Similarity Transformations

A similarity transformation is a linear change of coordinates. That is, the original $ N$-dimensional state vector $ {\underline{x}}(n)$ is recast in terms of a new coordinate basis. For any linear transformation of the coordinate basis, the transformed state vector $ \underline{{\tilde x}}(n)$ may be computed by means of a matrix multiply. Denoting the matrix of the desired one-to-one linear transformation by $ E$, we can express the change of coordinates as

$\displaystyle {\underline{x}}(n) \isdef E\underline{{\tilde x}}(n)

or $ \underline{{\tilde x}}(n)=E^{-1}{\underline{x}}(n)$, if we prefer, since the inverse of a one-to-one linear transformation always exists.

Let's now apply the linear transformation $ E$ to the general $ N$-dimensional state-space description in Eq.$ \,$(G.1). Substituting $ {\underline{x}}(n) \isdeftext E\underline{{\tilde x}}(n)$ in Eq.$ \,$(G.1) gives

$\displaystyle E\underline{{\tilde x}}(n+1)$ $\displaystyle =$ $\displaystyle A \, E\underline{{\tilde x}}(n) + B \underline{u}(n)$  
$\displaystyle \underline{y}(n)$ $\displaystyle =$ $\displaystyle C E\underline{{\tilde x}}(n) + D\underline{u}(n)$ (G.17)

Premultiplying the first equation above by $ E^{-1}$, we have
$\displaystyle \underline{{\tilde x}}(n+1)$ $\displaystyle =$ $\displaystyle \left(E^{-1}A E\right) \underline{{\tilde x}}(n) + \left(E^{-1}B\right) \underline{u}(n)$  
$\displaystyle \underline{y}(n)$ $\displaystyle =$ $\displaystyle \left(C E\right) \underline{{\tilde x}}(n) + D\underline{u}(n).$ (G.18)

$\displaystyle \tilde{A}$ $\displaystyle =$ $\displaystyle E^{-1}A E$  
$\displaystyle {\tilde B}$ $\displaystyle =$ $\displaystyle E^{-1}B$  
$\displaystyle {\tilde C}$ $\displaystyle =$ $\displaystyle C E$  
$\displaystyle {\tilde D}$ $\displaystyle =$ $\displaystyle D$ (G.19)

we can write
$\displaystyle \underline{{\tilde x}}(n+1)$ $\displaystyle =$ $\displaystyle \tilde{A}\, \underline{{\tilde x}}(n) + {\tilde B}\, \underline{u}(n)$  
$\displaystyle \underline{y}(n)$ $\displaystyle =$ $\displaystyle {\tilde C}\, \underline{{\tilde x}}(n) + {\tilde D}\, \underline{u}(n)$ (G.20)

The transformed system describes the same system as in Eq.$ \,$(G.1) relative to new state-variable coordinates. To verify that it's really the same system, from an input/output point of view, let's look at the transfer function using Eq.$ \,$(G.5):

{\tilde H}(z) &=& {\tilde D}+ {\tilde C}(zI - \tilde{A})^{-1}{...]^{-1} B\\
&=& D + C \left(zI - A\right)^{-1} B\\
&=& H(z)

Since the eigenvalues of $ A$ are the poles of the system, it follows that the eigenvalues of $ \tilde{A}=E^{-1}A E$ are the same. In other words, eigenvalues are unaffected by a similarity transformation. We can easily show this directly: Let $ \underline{e}$ denote an eigenvector of $ A$. Then by definition $ A\underline{e}=\lambda\underline{e}$, where $ \lambda$ is the eigenvalue corresponding to $ \underline{e}$. Define $ \underline{\tilde{e}}=E^{-1}\underline{e}$ as the transformed eigenvector. Then we have

$\displaystyle \tilde{A}\underline{\tilde{e}}= \tilde{A}(E^{-1}\underline{e}) = ...
...^{-1}A\underline{e}= E^{-1}\lambda\underline{e}= \lambda\underline{\tilde{e}}.

Thus, the transformed eigenvector is an eigenvector of the transformed $ A$ matrix, and the eigenvalue is unchanged.

The transformed Markov parameters, $ {\tilde C}\tilde{A}^n {\tilde B}$, are obviously the same also since they are given by the inverse $ z$ transform of the transfer function $ {\tilde H}(z)$. However, it is also easy to show this by direct calculation:

{\tilde h}(n) &=& {\tilde C}\tilde{A}^n{\tilde B}= (CE)(E^{-1}...
...ilde B}= C(EE^{-1}) A(EE^{-1}) \cdots A(EE^{-1}) B)\\
&=& CA^nB

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