# On the Delayed Scaled Consensus Problems
## 4 Scaled Consensus with Signal Transmission Delays
It follows from (9) and some basic algebras [15] that:  
\[\chi^{\prime}(0)= \mathrm{tr}\left(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L} \alpha)\cdot\left(I_{n}+\tilde{\tau}\mathrm{sgn}(\alpha)\mathcal{A}\alpha \right)\right)\] \[= \mathrm{tr}\left(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L} \alpha)\cdot\left(I_{n}+\tilde{\tau}\mathrm{sgn}(\alpha)\mathcal{D}\alpha \right)\right)\] \[= \mathrm{tr}\left(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L} \alpha)\cdot\left(I_{n}+\tilde{\tau}\mathrm{sgn}(\alpha)\mathcal{D}\alpha \right)\right),\]  
where in the last equality, we note that \(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L}\alpha)\cdot\mathrm{sgn}(\alpha) \mathcal{L}\alpha=\det(\mathrm{sgn}(\alpha)\ \mathcal{L}\alpha)I_{n}=\det(\mathrm{sgn}(\alpha))\det( \mathcal{L})\det(\alpha)I_{n}=0\). Let \(\tilde{\ell}_{ii}\) represent the diagonal entries of \(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L}\alpha)\). Then:  
\[\chi^{\prime}(0)=\sum_{i=1}^{n}\tilde{\ell}_{ii}(1+\tau|\alpha_{i}|d_{i}). \tag{10}\]  
As in Theorem 1, we know that zero is a simple eigenvalue of \(|\alpha|\mathcal{L}\) since \(\mathcal{G}\) contains a spanning tree. Since zero is also a simple eigenvalue of \(\mathrm{sgn}(\alpha)\mathcal{L}\alpha\) due to similarity, the matrix \(\mathrm{sgn}(\alpha)\mathcal{L}\alpha\) has rank \(n-1\) and a one-dimensional kernel spanned by \(v_{1}=(\alpha_{1}^{-1},\cdots,\alpha_{n}^{-1})^{\mathrm{T}}\). Therefore, \(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L}\alpha)\
eq 0\). Moreover, we have \(\mathrm{sgn}(\alpha)\mathcal{L}\alpha\cdot\mathrm{adj}(\mathrm{sgn}(\alpha) \mathcal{L}\alpha)=\det(\mathrm{sgn}(\alpha)\mathcal{L}\alpha)I_{n}=0\), which implies that the columns of \(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L}\alpha)\) belong to the kernel of \(\mathrm{sgn}(\alpha)\mathcal{L}\alpha\). Accordingly, \(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L}\alpha)=v_{1}u\) for some row vector \(u=(u_{1},\cdots,u_{n})\). Analogously, the equality \(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L}\alpha)\cdot\mathrm{sgn}(\alpha) \mathcal{L}\alpha=0\) implies that the rows of \(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L}\alpha)\) belong to \(\mathrm{span}\{w_{1}\}\), where \(w_{1}\in\mathbb{R}^{n}\) is defined in the proof of Theorem 1. Hence, for each \(1\leq i\leq n\), there exists some \(c_{i}\in\mathbb{R}\) such that:  
\[\left(\frac{u_{1}}{\alpha_{i}},\cdots,\frac{u_{n}}{\alpha_{i}}\right)=c_{i}w_{ 1}^{\mathrm{T}}. \tag{11}\]  
If \(c_{i}=0\), then the \(i\)-th row in the matrix (11) becomes zero, and \(u_{1}=\cdots=u_{n}=0\). This indicates \(\mathrm{adj}(\mathrm{sgn}(\alpha)\mathcal{L}\alpha)=0\), which is a contradiction. Therefore, \(c_{i}\
eq 0\) for all \(i\). We can show in the same way that \(u_{i}\
eq 0\) for all \(i\). Therefore, there exists some \(k\
eq 0\) such that \(c_{i}=\frac{k}{\alpha_{i}}\) and \(\tilde{\ell}_{ii}=\frac{kw_{1i}}{\alpha_{i}}\) for all \(1\leq i\leq n\), where we write \(w_{1}=(w_{11},\cdots,w_{1n})^{\mathrm{T}}\). In view of (10), we have:  
\[\chi^{\prime}(0)=k\sum_{i=1}^{n}\frac{w_{1i}}{\alpha_{i}}(1+\tilde{\tau}| \alpha_{i}|d_{i}). \tag{12}\]  
By definition, \(w_{1}^{\mathrm{T}}\mathrm{sgn}(\alpha)\mathcal{L}\alpha=0\). Writing \(w_{1}^{\mathrm{T}}\mathrm{sgn}(\alpha):=\tilde{w}_{1}^{\mathrm{T}}=(\tilde{w} _{11},\cdots,\tilde{w}_{1n})\), we obtain \(\tilde{w}_{1}^{\mathrm{T}}\mathcal{L}=0\). For a sufficiently small \(\varepsilon>0\), \(I_{n}-\varepsilon\mathcal{L}\) is a stochastic matrix, and \(\tilde{w}_{1}^{\mathrm{T}}(I-\varepsilon\mathcal{L})=\tilde{w}_{1}^{\mathrm{ T}}\). Based on the Perron-Frobenius theory, we know that \(\tilde{w}_{1}^{\mathrm{T}}\) (under an appropriate normalization) is the stationary distribution of the Markov chain described by the stochastic matrix. Hence, \(\tilde{w}_{1}^{\mathrm{T}}\
eq 0\) and has non-negative entries. Noting that \(\frac{w_{1i}}{\alpha_{i}}=\frac{\tilde{w}_{1i}\mathrm{sgn}(\alpha_{i})}{ \alpha_{i}}\) and \(1+\tau|\alpha_{i}|d_{i}>0\) for all \(i\), we see from (12) that \(\chi^{\prime}(0)\
eq 0\).