Questions: State Transformations and Similarity Transformations

Question 1

A control engineer has a state-space system that is unstable (A has eigenvalues with positive real parts). She applies a similarity transformation x̄ = Tx, obtaining Ā = TAT⁻¹. Is the transformed system stable?

• A. Yes — the transformation T can be chosen to place the eigenvalues of Ā in the left half-plane, stabilizing the system
• B. It depends — orthogonal transformations preserve eigenvalues, but other choices of T may alter them
• C. No — if the original system is unstable, any similarity transformation leaves it unstable
• D. Yes — transforming to diagonal form always produces a stable system because the modes decouple

Answer: (C)

Similarity transformations preserve eigenvalues. If Av = λv, then (TAT⁻¹)(Tv) = T(Av) = λ(Tv), so Ā has the same eigenvalues as A with transformed eigenvectors. Since stability is determined solely by whether eigenvalues have negative real parts, a similarity transformation can never stabilize or destabilize a system — it is merely a change of coordinates. Option A is the most tempting misconception: engineers think of transformations as tools to 'fix' systems, but pole placement (which does change eigenvalues) requires state feedback, not a coordinate change.

Question 2

Which of the following changes under a similarity transformation x̄ = Tx of a state-space system?

• A. The eigenvalues of the system matrix A
• B. The transfer function H(s) = C(sI − A)⁻¹B
• C. The controllability and observability of the system
• D. The specific numerical entries of the A, B, and C matrices

Answer: (D)

The individual numerical entries of A, B, and C change under a similarity transformation — Ā = TAT⁻¹, B̄ = TB, C̄ = CT⁻¹ are generally different matrices. What is preserved are the invariants: eigenvalues (poles, stability), the transfer function (input-output behavior), and structural properties like controllability and observability. The whole point of similarity transformations is that different canonical forms — controllable canonical, observable canonical, diagonal — have very different matrix entries while describing the exact same physical system.

Question 3

Two state-space representations of the same system related by a similarity transformation x̄ = Tx will have identical transfer functions.

• T. True
• F. False

Answer: True

The transfer function H(s) = C(sI−A)⁻¹B is a property of the physical system, not of the coordinate choice. Under the transformation, the new transfer function is C̄(sI−Ā)⁻¹B̄ = (CT⁻¹)(T(sI−A)⁻¹T⁻¹)(TB) = C(sI−A)⁻¹B. The T matrices cancel exactly, confirming invariance. This is why system identification — recovering system behavior from input-output data — gives a unique answer even though infinitely many valid state-space representations exist.

Question 4

Diagonalizing the system matrix A via a similarity transformation changes the poles of the system to the eigenvalues of the new diagonal matrix Λ, which may differ from the original poles.

• T. True
• F. False

Answer: False

The diagonal entries of Λ = TAT⁻¹ (when T contains the eigenvectors of A) are precisely the eigenvalues of A — not new values. Diagonalization reveals existing eigenvalues in explicit form; it does not create new ones. The poles of the system are the eigenvalues of A, and since similarity transformations preserve eigenvalues, the poles are identical before and after diagonalization. What changes is the representation: the diagonal form decouples each mode, simplifying analysis, but the underlying dynamics are unchanged.

Question 5

In the similarity transformation, the input matrix transforms as B̄ = TB but the output matrix transforms as C̄ = CT⁻¹. Why are these asymmetric — why not C̄ = TC?

Model answer: Under x̄ = Tx (so x = T⁻¹x̄), the state equation ẋ = Ax + Bu becomes T⁻¹ẋ̄ = AT⁻¹x̄ + Bu, which gives ẋ̄ = TAT⁻¹x̄ + TBu — so B̄ = TB. The output equation y = Cx becomes y = C(T⁻¹x̄) = (CT⁻¹)x̄ — so C̄ = CT⁻¹. The input matrix B maps from input space (unchanged) to state space (transformed), picking up T on the left. The output matrix C maps from state space (now expressed in new coordinates via T⁻¹) to output space, picking up T⁻¹ on the right.

The asymmetry reflects the distinct roles of inputs and outputs in the state equations. This transformation structure guarantees the transfer function is invariant: C̄(sI−Ā)⁻¹B̄ = (CT⁻¹)(T(sI−A)⁻¹T⁻¹)(TB) = C(sI−A)⁻¹B, with all T factors canceling exactly.