eq\Phi\), then, \(G^{L}\
eq 0\). As a result, \(x^{T}G^{L}x=\alpha^{2}\mathbf{1}_{n}^{T}G^{L}\mathbf{1}_{n}>0\), which contradicts \(x^{T}G^{L}x=0\). Therefore, such a nonzero vector \(x\) does not exist. Thus, for any nonzero vector \(x\), \(x^{T}Gx>0\), namely, \(G\) is positive-definite.  
When condition (\(i\)) is not satisfied such that \(\mathcal{G}\) is not connected, the graph can be divided into \(s\) connected sub-graphs \(\mathcal{G}_{1},\mathcal{G}_{2},\ldots,\mathcal{G}_{s}\). In that case, according to Proposition 1, we have the following corollary.  
The interaction matrix \(G\) is positive-definite, if the leaders set satisfies \(\mathcal{V}_{L}\geq s\), and \(\mathcal{V}_{IL}\
eq\Phi\), where \(i=\{1,2,\ldots,s\}\).  
Proof.: If a multi-agent system contains \(s\) connected sub-groups of agents, then, the interaction matrix is block diagonal, which has \(s\) blocks on the diagonal. Since each sub-group \(\mathcal{G}_{i}\), \(i=\{1,2,\ldots,s\}\) is connected and \(V_{IL}\
eq\Phi\), then, the block in the interaction matrix is positive definite. Hence, the interaction matrix is positive definite.  
The normalized interaction matrix is defined by  
\[\tilde{G}=(G^{D}+G^{L})^{-1}G \tag{2}\]  
The normalized interaction matrix \(\tilde{G}\) is not symmetric, but it has the following property.  
The normalized interaction matrix \(\tilde{G}\) has real eigenvalues, if \(\mathcal{G}\) is undirected.  
Proof.: Since \(\tilde{G}\) is a normalized interaction matrix, then, \(\tilde{G}=(G^{D}+G^{L})^{-1}\cdot G\). We recall that \(G\) is symmetric.  
Using \((G^{D}+G^{L})^{-\frac{1}{2}}\), we apply similarity transformation to \(\tilde{G}\), then, we obtain that \((G^{D}+G^{L})^{\frac{1}{2}}\cdot\tilde{G}\cdot(G^{D}+G^{L})^{-\frac{1}{2}}=(G^ {D}+G^{L})^{-\frac{1}{2}}\cdot\tilde{G}\cdot(G^{D}+G^{L})^{-\frac{1}{2}}\) is symmetric, because \(G^{D}+G^{L}\) are diagonal and \(G\) is symmetric.  
Since \(\tilde{G}\) is similar to \((G^{D}+G^{L})^{-\frac{1}{2}}\cdot G\cdot(G^{D}+G^{L})^{-\frac{1}{2}}\), the eigenvalues of \(\tilde{G}\) are real, moreover, they are equal to the eigenvalues of \((G^{D}+G^{L})^{-\frac{1}{2}}\cdot G\cdot(G^{D}+G^{L})^{-\frac{1}{2}}\).  
The normalized interaction matrix \(\tilde{G}\) is positive-definite, if conditions in Proposition 1 or Corollary 1 are satisfied.  
Proof.: According to the conditions in Proposition 1 or Corollary 1, we know that \(G\) is invertible. Since \(\tilde{G}\) is similar to \((G^{D}+G^{L})^{-\frac{1}{2}}\cdot\tilde{G}\cdot(G^{D}+G^{L})^{-\frac{1}{2}}\) according to Proposition 2, we obtain that \(\tilde{G}\) is also positive-definite.