32.2.1. Diagonalization
Spectral decomposition. If the spectral decomposition of matrix \(\mathbf{A} \in \mathbb{C}^{n,n}\) exists, there are vector basis \(\{ \mathbf{v}_i \}_{i=1:n}\), \(\{ \mathbf{w}_i \}_{i=1:n}\) of \(\mathbb{C}^n\) so that
\[\begin{split}\begin{aligned}
\mathbf{A} \mathbf{v}_i & = s_i \mathbf{v}_i \\
\mathbf{w}^*_i \mathbf{A} & = s_i \mathbf{w}^*_i \ ,
\end{aligned}\end{split}\]
being \(s_i\) the eigenvalues of the matrix, \(\mathbf{v}_i\) the right-eigenvectors and \(\mathbf{w}_i\) the left eigenvectors. Using matrix formalism
\[\begin{split}\begin{aligned}
\mathbf{A} \mathbf{V} & = \mathbf{V} \boldsymbol{\Lambda} \\
\mathbf{W} \mathbf{A} & = \boldsymbol{\Lambda} \mathbf{W} \ .
\end{aligned}\end{split}\]
Left- and right-eigenvectors associated with different eigenvalues are orthogonal.
\[\begin{split}\begin{aligned}
\mathbf{w}^*_k \mathbf{A} \mathbf{v}_i & = s_i \mathbf{w}^*_k \mathbf{v}_i \\
\mathbf{w}^*_k \mathbf{A} \mathbf{v}_i & = s_k \mathbf{w}^*_k \mathbf{v}_i \\
\end{aligned}\end{split}\]
implies (subtraction) \((s_i - s_k) \mathbf{w}^*_k \mathbf{v}_i = 0\). If \(s_i \ne s_k\), then \(\mathbf{w}^*_k \mathbf{v}_i = 0\).
If \(s_i, s_k \ne 0\), this also implies \(\mathbf{w}_k^* \mathbf{A} \mathbf{v}_i = 0\).
This relation when the indices are equal can be used for normalization condition
A unit orthogonal basis can be computed for eigenvalues with algebraic multiplicity \(> 1\)
If normalization condition reads \(\mathbf{w}_i^* \mathbf{v}_i = 1\) (no sum), \(\mathbf{W}^* \mathbf{V} = \mathbf{I}\), then
\[\mathbf{W}^* \mathbf{A} \mathbf{V} = \boldsymbol{\Lambda} \ .\]
Diagonalization. Let \(\mathbf{x} = \mathbf{V} \hat{\mathbf{x}}\) the representation of the state with the vector basis \(\mathbf{V}\), and the projection of the state dynamical equation onto \(\mathbf{W}\). The state-space representation of the system reads
\[\begin{split}\left\{\begin{aligned}
\dot{\hat{\mathbf{x}}} = \boldsymbol{\Lambda} \hat{\mathbf{x}} + \mathbf{W}^* \mathbf{B} \mathbf{u} \\
\mathbf{y} = \mathbf{C} \mathbf{V} \hat{\mathbf{x}} + \mathbf{D} \mathbf{u} \ ,
\end{aligned}\right. \end{split}\]
with \(\boldsymbol{\Lambda}\) diagonal. I.e. in components, with \(\hat{\mathbf{B}} := \mathbf{W}^* \mathbf{B}\),
\[\dot{\hat{x}}_i = \lambda_i \hat{x}_i + \sum_{j=1}^{m} \hat{B}_{ij} u_j = \lambda_i \hat{x}_i + \hat{\mathbf{b}}_i^* \mathbf{u} \ ,\]
or in Laplace domain, the \(i^{th}\) component in the diagonal representation reads
\[\hat{x}_i = \frac{1}{s - \lambda_i} \hat{\mathbf{b}}_i^* \mathbf{u} \ .\]
the state in the original coordinates reads
\[\mathbf{x} = \mathbf{V} \hat{\mathbf{x}} = \sum_{i=1}^{n} \mathbf{v}_i \hat{x}_i = \sum_{i=1}^{n} \mathbf{v}_i \frac{1}{s - \lambda_i} \hat{\mathbf{b}}_i^* \mathbf{u} \ .\]
and the output reads
\[\begin{split}\begin{aligned}
\mathbf{y}
& = \mathbf{C} \mathbf{V} \hat{\mathbf{x}} + \mathbf{D} \mathbf{u} = \\
& = \hat{\mathbf{C}} \hat{\mathbf{x}} + \mathbf{D} \mathbf{u} = \\
& = \sum_{i=1}^{n} \hat{\mathbf{c}}_i \hat{x}_i + \mathbf{D} \mathbf{u} = \\
& = \left[ \sum_{i=1}^{n} \frac{1}{s-\lambda_i} \hat{\mathbf{c}}_i \hat{\mathbf{b}}^*_i + \mathbf{D} \right] \mathbf{u} = \\
& = \mathbf{H}(s) \mathbf{u} \ .
\end{aligned}\end{split}\]
Transfer function. Thus, the transfer function between the input and the output can be written as
\[\begin{aligned}
\hat{\mathbf{H}}(s) = \sum_{i=1}^{n} \frac{1}{s-\lambda_i} \hat{\mathbf{c}}_i \hat{\mathbf{b}}^*_i + \mathbf{D} \ .
\end{aligned}\]
This is a linear combination of factors \(\frac{1}{s - \lambda_i}\). Thus, the denominators of these fractions appear in the denominatore of the transfer function if both \(\mathbf{b}_i \ne \mathbf{0}\) and \(\mathbf{c}_i \ne \mathbf{0}\). todo Further cancellations may occur. It’s better to use observability or controllability vector basis or Kalman decomposition to link cancellations and observability/controllability
32.2.2. Kalman decomposition
\[\begin{split}\mathbf{x} = \mathbf{T} \mathbf{x} = \begin{bmatrix} \ \mathbf{U}_1 & \mathbf{U}_2 & \mathbf{U}_3 & \mathbf{U}_4 \ \end{bmatrix} \mathbf{z} = \begin{bmatrix} \ \mathbf{U}_1 & \mathbf{U}_2 & \mathbf{U}_3 & \mathbf{U}_4 \ \end{bmatrix} \begin{bmatrix} \mathbf{z}_1 \\ \mathbf{z}_2 \\ \mathbf{z}_3 \\ \mathbf{z}_4 \end{bmatrix} \ ,\end{split}\]
Kalman decomposition of a linear system reads
\[\begin{split}\left\{\begin{aligned}
\frac{d}{dt} \begin{bmatrix} \mathbf{z}_1 \\ \mathbf{z}_2 \\ \mathbf{z}_3 \\ \mathbf{z}_4 \end{bmatrix} & =
\begin{bmatrix}
\mathbf{A}_{11} & \mathbf{A}_{12} & \mathbf{A}_{13} & \mathbf{A}_{14} \\
\mathbf{0} & \mathbf{A}_{22} & \mathbf{0} & \mathbf{A}_{24} \\
\mathbf{0} & \mathbf{0} & \mathbf{A}_{33} & \mathbf{A}_{34} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{A}_{44} \\
\end{bmatrix}
\begin{bmatrix} \mathbf{z}_1 \\ \mathbf{z}_2 \\ \mathbf{z}_3 \\ \mathbf{z}_4 \end{bmatrix} +
\begin{bmatrix} \mathbf{B}_1 \\ \mathbf{B}_2 \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix} \mathbf{u} \\
\mathbf{y} & = \begin{bmatrix} \ \mathbf{0} & \mathbf{C}_2 & \mathbf{0} & \mathbf{C}_4 \end{bmatrix}
\begin{bmatrix} \mathbf{z}_1 \\ \mathbf{z}_2 \\ \mathbf{z}_3 \\ \mathbf{z}_4 \end{bmatrix} +
\mathbf{D} \mathbf{u}
\end{aligned}\right.\end{split}\]
32.2.2.1. Transfer function
Asymptotical stable systems have the free response decaying to zero after a certain time range, so that only the forced response survives. Forced response in Laplace domain reads
\[\begin{split}\begin{aligned}
( s \mathbf{I} - \mathbf{A}_{44}) \mathbf{z}_4 & = \mathbf{0} \\
\mathbf{z}_3 & = ( s \mathbf{I} - \mathbf{A}_{33})^{-1} \mathbf{A}_{34} \mathbf{z}_4 \\
\mathbf{z}_2 & = ( s \mathbf{I} - \mathbf{A}_{22})^{-1} \left( \mathbf{A}_{24} \mathbf{z}_4 + \mathbf{B}_2 \mathbf{u} \right) \\
\mathbf{z}_1 & = ( s \mathbf{I} - \mathbf{A}_{11})^{-1} \left( \mathbf{A}_{12} \mathbf{z}_2 + \mathbf{A}_{13} \mathbf{z}_3 + \mathbf{A}_{14} \mathbf{z}_4 + \mathbf{B}_1 \mathbf{u} \right) \\
\end{aligned}\end{split}\]
and thus
\[\begin{split}\begin{aligned}
\mathbf{z}_4 & = \mathbf{0} \\
\mathbf{z}_3 & = \mathbf{0} \\
\mathbf{z}_2 & = ( s \mathbf{I} - \mathbf{A}_{22})^{-1} \mathbf{B}_2 \mathbf{u} \\
\mathbf{z}_1 & = ( s \mathbf{I} - \mathbf{A}_{11})^{-1} \left[ \mathbf{A}_{12} ( s \mathbf{I} - \mathbf{A}_{22})^{-1} \mathbf{B}_2 + \mathbf{B}_1 \right] \mathbf{u}
\end{aligned}\end{split}\]
and the output reads
\[\mathbf{y} = \mathbf{C}_2 ( s \mathbf{I} - \mathbf{A}_{22})^{-1} \mathbf{B}_2 \mathbf{u} \ .\]
From the input-output relation, it’s clear that only the reachable and observable part of the system contributes to the transfer function, i.e. to the input-output relation for asymptotically stable systems after the free response due to non-zero initial conditions has decayed.
