## 3 First Step: Exact Placement of the Dominant Poles  
### Formulation for Symmetric Systems  
The method is first introduced with reference to a system with symmetric matrices. To understand the genesis of the method, let us consider the case of complete pole placement, i.e., \(p=2N\) as proposed in [1]. The gains ensuring exact placement of all the poles are those that solves the following linear system  
\[\left[\begin{array}{cc}\mu_{1}\boldsymbol{t}_{1}^{T}&\boldsymbol{t}_{1}^{T} \\ \mu_{2}\boldsymbol{t}_{2}^{T}&\boldsymbol{t}_{2}^{T}\\ \vdots&\vdots\\ \mu_{p}\boldsymbol{t}_{p}^{T}&\boldsymbol{t}_{p}^{T}\end{array}\right]\left( \begin{array}{c}\boldsymbol{f}\\ \boldsymbol{g}\end{array}\right)=\left\{\begin{array}{c}-1\\ -1\\ \vdots\\ -1\end{array}\right\}, \tag{3}\]  
where it is defined that  
\[\boldsymbol{t}_{i}=\mathbf{H}(\mu_{i})\boldsymbol{b}, \tag{4}\]  
with \(\mathbf{H}\) being the receptance matrix of the open-loop system,  
\[\mathbf{H}(\lambda)=\left[\lambda^{2}\mathbf{M}+\lambda\mathbf{C}+\mathbf{K} \right]^{-1}. \tag{5}\]If \(p=2N\), the coefficient matrix of the linear system in Equation (3), henceforth denoted as \(\mathbf{T}\)  
\[\mathbf{T}=\left[\begin{array}{cc}\mu_{1}\mathbf{\boldsymbol{t}}_{1}^{T},& \mathbf{\boldsymbol{t}}_{1}^{T}\\ \mu_{2}\mathbf{\boldsymbol{t}}_{2}^{T}&\mathbf{\boldsymbol{t}}_{2}^{T}\\ \vdots&\vdots\\ \mu_{p}\mathbf{\boldsymbol{t}}_{p}^{T}&\mathbf{\boldsymbol{t}}_{p}^{T}\end{array} \right] \tag{6}\]  
is a square matrix. Whenever the system is controllable, i.e., \(\text{rank}\Big{[}\lambda_{i}^{2}\mathbf{M}+\lambda_{i}\mathbf{C}+\mathbf{K} \big{|}\boldsymbol{b}\Big{]}=N\) for all the open-loop poles \(\lambda_{i}\left(i=1,\ldots,2N\right)\) and \(\mu_{1},\ldots,\mu_{2N}\in\mathbb{C}\) is self-conjugate, matrix \(\mathbf{T}\) is invertible, hence the solution of Equation (3) exists, it is real and unique. A linear system can be formulated in the case of higher-rank control too (see e.g., [18]).  
In contrast, if \(p<2N\) then matrix \(\mathbf{T}\) is rectangular and Equation (3) is an underdetermined linear system. Hence, infinite solutions can be found. The parametrization of the set of solutions is helpful to perform RPP, as required in the second stage, while accomplishing a secondary task, such as reducing the controller gains. Without loss of generality, it can be assumed that \(\text{rank}(\mathbf{T})=p\). Hence, the solutions can be written as  
\[\left\{\begin{array}{c}\boldsymbol{f}\\ \boldsymbol{g}\end{array}\right\}=\left\{\begin{array}{c}\boldsymbol{f}_{0} \\ \boldsymbol{g}_{0}\end{array}\right\}+\mathbf{V}\boldsymbol{k}_{r}, \tag{7}\]  
where \(\boldsymbol{k}_{0}=\left\{\begin{array}{c}\boldsymbol{f}_{0}\\ \boldsymbol{g}_{0}\end{array}\right\}\in\mathbb{R}^{2N}\) is a particular solution of Equation (7), \(\mathbf{V}\) is a \(2N\times(2N-p)\) matrix whose columns span \(\text{ker}(\mathbf{T})\) and \(\boldsymbol{k}_{r}\in\mathbb{R}^{2N-p}\) is a vector of parameters that are to be properly determined.  
The particular solution \(\boldsymbol{k}_{0}\) can be calculated either through a least-squares problem, to compute the minimum norm solution, or by imposing arbitrary values to \(2N-p\) entries of the gain vector (such as 0, for example). Additionally, \(2N-p\) rows can be wisely added to \(\mathbf{T}\). If a least-squares solution is sought, the analytic solution is  
\[\left\{\begin{array}{c}\boldsymbol{f}_{0}\\ \boldsymbol{g}_{0}\end{array}\right\}=\left(\mathbf{T}^{T}\mathbf{T}\right)^ {-1}\mathbf{T}^{T}\left\{\begin{array}{c}-1\\ -1\\ \vdots\\ -1\end{array}\right\}, \tag{8}\]  
although numerical solutions of the minimum-norm problem can be exploited too.  
Regardless of how \(\left\{\begin{array}{c}\boldsymbol{f}_{0}\\ \boldsymbol{g}_{0}\end{array}\right\}\) is computed, and for arbitrary choices of \(\boldsymbol{k}_{r}\in\mathbb{R}^{2N-p}\), the gains \(\boldsymbol{f}\), \(\boldsymbol{g}\) solving Equation (7) lead to a controlled system that has the poles \(\mu_{1},\ldots,\mu_{p}\).