A state-space model (continuous-time) is written as \( \dot{\mathbf{x}}(t)=\mathbf{A}\mathbf{x}(t)+\mathbf{B}\mathbf{u}(t) \), \( \mathbf{y}(t)=\mathbf{C}\mathbf{x}(t)+\mathbf{D}\mathbf{u}(t) \), with \( \mathbf{x}(t)\in\mathbb{R}^n \). The state \( \mathbf{x}(t) \) is not unique: different choices of coordinates can represent the same internal “information” needed to predict future behavior.

In practice, we transform state coordinates to:

• obtain a desired canonical form (controllable/observable/diagonal) introduced in Lesson 3,
• improve numerical conditioning (scaling/balancing),
• connect different modeling pipelines (physics-based vs. canonical realizations),
• prove two models are equivalent (same input–output map) even if matrices differ.

flowchart TD
  P["Physical laws / ODEs"] --> M["Choose states x(t)"]
  M --> S1["State-space (A,B,C,D)"]
  S1 --> T["Pick invertible T (change of coordinates)"]
  T --> S2["Transformed (Abar,Bbar,Cbar,D)"]
  S2 --> IO["Same input-output behavior"]
  S2 --> F["Choose form: controllable / observable / diagonal"]
        

Let \( \mathbf{T}\in\mathbb{R}^{n\times n} \) be invertible and define a new state coordinate \( \mathbf{z}(t) \) by \( \mathbf{x}(t)=\mathbf{T}\mathbf{z}(t) \). This is a change of basis in the state space.

2.1 Continuous-time derivation

Differentiate \( \mathbf{x}(t)=\mathbf{T}\mathbf{z}(t) \) (with constant \( \mathbf{T} \)): \( \dot{\mathbf{x}}(t)=\mathbf{T}\dot{\mathbf{z}}(t) \). Substitute into \( \dot{\mathbf{x}}=\mathbf{A}\mathbf{x}+\mathbf{B}\mathbf{u} \):

\[ \mathbf{T}\dot{\mathbf{z}}(t) = \mathbf{A}\mathbf{T}\mathbf{z}(t) + \mathbf{B}\mathbf{u}(t) \;\;\Rightarrow\;\; \dot{\mathbf{z}}(t) = \underbrace{\mathbf{T}^{-1}\mathbf{A}\mathbf{T}}_{\mathbf{\bar A}}\mathbf{z}(t) + \underbrace{\mathbf{T}^{-1}\mathbf{B}}_{\mathbf{\bar B}}\mathbf{u}(t) \]

For the output equation, \( \mathbf{y}(t)=\mathbf{C}\mathbf{x}(t)+\mathbf{D}\mathbf{u}(t)=\mathbf{C}\mathbf{T}\mathbf{z}(t)+\mathbf{D}\mathbf{u}(t) \), so define \( \mathbf{\bar C}=\mathbf{C}\mathbf{T} \) and \( \mathbf{\bar D}=\mathbf{D} \).

2.2 Discrete-time derivation

For a discrete-time model \( \mathbf{x}_{k+1}=\mathbf{A}\mathbf{x}_k+\mathbf{B}\mathbf{u}_k \), \( \mathbf{y}_k=\mathbf{C}\mathbf{x}_k+\mathbf{D}\mathbf{u}_k \), define \( \mathbf{x}_k=\mathbf{T}\mathbf{z}_k \):

\[ \mathbf{T}\mathbf{z}_{k+1}=\mathbf{A}\mathbf{T}\mathbf{z}_k+\mathbf{B}\mathbf{u}_k \;\;\Rightarrow\;\; \mathbf{z}_{k+1}=\mathbf{T}^{-1}\mathbf{A}\mathbf{T}\,\mathbf{z}_k + \mathbf{T}^{-1}\mathbf{B}\,\mathbf{u}_k \]

The transformation rules are identical: \( \mathbf{\bar A}=\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \), \( \mathbf{\bar B}=\mathbf{T}^{-1}\mathbf{B} \), \( \mathbf{\bar C}=\mathbf{C}\mathbf{T} \), \( \mathbf{\bar D}=\mathbf{D} \).

2.3 Equivalence as a precise definition

Two realizations \( (\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D}) \) and \( (\mathbf{\bar A},\mathbf{\bar B},\mathbf{\bar C},\mathbf{\bar D}) \) are state-coordinate equivalent if there exists an invertible \( \mathbf{T} \) such that:

\[ \mathbf{\bar A}=\mathbf{T}^{-1}\mathbf{A}\mathbf{T},\quad \mathbf{\bar B}=\mathbf{T}^{-1}\mathbf{B},\quad \mathbf{\bar C}=\mathbf{C}\mathbf{T},\quad \mathbf{\bar D}=\mathbf{D}. \]

3.1 Matrix exponential similarity identity

The continuous-time state transition matrix is \( e^{\mathbf{A}t} \). Similarity transformations preserve it in a specific way.

Theorem (Similarity of matrix exponentials): For invertible \( \mathbf{T} \) and any real \( t \),

\[ e^{(\mathbf{T}^{-1}\mathbf{A}\mathbf{T})t} = \mathbf{T}^{-1}e^{\mathbf{A}t}\mathbf{T}. \]

Proof: Use the power-series definition \( e^{\mathbf{M}}=\sum_{k=0}^{\infty}\frac{1}{k!}\mathbf{M}^k \). Since \( (\mathbf{T}^{-1}\mathbf{A}\mathbf{T})^k=\mathbf{T}^{-1}\mathbf{A}^k\mathbf{T} \),

\[ e^{(\mathbf{T}^{-1}\mathbf{A}\mathbf{T})t} = \sum_{k=0}^{\infty}\frac{t^k}{k!}(\mathbf{T}^{-1}\mathbf{A}\mathbf{T})^k = \sum_{k=0}^{\infty}\frac{t^k}{k!}\mathbf{T}^{-1}\mathbf{A}^k\mathbf{T} = \mathbf{T}^{-1}\left(\sum_{k=0}^{\infty}\frac{t^k}{k!}\mathbf{A}^k\right)\mathbf{T} = \mathbf{T}^{-1}e^{\mathbf{A}t}\mathbf{T}. \]

Therefore state trajectories map consistently by \( \mathbf{x}(t)=\mathbf{T}\mathbf{z}(t) \).

3.2 Input–output equivalence and transfer function invariance

For continuous-time LTI systems (zero initial condition), the transfer function matrix is:

\[ \mathbf{G}(s)=\mathbf{C}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{B}+\mathbf{D}. \]

Theorem (Transfer function invariance): If \( (\mathbf{\bar A},\mathbf{\bar B},\mathbf{\bar C},\mathbf{\bar D}) \) is obtained by a similarity transform, then \( \mathbf{\bar G}(s)=\mathbf{G}(s) \).

Proof: Note the resolvent identity: \( s\mathbf{I}-\mathbf{\bar A}=s\mathbf{I}-\mathbf{T}^{-1}\mathbf{A}\mathbf{T}=\mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})\mathbf{T} \), hence \( (s\mathbf{I}-\mathbf{\bar A})^{-1}=\mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{T} \). Then:

\[ \mathbf{\bar G}(s) = \mathbf{\bar C}(s\mathbf{I}-\mathbf{\bar A})^{-1}\mathbf{\bar B}+\mathbf{\bar D} = (\mathbf{C}\mathbf{T})\left[\mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{T}\right](\mathbf{T}^{-1}\mathbf{B})+\mathbf{D} = \\ \mathbf{C}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{B}+\mathbf{D} = \mathbf{G}(s). \]

3.3 Eigenvalues (poles) are invariant

Similar matrices have the same characteristic polynomial and thus the same eigenvalues. This is fundamental because eigenvalues of \( \mathbf{A} \) govern natural modes.

Theorem (Characteristic polynomial invariance):

\[ \det(\lambda\mathbf{I}-\mathbf{\bar A})=\det(\lambda\mathbf{I}-\mathbf{A}). \]

Proof:

\[ \det(\lambda\mathbf{I}-\mathbf{\bar A}) = \det(\lambda\mathbf{I}-\mathbf{T}^{-1}\mathbf{A}\mathbf{T}) = \det\!\left(\mathbf{T}^{-1}(\lambda\mathbf{I}-\mathbf{A})\mathbf{T}\right) = \\ \det(\mathbf{T}^{-1})\det(\lambda\mathbf{I}-\mathbf{A})\det(\mathbf{T}) = \det(\lambda\mathbf{I}-\mathbf{A}). \]

Consequently, stability conclusions based on eigenvalues (introduced later formally) do not depend on the chosen state coordinates.

3.4 Controllability and observability ranks are invariant

For a continuous-time (or discrete-time) system, define the controllability matrix (SISO form) as \( \mathcal{C}=[\mathbf{B}\;\;\mathbf{A}\mathbf{B}\;\;\cdots\;\;\mathbf{A}^{n-1}\mathbf{B}] \) and the observability matrix as \( \mathcal{O}=\begin{bmatrix}\mathbf{C}\\ \mathbf{C}\mathbf{A}\\ \vdots\\ \mathbf{C}\mathbf{A}^{n-1}\end{bmatrix} \).

Theorem (Rank invariance): \( \operatorname{rank}(\mathcal{\bar C})=\operatorname{rank}(\mathcal{C}) \) and \( \operatorname{rank}(\mathcal{\bar O})=\operatorname{rank}(\mathcal{O}) \).

Proof (controllability): Since \( \mathbf{\bar A}=\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \) and \( \mathbf{\bar B}=\mathbf{T}^{-1}\mathbf{B} \), we have by induction \( \mathbf{\bar A}^k\mathbf{\bar B}=\mathbf{T}^{-1}\mathbf{A}^k\mathbf{B} \). Therefore

\[ \mathcal{\bar C}= [\mathbf{\bar B}\;\;\mathbf{\bar A}\mathbf{\bar B}\;\;\cdots\;\;\mathbf{\bar A}^{n-1}\mathbf{\bar B}] = \mathbf{T}^{-1}[\mathbf{B}\;\;\mathbf{A}\mathbf{B}\;\;\cdots\;\;\mathbf{A}^{n-1}\mathbf{B}] = \mathbf{T}^{-1}\mathcal{C}. \]

Multiplication by an invertible matrix does not change rank, hence \( \operatorname{rank}(\mathcal{\bar C})=\operatorname{rank}(\mathcal{C}) \). The observability proof is analogous because \( \mathcal{\bar O}=\mathcal{O}\mathbf{T} \).

Suppose you are given two state-space models of the same order: \( (\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D}) \) and \( (\mathbf{A}_2,\mathbf{B}_2,\mathbf{C}_2,\mathbf{D}_2) \). A practical goal is to decide whether there exists an invertible \( \mathbf{T} \) such that \( \mathbf{A}_2=\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \), \( \mathbf{B}_2=\mathbf{T}^{-1}\mathbf{B} \), \( \mathbf{C}_2=\mathbf{C}\mathbf{T} \), \( \mathbf{D}_2=\mathbf{D} \).

4.1 Similarity is an equivalence relation

Define the relation \( \sim \) on realizations by \( (\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})\sim(\mathbf{A}_2,\mathbf{B}_2,\mathbf{C}_2,\mathbf{D}_2) \) if such an invertible \( \mathbf{T} \) exists. Then:

• Reflexive: choose \( \mathbf{T}=\mathbf{I} \).
• Symmetric: if \( \mathbf{T} \) works, then \( \mathbf{T}^{-1} \) maps back.
• Transitive: if \( \mathbf{T}_1 \) maps 1→2 and \( \mathbf{T}_2 \) maps 2→3, then \( \mathbf{T}_1\mathbf{T}_2 \) maps 1→3.

4.2 Recovering \( \mathbf{T} \) using controllability (SISO, full rank)

For SISO systems (one input), the SISO controllability matrix is square: \( \mathcal{C}=[\mathbf{B}\;\;\mathbf{A}\mathbf{B}\;\;\cdots\;\;\mathbf{A}^{n-1}\mathbf{B}]\in\mathbb{R}^{n\times n} \). If \( \det(\mathcal{C})\neq 0 \), the realization is controllable.

If the two realizations are related by similarity, then as shown in Section 3.4: \( \mathcal{C}_2=\mathbf{T}^{-1}\mathcal{C} \). If both are controllable (so both matrices are invertible), then

\[ \mathcal{C}_2=\mathbf{T}^{-1}\mathcal{C} \;\;\Rightarrow\;\; \mathbf{T}=\mathcal{C}\mathcal{C}_2^{-1}. \]

After computing \( \mathbf{T} \), you must verify consistency: \( \mathbf{A}_2 \stackrel{?}{=} \mathbf{T}^{-1}\mathbf{A}\mathbf{T} \), \( \mathbf{B}_2 \stackrel{?}{=} \mathbf{T}^{-1}\mathbf{B} \), \( \mathbf{C}_2 \stackrel{?}{=} \mathbf{C}\mathbf{T} \), \( \mathbf{D}_2 \stackrel{?}{=} \mathbf{D} \).

4.3 Recovering \( \mathbf{T} \) using observability (SISO, full rank)

The SISO observability matrix is \( \mathcal{O}=\begin{bmatrix}\mathbf{C}\\ \mathbf{C}\mathbf{A}\\ \vdots\\ \mathbf{C}\mathbf{A}^{n-1}\end{bmatrix}\in\mathbb{R}^{n\times n} \). If \( \det(\mathcal{O})\neq 0 \), the realization is observable. Under similarity, \( \mathcal{O}_2=\mathcal{O}\mathbf{T} \), hence if both are observable:

\[ \mathcal{O}_2=\mathcal{O}\mathbf{T} \;\;\Rightarrow\;\; \mathbf{T}=\mathcal{O}^{-1}\mathcal{O}_2. \]

4.4 Minimal realizations and uniqueness up to similarity

A realization is called minimal if it has the smallest possible state dimension among all realizations that produce the same transfer function. In this chapter, a standard sufficient characterization is: \( \text{minimal} \iff \text{controllable and observable} \).

Theorem (Uniqueness of minimal realizations up to similarity): If two realizations are both minimal and have the same transfer function, then they are similar (state-coordinate equivalent).

This theorem explains why canonical forms in Lesson 3 are powerful: they represent an entire equivalence class of models sharing the same input–output behavior.

flowchart TD
  I["Given (A,B,C,D) and (A2,B2,C2,D2)"] --> D["Check D2 == D"]
  D --> R["Compute invariants: eigenvalues(A) vs eigenvalues(A2)"]
  R --> C1["If SISO and controllable: build Cmat, Cmat2"]
  R --> O1["If SISO and observable: build Omat, Omat2"]
  C1 --> T1["Candidate T = Cmat * inv(Cmat2)"]
  O1 --> T2["Candidate T = inv(Omat) * Omat2"]
  T1 --> V["Verify: A2=inv(T)AT, B2=inv(T)B, C2=CT"]
  T2 --> V
  V --> OK["Equivalent (similar)"]
  V --> NO["Not similar (or numerically ill-conditioned)"]
        

4.5 Numerical note: conditioning

In practice, matrices like \( \mathcal{C} \) or \( \mathcal{O} \) may be ill-conditioned even when invertible. This can make computed \( \mathbf{T} \) numerically unstable. Robust workflows therefore:

• use rank-revealing factorizations (QR/SVD) to test controllability/observability,
• verify equivalence by checking norms (e.g., \( \|\mathbf{A}_2-\mathbf{T}^{-1}\mathbf{A}\mathbf{T}\| \)),
• compare input–output responses (impulse/step) as a sanity check.

Common Python libraries for state-space work in system dynamics include \( \texttt{numpy} \), \( \texttt{scipy.linalg} \), \( \texttt{scipy.signal} \), and the Control Systems library \( \texttt{control} \) (often installed as python-control).

import numpy as np
import scipy.linalg as la

# Example SISO system (controllable and observable)
A = np.array([[0.0, 1.0],
              [-2.0, -3.0]])
B = np.array([[0.0],
              [1.0]])
C = np.array([[1.0, 0.0]])
D = np.array([[0.0]])

# Random invertible similarity transform T
rng = np.random.default_rng(4)
T = rng.normal(size=(2, 2))
while abs(np.linalg.det(T)) < 1e-2:
    T = rng.normal(size=(2, 2))

A2 = la.inv(T) @ A @ T
B2 = la.inv(T) @ B
C2 = C @ T
D2 = D.copy()

# Invariants: eigenvalues
lam_A = np.linalg.eigvals(A)
lam_A2 = np.linalg.eigvals(A2)
print("eig(A) :", lam_A)
print("eig(A2):", lam_A2)

# Build SISO controllability matrices (square here because n=2)
Cmat  = np.hstack([B, A @ B])
Cmat2 = np.hstack([B2, A2 @ B2])

# Recover candidate T from controllability
T_hat = Cmat @ la.inv(Cmat2)

# Verify similarity relations numerically
def rel_err(X, Y):
    return la.norm(X - Y) / max(1.0, la.norm(X))

print("relerr(A2, inv(T_hat) A T_hat) =", rel_err(A2, la.inv(T_hat) @ A @ T_hat))
print("relerr(B2, inv(T_hat) B)       =", rel_err(B2, la.inv(T_hat) @ B))
print("relerr(C2, C T_hat)            =", rel_err(C2, C @ T_hat))
print("relerr(D2, D)                  =", rel_err(D2, D))

# Optional: verify transfer function invariance by comparing Markov parameters:
# h[1] = C B, h[2] = C A B, ...
H1  = C @ B
H2  = C @ A @ B
H1_2 = C2 @ B2
H2_2 = C2 @ A2 @ B2
print("Markov check h1:", H1, H1_2)
print("Markov check h2:", H2, H2_2)
      

The Markov parameter check (\( \mathbf{C}\mathbf{A}^{k-1}\mathbf{B} \)) is a lightweight way to confirm that the impulse response coefficients match (for discrete-time) and similarly reflects invariance in continuous-time expansions.

MATLAB Control System Toolbox supports state-space objects via ss. A similarity transformation can be implemented directly using matrix operations, and input–output equivalence can be checked by comparing transfer functions or time responses.

% Example SISO system
A = [0 1; -2 -3];
B = [0; 1];
C = [1 0];
D = 0;

sys  = ss(A,B,C,D);

% Choose an invertible similarity transform T
T = [2 1; 1 1];
A2 = inv(T)*A*T;
B2 = inv(T)*B;
C2 = C*T;
D2 = D;

sys2 = ss(A2,B2,C2,D2);

% Invariants
eigA  = eig(A);
eigA2 = eig(A2);

disp(eigA); disp(eigA2);

% Transfer-function equality (symbolic equality may not be exact numerically)
G1 = tf(sys);
G2 = tf(sys2);

% Compare frequency response numerically
w = logspace(-2,2,200);
mag1 = squeeze(abs(freqresp(sys, w)));
mag2 = squeeze(abs(freqresp(sys2, w)));
disp(max(abs(mag1 - mag2)));

% Compare step responses
t = linspace(0,10,2000);
y1 = step(sys, t);
y2 = step(sys2, t);
disp(max(abs(y1 - y2)));

% If available, ss2ss can perform coordinate changes directly:
% sys2b = ss2ss(sys, T);  % interprets transformation consistent with x = T*z
      

In Simulink, you can use the State-Space block with matrices \( \mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D} \). If you build two blocks with \( (A,B,C,D) \) and \( (A2,B2,C2,D) \) and drive them with the same input, their outputs should coincide (up to numerical solver tolerances).

% Programmatically create a simple Simulink model with two State-Space blocks
model = 'ss_similarity_demo';
new_system(model); open_system(model);

add_block('simulink/Sources/Step', [model '/u']);
add_block('simulink/Continuous/State-Space', [model '/Sys1']);
add_block('simulink/Continuous/State-Space', [model '/Sys2']);
add_block('simulink/Sinks/Scope', [model '/Scope']);

set_param([model '/Sys1'], 'A', mat2str(A),  'B', mat2str(B),  'C', mat2str(C),  'D', mat2str(D));
set_param([model '/Sys2'], 'A', mat2str(A2), 'B', mat2str(B2), 'C', mat2str(C2), 'D', mat2str(D2));

add_line(model, 'u/1', 'Sys1/1');
add_line(model, 'u/1', 'Sys2/1');
add_line(model, 'Sys1/1', 'Scope/1');
add_line(model, 'Sys2/1', 'Scope/2');

set_param(model, 'StopTime', '10');
save_system(model);
      

In C++, a common workflow is to use Eigen for linear algebra, then implement simulation (Euler / Runge–Kutta) to compare outputs. Below is a compact Euler simulation demonstrating output equivalence after a similarity transform.

#include <iostream>
#include <Eigen/Dense>

int main() {
  using Eigen::MatrixXd;
  using Eigen::VectorXd;

  // System matrices (SISO)
  MatrixXd A(2,2), B(2,1), C(1,2), D(1,1);
  A << 0.0, 1.0,
      -2.0, -3.0;
  B << 0.0,
       1.0;
  C << 1.0, 0.0;
  D << 0.0;

  // Similarity transform
  MatrixXd T(2,2);
  T << 2.0, 1.0,
       1.0, 1.0;

  MatrixXd Tinv = T.inverse();
  MatrixXd A2 = Tinv * A * T;
  MatrixXd B2 = Tinv * B;
  MatrixXd C2 = C * T;
  MatrixXd D2 = D;

  // Euler simulation parameters
  double dt = 0.001;
  int N = 5000;
  VectorXd x = VectorXd::Zero(2);
  VectorXd z = VectorXd::Zero(2);

  // Step input u(t)=1
  double max_abs_diff = 0.0;

  for (int k = 0; k < N; ++k) {
    double u = 1.0;

    // Original system
    VectorXd xdot = A * x + B * u;
    x = x + dt * xdot;
    double y = (C * x)(0) + (D * u)(0);

    // Transformed system in z-coordinates
    VectorXd zdot = A2 * z + B2 * u;
    z = z + dt * zdot;
    double y2 = (C2 * z)(0) + (D2 * u)(0);

    double diff = std::abs(y - y2);
    if (diff > max_abs_diff) max_abs_diff = diff;
  }

  std::cout << "max |y - y2| over simulation = " << max_abs_diff << std::endl;
  return 0;
}
      

In Java, EJML (Efficient Java Matrix Library) is a widely used linear algebra library for control-oriented computations. The example below computes the similarity-transformed matrices and checks algebraic consistency.

import org.ejml.simple.SimpleMatrix;

public class SimilarityDemo {
  static double relErr(SimpleMatrix X, SimpleMatrix Y) {
    double denom = Math.max(1.0, X.normF());
    return X.minus(Y).normF() / denom;
  }

  public static void main(String[] args) {
    SimpleMatrix A = new SimpleMatrix(new double[][]{
      {0.0, 1.0},
      {-2.0, -3.0}
    });
    SimpleMatrix B = new SimpleMatrix(new double[][]{
      {0.0},
      {1.0}
    });
    SimpleMatrix C = new SimpleMatrix(new double[][]{
      {1.0, 0.0}
    });
    SimpleMatrix D = new SimpleMatrix(new double[][]{
      {0.0}
    });

    SimpleMatrix T = new SimpleMatrix(new double[][]{
      {2.0, 1.0},
      {1.0, 1.0}
    });
    SimpleMatrix Tinv = T.invert();

    SimpleMatrix A2 = Tinv.mult(A).mult(T);
    SimpleMatrix B2 = Tinv.mult(B);
    SimpleMatrix C2 = C.mult(T);
    SimpleMatrix D2 = D.copy();

    System.out.println("relerr(A2, inv(T)AT) = " + relErr(A2, Tinv.mult(A).mult(T)));
    System.out.println("relerr(B2, inv(T)B)  = " + relErr(B2, Tinv.mult(B)));
    System.out.println("relerr(C2, CT)       = " + relErr(C2, C.mult(T)));
    System.out.println("relerr(D2, D)        = " + relErr(D2, D));
  }
}
      

Mathematica can verify invariance symbolically by computing \( \mathbf{G}(s)=\mathbf{C}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{B}+\mathbf{D} \) and comparing it before and after transformation.

(* Define system matrices *)
A = { {0, 1}, {-2, -3} };
B = { {0}, {1} };
C = { {1, 0} };
D = { {0} };

(* Similarity transform *)
T = { {2, 1}, {1, 1} };
A2 = Inverse[T].A.T;
B2 = Inverse[T].B;
C2 = C.T;
D2 = D;

(* Transfer function G(s) = C (sI - A)^(-1) B + D *)
s = Symbol["s"];
I2 = IdentityMatrix[2];

G1 = Simplify[C.Inverse[s I2 - A].B + D];
G2 = Simplify[C2.Inverse[s I2 - A2].B2 + D2];

Print["G1(s) = ", G1];
Print["G2(s) = ", G2];
Print["Simplify[G1 - G2] = ", Simplify[G1 - G2]];

(* Eigenvalue invariance *)
Print["Eigenvalues(A)  = ", Eigenvalues[A]];
Print["Eigenvalues(A2) = ", Eigenvalues[A2]];
      

Problem 1 (Derive the transformed model): Consider the continuous-time system \( \dot{\mathbf{x}}=\mathbf{A}\mathbf{x}+\mathbf{B}\mathbf{u} \), \( \mathbf{y}=\mathbf{C}\mathbf{x}+\mathbf{D}\mathbf{u} \). Let \( \mathbf{x}=\mathbf{T}\mathbf{z} \) for an invertible \( \mathbf{T} \). Derive formulas for \( \mathbf{\bar A},\mathbf{\bar B},\mathbf{\bar C},\mathbf{\bar D} \).

Solution: Substitute \( \mathbf{x}=\mathbf{T}\mathbf{z} \) and \( \dot{\mathbf{x}}=\mathbf{T}\dot{\mathbf{z}} \):

\[ \mathbf{T}\dot{\mathbf{z}}=\mathbf{A}\mathbf{T}\mathbf{z}+\mathbf{B}\mathbf{u} \;\;\Rightarrow\;\; \dot{\mathbf{z}}=\mathbf{T}^{-1}\mathbf{A}\mathbf{T}\mathbf{z}+\mathbf{T}^{-1}\mathbf{B}\mathbf{u}. \]

Output: \( \mathbf{y}=\mathbf{C}\mathbf{T}\mathbf{z}+\mathbf{D}\mathbf{u} \). Therefore \( \mathbf{\bar A}=\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \), \( \mathbf{\bar B}=\mathbf{T}^{-1}\mathbf{B} \), \( \mathbf{\bar C}=\mathbf{C}\mathbf{T} \), \( \mathbf{\bar D}=\mathbf{D} \).

Problem 2 (Prove transfer function invariance): Prove that \( \mathbf{G}(s)=\mathbf{C}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{B}+\mathbf{D} \) is unchanged by similarity transformations.

Solution: Using \( s\mathbf{I}-\mathbf{\bar A}=\mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})\mathbf{T} \),

\[ (s\mathbf{I}-\mathbf{\bar A})^{-1}=\mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{T}. \]

Then substitute \( \mathbf{\bar B}=\mathbf{T}^{-1}\mathbf{B} \), \( \mathbf{\bar C}=\mathbf{C}\mathbf{T} \), \( \mathbf{\bar D}=\mathbf{D} \):

\[ \mathbf{\bar G}(s)=(\mathbf{C}\mathbf{T})\left[\mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{T}\right](\mathbf{T}^{-1}\mathbf{B})+\mathbf{D} =\mathbf{C}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{B}+\mathbf{D}=\mathbf{G}(s). \]

Problem 3 (Recover \( \mathbf{T} \) from controllability, SISO): Let two SISO realizations of order \( n \) be related by similarity, and assume both are controllable. Show that \( \mathbf{T}=\mathcal{C}\mathcal{C}_2^{-1} \), where \( \mathcal{C}=[\mathbf{B}\;\mathbf{A}\mathbf{B}\;\cdots\;\mathbf{A}^{n-1}\mathbf{B}] \) and \( \mathcal{C}_2=[\mathbf{B}_2\;\mathbf{A}_2\mathbf{B}_2\;\cdots\;\mathbf{A}_2^{n-1}\mathbf{B}_2] \).

Solution: From Section 3.4, for SISO similarity transforms,

\[ \mathcal{C}_2 = \mathbf{T}^{-1}\mathcal{C}. \]

If both systems are controllable, then \( \det(\mathcal{C})\neq 0 \) and \( \det(\mathcal{C}_2)\neq 0 \). Multiply both sides by \( \mathcal{C}_2^{-1} \) on the right:

\[ \mathcal{C}_2\mathcal{C}_2^{-1}=\mathbf{T}^{-1}\mathcal{C}\mathcal{C}_2^{-1} \;\;\Rightarrow\;\; \mathbf{I}=\mathbf{T}^{-1}\mathcal{C}\mathcal{C}_2^{-1} \;\;\Rightarrow\;\; \mathbf{T}=\mathcal{C}\mathcal{C}_2^{-1}. \]

Problem 4 (Rank invariance test): Show that if \( \operatorname{rank}(\mathcal{C}) < n \) for one realization, then it must also satisfy \( \operatorname{rank}(\mathcal{C}_2) < n \) for any similar realization.

Solution: Similarity implies \( \mathcal{C}_2=\mathbf{T}^{-1}\mathcal{C} \). Since \( \mathbf{T}^{-1} \) is invertible, left multiplication cannot change rank: \( \operatorname{rank}(\mathcal{C}_2)=\operatorname{rank}(\mathcal{C}) \). Therefore if one is rank-deficient, so is the other.

Problem 5 (Eigenvalue-based necessary condition): Two realizations are claimed to be similar. Provide a necessary condition using eigenvalues and justify it.

Solution: A necessary condition is that \( \mathbf{A} \) and \( \mathbf{A}_2 \) have the same characteristic polynomial (hence identical eigenvalues with multiplicity). This follows from \( \mathbf{A}_2=\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \) and the determinant identity:

\[ \det(\lambda\mathbf{I}-\mathbf{A}_2)=\det(\lambda\mathbf{I}-\mathbf{A}). \]

If eigenvalues differ, the realizations cannot be related by a similarity transformation (and therefore cannot be state-coordinate equivalent).

We defined similarity (state-coordinate) transformations and proved that they preserve the input–output map, transfer function, eigenvalues, and controllability/observability ranks. We also provided constructive SISO formulas to recover the transformation matrix \( \mathbf{T} \) from controllability or observability matrices and implemented equivalence verification workflows across Python, MATLAB/Simulink, C++, Java, and Mathematica. These results justify why canonical forms (Lesson 3) represent entire equivalence classes of realizations, and they set up the next lesson connecting state-space models to transfer functions.