(system-theory:kalman-decomposition)= # Kalman decomposition [Controllability](controllability-reachability) and [Non-observability](observability-detectability) properties define two sub-spaces of the state space, $\mathbb{R}^n$. The complements of these subpsaces define subspaces as well. **Kalman decomposition** applies a coordinate transformation in order to make explicit the existence of: * $X_1$, non-observable and reachable sub-space, $X_1 := X_{\overline{o}} \cap X_r$ * $X_2$, the complement of $X_1$ w.r.t. the reachable sub-space, i.e. $X_1 \cup X_2 = X_r$ * $X_3$, the complement of $X_1$ w.r.t. the non-observable sub-space, i.e. $X_1 \cup X_3 = X_{\overline{o}}$ * $X_4$, the complement of $X_1$, $X_2$, $X_3$ w.r.t. the state-space $\mathbb{R}^n$. $A$**-invariance** of $\text{Ran}(\mathbf{W}_c)$ and $\text{Ker}(\mathbf{W}_o)$ is exploited. In this section, [singular value decomposition](math:svd) of the Gramians of [observability](observability-detectability) and [controllability](controllability-reachability) helps in the decomposition of a LTI system in its non-observable/observable and controllable/non-controllable parts. As shown in the section about [coordinate transformation](system-theory:coordinate:transformation), in the [Kalman decomposition subsection](system-theory:coordinate:transformation:tf), only the observable and controllable parts appear in the input-output transfer function of the system. **todo** *SVD may help or not, but that's not the point here. We're supposed to know how to find an orthonormal basis, and if we can't remember how to do it, we're supposed to find a the method we like the most to do that. Orthonormal basis are not even strictly required here. Moreover, SVD of symmetric (semi) definite positive matrices coincides with spectral decomposition...* **todo** *Infinite-horizon time, or steady-state conditions, are discussed here?* (system-theory:kalman-decomposition:controllability)= ## Controllability ```{dropdown} Controllability :open: Let $\mathbf{W}_c$ the controllability Gramian of a LTI system. Controllable states belong to the range of $W_c$. Controllable states define a sub-space of $\mathbb{R}^n$, being $\mathbf{x} \in \mathbb{R}^n$. Singular value decomposition (or spectral decomposition) of the semi-poisitive definite symmetric Gramian matrix, $$\mathbf{W}_C = \mathbf{U}_C \boldsymbol\Sigma_C \mathbf{U}_C^* \ ,$$ provides a decomposition of $\mathbb{R}^n$ into the controllable sub-space and its orthogonal complement, with $$\mathbf{W}_C = [ \, \mathbf{U}_c \, | \, \mathbf{U}_{\overline{c}} \, ] \begin{bmatrix} \boldsymbol\Sigma_c & \cdot \\ \cdot & \cdot \end{bmatrix} \begin{bmatrix} \mathbf{U}_c^* \\ \mathbf{U}_{\overline{c}} \end{bmatrix} \ ,$$ with the columns of $\mathbf{U}_c$ as a base of the controllable sub-space. Introducing a change of variable of the state $$\mathbf{x} = \mathbf{U}_C \mathbf{x}_1 \ ,$$ the LTI system becomes $$\begin{aligned} \dot{\mathbf{x}}_1 & = \mathbf{U}_C^* \mathbf{A} \mathbf{U}_C \mathbf{x}_1 + \mathbf{U}_C^* \mathbf{B} \mathbf{u} \\ \mathbf{y} & = \mathbf{C} \mathbf{U}_C \mathbf{x}_1 + \mathbf{D} \mathbf{u} \ . \end{aligned}$$ and thus $$\begin{aligned} \frac{d}{dt} \begin{bmatrix} \mathbf{x}_{1,c} \\ \mathbf{x}_{1,\overline{c}} \end{bmatrix} & = \begin{bmatrix} \mathbf{U}_c^* \mathbf{A} \mathbf{U}_c & \mathbf{U}_c^* \mathbf{A} \mathbf{U}_{\overline{c}} \\ \mathbf{0} & \mathbf{U}_{\overline{c}}^* \mathbf{A} \mathbf{U}_{\overline{c}} \end{bmatrix} \begin{bmatrix} \mathbf{x}_{1,c} \\ \mathbf{x}_{1,\overline{c}} \end{bmatrix} + \begin{bmatrix} \mathbf{U}_c^* \mathbf{B} \\ \mathbf{0} \end{bmatrix} \mathbf{u} \\ \mathbf{y} & = \begin{bmatrix} \mathbf{C} \mathbf{U}_c & \mathbf{C} \mathbf{U}_{\overline{c}} \end{bmatrix} \begin{bmatrix} \mathbf{x}_{1,c} \\ \mathbf{x}_{1, \overline{c}} \end{bmatrix} + \mathbf{D} \mathbf{u} \ . \end{aligned}$$ ``` ```{dropdown} Properties :open: The columns of the matrix $\mathscr{C} = [ \, \mathbf{B} \, | \, \mathbf{A} \mathbf{B} \, | \, \dots \, | \, \mathbf{A}^{n-1} \mathbf{B} \, ]$ are linear combinations of the range of $\mathbf{W}_C$, i.e. of the columns of $\mathbf{U}_c$, and thus they're orthogonal w.r.t. the columns of $\mathbf{U}_{\overline{c}}$. It follows that $$\begin{aligned} \mathbf{U}_{\overline{c}}^* \mathbf{B} & = \mathbf{0} \\ \mathbf{U}_{\overline{c}}^* \mathbf{A}^k \mathbf{B} & = \mathbf{0} \quad , \quad \forall k \in \mathbb{N} \\ \end{aligned}$$ as $\mathbf{U}^*_{\overline{c}} \mathbf{A}^k \mathbf{B} = \mathbf{U}_{\overline{c}}^* \mathbf{U}_c \boldsymbol\alpha$ and $\mathbf{U}_{\overline{c}}^* \mathbf{U}_c = \mathbf{0}$ ``` (system-theory:kalman-decomposition:observability)= ## Observability ```{dropdown} Observability :open: Let $\mathbf{W}_o$ the observability Gramian of a LTI system. Non-observable states belong to the kernel of $W_o$. Non-observable states define a sub-space of $\mathbb{R}^n$, being $\mathbf{x} \in \mathbb{R}^n$. Singular value decomposition (or spectral decomposition) of the semi-poisitive definite symmetric Gramian matrix, $$\mathbf{W}_O = \mathbf{U}_O \boldsymbol\Sigma_O \mathbf{U}_O^* \ ,$$ provides a decomposition of $\mathbb{R}^n$ into the non-observable sub-space and its orthogonal complement, with $$\mathbf{W}_O = [ \, \mathbf{U}_o \, | \, \mathbf{U}_{\overline{o}} \, ] \begin{bmatrix} \boldsymbol\Sigma_o & \cdot \\ \cdot & \cdot \end{bmatrix} \begin{bmatrix} \mathbf{U}_o^* \\ \mathbf{U}_{\overline{o}} \end{bmatrix} \ ,$$ with the columns of $\mathbf{U}_{\overline{o}}$ as a base of the non-observable sub-space. Introducing a change of variable of the state $$\mathbf{x} = \mathbf{U}_o \mathbf{x}_1 \ ,$$ the LTI system becomes $$\begin{aligned} \dot{\mathbf{x}}_1 & = \mathbf{U}_O^* \mathbf{A} \mathbf{U}_O \mathbf{x}_1 + \mathbf{U}_O^* \mathbf{B} \mathbf{u} \\ \mathbf{y} & = \mathbf{C} \mathbf{U}_O \mathbf{x}_1 + \mathbf{D} \mathbf{u} \ . \end{aligned}$$ and thus $$\begin{aligned} \frac{d}{dt} \begin{bmatrix} \mathbf{x}_{1,o} \\ \mathbf{x}_{1,\overline{o}} \end{bmatrix} & = \begin{bmatrix} \mathbf{U}_o^* \mathbf{A} \mathbf{U}_o & \mathbf{0} \\ \mathbf{U}_{\overline{o}}^* \mathbf{A} \mathbf{U}_{o} & \mathbf{U}_{\overline{o}}^* \mathbf{A} \mathbf{U}_{\overline{o}} \end{bmatrix} \begin{bmatrix} \mathbf{x}_{1,o} \\ \mathbf{x}_{1,\overline{o}} \end{bmatrix} + \begin{bmatrix} \mathbf{U}_o^* \mathbf{B} \\ \mathbf{U}_{\overline{o}}^* \mathbf{B} \end{bmatrix} \mathbf{u} \\ \mathbf{y} & = \begin{bmatrix} \mathbf{C} \mathbf{U}_o & \mathbf{0} \end{bmatrix} \begin{bmatrix} \mathbf{x}_{1,o} \\ \mathbf{x}_{1, \overline{o}} \end{bmatrix} + \mathbf{D} \mathbf{u} \ , \end{aligned}$$ as $\mathbf{U}_o^* \mathbf{A} \mathbf{U}_{\overline{o}} = \mathbf{0}$ (see below). ``` ```{dropdown} Properties :open: The kernel of the observability Gramian is the sub-space of vectors $\mathbf{v}$ so that $\mathscr{O} \mathbf{v} = \mathbf{0}$, with $$\mathscr{O} = \begin{bmatrix} \mathbf{C} \\ \mathbf{C} \mathbf{A} \\ \dots \\ \mathbf{C} \mathbf{A}^{n-1} \end{bmatrix} \ .$$ It follows that $$\begin{aligned} \mathbf{C} \mathbf{U}_{\overline{o}} & = \mathbf{0} \\ \mathbf{C} \mathbf{A}^k \mathbf{U}_{\overline{o}} & = \mathbf{0} \quad , \quad \forall k \in \mathbb{N} \\ \end{aligned}$$ so that the rows of $\mathbf{C}$ and the rows of $\mathbf{C} \mathbf{A}^k$ are linear combinations of the columns of $\mathbf{U}_o$, i.e. the columns of $\left( \mathbf{C} \mathbf{A}^k \right)^*$ are linear combinations of the columns of $\mathbf{U}_o$, $$\left( \mathbf{C} \mathbf{A}^k \right)^* = \mathbf{U}_o \boldsymbol\alpha \ , $$ or $$\mathbf{C} \mathbf{A}^k = \boldsymbol\alpha^* \mathbf{U}_o^* \ .$$ As the property holds for every $k \in \mathbb{N}$, the following holds as well $$\mathbf{U}_o^* \mathbf{A} \mathbf{U}_{\overline{o}} \ .$$ ``` (system-theory:kalman-decomposition:kalman)= ## Kalman decomposition Let * $\mathbf{U}_1$ spanning the **reachable** and **non-observable sub-psace**, $R \cap \overline{O}$ * $\mathbf{U}_2$ complementing $\mathbf{U}_1$ to get the reachable sub-sspace $R$ * $\mathbf{U}_3$ complementing $\mathbf{U}_3$ to get the non-observable sub-space $O$ * $\mathbf{U}_4$ complementing $\mathbf{U}_1$, $\mathbf{U}_2$, $\mathbf{U}_3$ to get $\mathbb{R}^n$ The columns of $\mathbf{U}_1$ and $\mathbf{U}_3$ form a vector basis of the non-observable sub-space. The columns of $\mathbf{U}_1$ and $\mathbf{U}_2$ form a vector basis of the reachable subspace. A coordinate transfromation of the form $$\mathbf{x} = \mathbf{T} \mathbf{x} = \begin{bmatrix} \ \mathbf{U}_1 & \mathbf{U}_2 & \mathbf{U}_3 & \mathbf{U}_4 \ \end{bmatrix} \mathbf{z} = \begin{bmatrix} \ \mathbf{U}_1 & \mathbf{U}_2 & \mathbf{U}_3 & \mathbf{U}_4 \ \end{bmatrix} \begin{bmatrix} \mathbf{z}_1 \\ \mathbf{z}_2 \\ \mathbf{z}_3 \\ \mathbf{z}_4 \end{bmatrix} \ ,$$ produce a transformed system with the following structure $$\left\{\begin{aligned} \frac{d}{dt} \begin{bmatrix} \mathbf{z}_1 \\ \mathbf{z}_2 \\ \mathbf{z}_3 \\ \mathbf{z}_4 \end{bmatrix} & = \begin{bmatrix} \mathbf{A}_{11} & \mathbf{A}_{12} & \mathbf{A}_{13} & \mathbf{A}_{14} \\ \mathbf{0} & \mathbf{A}_{22} & \mathbf{0} & \mathbf{A}_{24} \\ \mathbf{0} & \mathbf{0} & \mathbf{A}_{33} & \mathbf{A}_{34} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{A}_{44} \\ \end{bmatrix} \begin{bmatrix} \mathbf{z}_1 \\ \mathbf{z}_2 \\ \mathbf{z}_3 \\ \mathbf{z}_4 \end{bmatrix} + \begin{bmatrix} \mathbf{B}_1 \\ \mathbf{B}_2 \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix} \mathbf{u} \\ \mathbf{y} & = \begin{bmatrix} \ \mathbf{0} & \mathbf{C}_2 & \mathbf{0} & \mathbf{C}_4 \end{bmatrix} \begin{bmatrix} \mathbf{z}_1 \\ \mathbf{z}_2 \\ \mathbf{z}_3 \\ \mathbf{z}_4 \end{bmatrix} + \mathbf{D} \mathbf{u} \end{aligned}\right.$$ ````{dropdown} Details :open: **Columns of $\ \mathbf{T}$.** Here the columns of $\mathbf{U}_1$ are assumed to be a set of unit normal vectors, $\mathbf{U}_1^* \mathbf{U}_1 = \mathbf{I}$. Without this assumption, Gram-Schmidt orthogonalization process ensure a orthogonal basis $\boldsymbol\Phi_1$ exists and the basis in $\mathbf{U}_1$ can be written as a linear combination of the columns of $\boldsymbol\Phi_1$, i.e. $\mathbf{U}_1 = \boldsymbol\Phi_1 \boldsymbol\alpha_1$, with non-singular $\boldsymbol\alpha_1$. Columns of $\mathbf{U}_2$ are linearly independent from the columns of $\mathbf{U}_1$ to form a basis of the reachable sub-space. Columns $\mathbf{U}_3$ are linearly independent from the columns of $\mathbf{U}_1$ to form a basis of the non-observable sub-space. In general, they're not unit orthogonal but a unit orthogonal basis can be found, to get $$\begin{aligned} \mathbf{U}_2 & = \boldsymbol\Phi_2 \boldsymbol\alpha_2 + \mathbf{U}_1 \boldsymbol\alpha_{12} \\ \mathbf{U}_3 & = \boldsymbol\Phi_3 \boldsymbol\alpha_3 + \mathbf{U}_1 \boldsymbol\alpha_{13} \ . \end{aligned}$$ In general, the columns of $\boldsymbol\Phi_2$ and $\boldsymbol\Phi_3$ are not mutually orthogonal, as the reachable sub-space and the non-observable sub-space are not orthogonal, $\boldsymbol\Phi_2^* \boldsymbol\Phi_3 \ne \mathbf{0}$. The columns of the complement $\mathbf{U}_4$ are assumed to be unit normal vectors, orthogonal w.r.t. the columns of $\mathbf{U}_1$, $\mathbf{U}_2$, $\mathbf{U}_3$ (and of $\boldsymbol\Phi_2$, $\boldsymbol\Phi_3$). **Recast $\ \mathbf{T}$.** $$ \mathbf{T} = \begin{bmatrix} \ \mathbf{U}_1 & \boldsymbol\Phi_2 & \boldsymbol\Phi_3 & \mathbf{U}_4 \ \end{bmatrix} \begin{bmatrix} \mathbf{I} & \boldsymbol\alpha_{12} & \boldsymbol\alpha_{13} & \cdot \\ \cdot & \boldsymbol\alpha_{2 } & \cdot & \cdot \\ \cdot & \cdot & \boldsymbol\alpha_{3 } & \cdot \\ \cdot & \cdot & \cdot & \mathbf{I} \end{bmatrix} = \boldsymbol\Phi \boldsymbol\alpha $$ **Inverse $\ \mathbf{T}^{-1}$.** $$\mathbf{T}^{-1} = \boldsymbol\alpha^{-1} \boldsymbol\Phi^{-1} \ ,$$ with $$\boldsymbol\alpha^{-1} & = \begin{bmatrix} \mathbf{I} &-\boldsymbol\alpha_{12}\boldsymbol\alpha_{2 }^{-1} &-\boldsymbol\alpha_{13}\boldsymbol\alpha_{3 }^{-1} & \cdot \\ \cdot & \boldsymbol\alpha_{2 }^{-1} & \cdot & \cdot \\ \cdot & \cdot & \boldsymbol\alpha_{3 }^{-1} & \cdot \\ \cdot & \cdot & \cdot & \mathbf{I} \end{bmatrix}$$ $$\boldsymbol\Phi^{-1} & = \begin{bmatrix} \mathbf{U}_1^* \\ \left[ \left( \mathbf{I} - \boldsymbol\Phi_3 \boldsymbol\Phi_3^* \right) \boldsymbol\Phi_2 \left( \mathbf{I} - \boldsymbol\Phi_2^* \boldsymbol\Phi_3 \boldsymbol\Phi_3^* \boldsymbol\Phi_2 \right)^{-1} \right]^* \\ \left[ \left( \mathbf{I} - \boldsymbol\Phi_2 \boldsymbol\Phi_2^* \right) \boldsymbol\Phi_3 \left( \mathbf{I} - \boldsymbol\Phi_3^* \boldsymbol\Phi_2 \boldsymbol\Phi_2^* \boldsymbol\Phi_3 \right)^{-1} \right]^* \\ \mathbf{U}_4^* \end{bmatrix} $$ ```{dropdown} Details :open: ``` **Transformed matrices: matrix $\hat{\mathbf{C}}$.** $$\hat{\mathbf{C}} = \mathbf{C} \mathbf{T} = \mathbf{C} \boldsymbol\Phi \boldsymbol\alpha$$ $$\mathbf{C} \boldsymbol\Phi = \mathbf{C} \begin{bmatrix} \ \mathbf{U}_1 & \boldsymbol\Phi_2 & \boldsymbol\Phi_3 & \mathbf{U}_4 \ \end{bmatrix} = \begin{bmatrix} \ \mathbf{0} & \mathbf{C} \boldsymbol\Phi_2 & \mathbf{0} & \mathbf{C} \mathbf{U}_4 \ \end{bmatrix} \ ,$$ as $\mathbf{C} \mathbf{U}_1 = \mathbf{0}$, $\mathbf{C} \boldsymbol\Phi_3 = \mathbf{0}$. Looking at the structures of the matrices $\mathbf{C} \boldsymbol\Phi$ and $\boldsymbol\alpha$, $$ \begin{bmatrix} \cdot & \ast & \cdot & \ast \end{bmatrix} \begin{bmatrix} \ast & \ast & \ast & \cdot \\ \cdot & \ast & \cdot & \cdot \\ \cdot & \cdot & \ast & \cdot \\ \cdot & \cdot & \cdot & \ast \\ \end{bmatrix} = \begin{bmatrix} \cdot & \ast & \cdot & \ast \end{bmatrix} \ , $$ it's easy to show that this multiplication preserves the structure of the matrix $\mathbf{C}\boldsymbol\Phi$. Explicitly evaluating the non-zero blocks, the transformed matrix reads $$\hat{\mathbf{C}} = \begin{bmatrix} \ \mathbf{0} & \mathbf{C} \boldsymbol\Phi_2 \boldsymbol\alpha_2 & \mathbf{0} & \mathbf{C}\mathbf{U}_4 \ \end{bmatrix}$$ **Transformed matrices: matrix $\hat{\mathbf{B}}$.** $$\hat{\mathbf{B}} = \mathbf{T}^{-1} \mathbf{B} = \boldsymbol\alpha^{-1} \boldsymbol\Phi^{-1} \mathbf{B}$$ $$ \boldsymbol\Phi^{-1} \mathbf{B} = \begin{bmatrix} \mathbf{U}_1^* \\ \left[ \left( \mathbf{I} - \boldsymbol\Phi_3 \boldsymbol\Phi_3^* \right) \boldsymbol\Phi_2 \left( \mathbf{I} - \boldsymbol\Phi_2^* \boldsymbol\Phi_3 \boldsymbol\Phi_3^* \boldsymbol\Phi_2 \right)^{-1} \right]^* \\ \left[ \left( \mathbf{I} - \boldsymbol\Phi_2 \boldsymbol\Phi_2^* \right) \boldsymbol\Phi_3 \left( \mathbf{I} - \boldsymbol\Phi_3^* \boldsymbol\Phi_2 \boldsymbol\Phi_2^* \boldsymbol\Phi_3 \right)^{-1} \right]^* \\ \mathbf{U}_4^* \end{bmatrix} \mathbf{B} $$ as $\mathbf{U}_1^* \mathbf{B} = \mathbf{0}$, $\left[ \left( \mathbf{I} - \boldsymbol\Phi_2 \boldsymbol\Phi_2^* \right) \boldsymbol\Phi_3 \right]^* \mathbf{B} = \mathbf{0}$. The last relation holds, as $\left( \mathbf{I} - \boldsymbol\Phi_2 \boldsymbol\Phi_2^* \right) \boldsymbol\Phi_3$ is the orthogonal projection of $\boldsymbol\Phi_3$ perpendicular to the vectors $\boldsymbol\Phi_2$. Since this projection is perpendicular both to $\mathbf{U}_1$ and to $\boldsymbol\Phi_2$ (that generates the reachable sub-space), it follows that $\left[ \left( \mathbf{I} - \boldsymbol\Phi_2 \boldsymbol\Phi_2^* \right) \boldsymbol\Phi_3 \right]^* \mathbf{B} = \mathbf{0}$. The same argument gives $\mathbf{U}_4^* \mathbf{B} = \mathbf{0}$, and thus $$\boldsymbol\Phi^{-1} \mathbf{B} = \begin{bmatrix} \ \ast \ \\ \ \ast \ \\ \ \cdot \ \\ \ \cdot \ \end{bmatrix} \ .$$ Looking at the structure of the matrices $\boldsymbol\alpha^{-1}$ and $\boldsymbol\Phi^{-1} \mathbf{B}$, it's easy to show that their matrix multiplication preserves the structure $$ \begin{bmatrix} \ast & \ast & \ast & \cdot \\ \cdot & \ast & \cdot & \cdot \\ \cdot & \cdot & \ast & \cdot \\ \cdot & \cdot & \cdot & \ast \\ \end{bmatrix} \begin{bmatrix} \ \ast \ \\ \ \ast \ \\ \ \cdot \ \\ \ \cdot \ \end{bmatrix} = \begin{bmatrix} \ \ast \ \\ \ \ast \ \\ \ \cdot \ \\ \ \cdot \ \end{bmatrix} \ . $$ **Transformed matrices: matrix $\hat{\mathbf{A}}$.** The structure of the matrix arises from $\mathbf{A}$-invariance of reachability and controllability sub-spaces and from the properties $$\begin{aligned} \overline{\mathbf{r}}^* \mathbf{A} \mathbf{r} & = 0 \\ \mathbf{o}^* \mathbf{A} \overline{\mathbf{o}} & = 0 \ , \end{aligned}$$ and recalling that $\mathbf{U}_1: \, r \overline{o}$, $(\mathbf{I}-\boldsymbol\Phi_3 \boldsymbol\Phi_3)^* \boldsymbol\Phi_2: r o \, $, $(\mathbf{I}-\boldsymbol\Phi_2 \boldsymbol\Phi_2)^* \boldsymbol\Phi_3: \, \overline{r} \overline{o}$, $\mathbf{U}_4: \, \overline{r} o$. Focusing on $\boldsymbol\Phi^{-1} \mathbf{A} \boldsymbol\Phi$ first $$\begin{aligned} \boldsymbol\Phi^{-1} \mathbf{A} \boldsymbol\Phi & = \begin{bmatrix} \mathbf{U}_1^* \\ \left[ \left( \mathbf{I} - \boldsymbol\Phi_3 \boldsymbol\Phi_3^* \right) \boldsymbol\Phi_2 \left( \mathbf{I} - \boldsymbol\Phi_2^* \boldsymbol\Phi_3 \boldsymbol\Phi_3^* \boldsymbol\Phi_2 \right)^{-1} \right]^* \\ \left[ \left( \mathbf{I} - \boldsymbol\Phi_2 \boldsymbol\Phi_2^* \right) \boldsymbol\Phi_3 \left( \mathbf{I} - \boldsymbol\Phi_3^* \boldsymbol\Phi_2 \boldsymbol\Phi_2^* \boldsymbol\Phi_3 \right)^{-1} \right]^* \\ \mathbf{U}_4^* \end{bmatrix} \mathbf{A} \begin{bmatrix} \ \mathbf{U}_1 & \boldsymbol\Phi_2 & \boldsymbol\Phi_3 & \mathbf{U}_4 \ \end{bmatrix} = \\ & = \begin{bmatrix} \ast & \ast & \ast & \ast \\ \cdot & \ast & \cdot & \ast \\ \cdot & \cdot & \ast & \ast \\ \cdot & \cdot & \cdot & \ast \end{bmatrix} \end{aligned}$$ Pre- and post-multiplication by $\boldsymbol\alpha^{-1}$ and $\boldsymbol\alpha$ preserves the structure of $\boldsymbol\Phi^{-1} \mathbf{A} \boldsymbol\Phi$ to have $$ \hat{\mathbf{A}} = \begin{bmatrix} \mathbf{A}_{11} & \mathbf{A}_{12} & \mathbf{A}_{13} & \mathbf{A}_{14} \\ \mathbf{0} & \mathbf{A}_{22} & \mathbf{0} & \mathbf{A}_{24} \\ \mathbf{0} & \mathbf{0} & \mathbf{A}_{33} & \mathbf{A}_{34} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{A}_{44} \\ \end{bmatrix} \ . $$ ````