# On the Delayed Scaled Consensus Problems
## 3 Undelayed Scaled Consensus Process  
In this section, we first consider the undelayed scaled consensus problem. Given the scales \((\alpha_{1},\cdots,\alpha_{n})\) with nonzero \(\alpha_{i}\)'s, we apply the nearest-neighbor rules to propose the distributed strategy for agent \(i\) as:  
\[u_{i}(t)=\mathrm{sgn}(\alpha_{i})\sum_{j=1}^{n}a_{ij}(\alpha_{i}x_{j}(t)- \alpha_{i}x_{i}(t)), \tag{2}\]  
where \(\mathrm{sgn}(\cdot)\) is the signum function. Let \(\mathcal{D}=\mathrm{diag}(d_{1},\cdots,d_{n})\in\mathbb{R}^{n\times n}\) be the degree diagonal matrix. Then, we have \(\mathcal{L}=\mathcal{D}-\mathcal{A}\). Moreover, for ease of presentation, we define \(\alpha:=\mathrm{diag}(\alpha_{1},\cdots,\alpha_{n})\in\mathbb{R}^{n\times n}\), \(|\alpha|:=\mathrm{diag}(|\alpha_{1}|,\cdots,|\alpha_{n}|)\in\mathbb{R}^{n\times n}\), \(\alpha^{-1}:=\mathrm{diag}(\alpha_{1}^{-1},\cdots,\alpha_{n}^{-1})\in\mathbb{R} ^{n\times n}\)and \(\text{sgn}(\alpha):=\text{diag}(\text{sgn}(\alpha_{1}),\cdots,\text{sgn}(\alpha_{n}) )\in\mathbb{R}^{n\times n}\). With these notations, the switched multi-agent system (1) with Protocol (2) can be recast as:  
\[\dot{x}(t)=-\text{sgn}(\alpha)\mathcal{L}\alpha x(t). \tag{3}\]  
The following result gives a concise characterization for the scaled consensus process governed by (3). Theorem 1: _The multi-agent system (3) achieves scaled consensus to \((\alpha_{1},\cdots,\alpha_{n})\) if and only if \(\mathcal{G}\) has a spanning tree._  
Proof: Sufficiency: It follows from (3) that \(x(t)=\exp(-\text{sgn}(\alpha)\mathcal{L}\alpha t)x(0)\) for \(t\geq 0\) using the matrix exponential. We first consider the matrix \(|\alpha|\mathcal{L}\). Clearly, it has zero row sums, and hence, zero is always an eigenvalue of \(|\alpha|\mathcal{L}\). By applying the Gershgorin disk theorem [15], we see that all eigenvalues of \(|\alpha|\mathcal{L}\) are in the right half plane, and its nonzero eigenvalues have strictly positive real parts (cf. [16]). Since \(\mathcal{G}\) has a spanning tree, \(\lambda_{1}=0\) is a simple eigenvalue of the Laplacian \(\mathcal{L}\), as well as of \(|\alpha|\mathcal{L}\) recalling that \(|\alpha|\) is a positive definite diagonal matrix [17]. Noting that \(\alpha^{-1}|\alpha|\mathcal{L}\alpha=\text{sgn}(\alpha)\mathcal{L}\alpha\), we see that the state matrix \(-\text{sgn}(\alpha)\mathcal{L}\alpha\) is similar to \(-|\alpha|\mathcal{L}\). Hence, all eigenvalues of \(-\text{sgn}(\alpha)\mathcal{L}\alpha\), denoted by \(\lambda_{1}=0,\lambda_{2},\cdots,\lambda_{n}\), are in the left half plane, and \(\lambda_{2},\cdots,\lambda_{n}\) have strictly negative real parts. It is direct to see that \(\exp(-\text{sgn}(\alpha)\mathcal{L}\alpha t)\) is a normal matrix, and its eigenvalues are \(e^{\lambda_{1}t},\cdots,e^{\lambda_{n}t}\). Therefore, \(\exp(-\text{sgn}(\alpha)\mathcal{L}\alpha t)\) possesses a complete set of eigenvectors, denoted by \(\{v_{1}=(\alpha_{1}^{-1},\cdots,\alpha_{n}^{-1})^{\text{T}},v_{2},\cdots,v_{n}\}\), which are exactly the corresponding eigenvectors of \(-\text{sgn}(\alpha)\mathcal{L}\alpha\)[15]. That is, \(-\text{sgn}(\alpha)\mathcal{L}\alpha v_{i}=\lambda_{i}v_{i}\) for \(i=1,\cdots,n\). Let \(\{w_{1},\cdots,w_{n}\}\) be the set of linearly independent eigenvectors of \((\exp(-\text{sgn}(\alpha)\mathcal{L}\alpha t))^{\text{T}}\), such that \(w_{i}^{\text{T}}v_{j}=\delta_{ij}\), where \(\delta_{ij}\) is the Kronecker delta for \(i,j=1,\cdots,n\). Hence, \(x(t)=\sum_{i=1}^{n}(w_{i}^{\text{T}}x(t))v_{i}\), and furthermore,  
\[x(t) =\exp(-\text{sgn}(\alpha)\mathcal{L}\alpha t)x(0)=\sum_{i=1}^{n}w_ {i}^{\text{T}}\exp(-\text{sgn}(\alpha)\mathcal{L}\alpha t)x(0)v_{i}\] \[=\sum_{i=1}^{n}e^{\lambda_{i}t}w_{i}^{\text{T}}x(0)v_{i}\to w _{1}^{\text{T}}x(0)v_{1} \tag{4}\]  
as \(t\) tends to infinity. Recalling \(v_{1}=(\alpha_{1}^{-1},\cdots,\alpha_{n}^{-1})^{\text{T}}\), we see that the multi-agent system (3) achieves scaled consensus to \((\alpha_{1},\cdots,\alpha_{n})\). Necessity: Suppose that \(\mathcal{G}\) does not contain a spanning tree.