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

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

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

$$E\underline{{\tilde x}}(n+1) = A \, E\underline{{\tilde x}}(n) + B \underline{u}(n)$$

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

Premultiplying the first equation above by $E^{-1}$ , we have

$$\underline{{\tilde x}}(n+1) = \left(E^{-1}A E\right) \underline{{\tilde x}}(n) + \left(E^{-1}B\right) \underline{u}(n)$$

$$\underline{y}(n) = \left(C E\right) \underline{{\tilde x}}(n) + D\underline{u}(n).$$ (G.18)

Defining

$$\tilde{A} = E^{-1}A E$$

$${\tilde B} = E^{-1}B$$

$${\tilde C} = C E$$

$${\tilde D} = D$$ (G.19)

we can write

$$\underline{{\tilde x}}(n+1) = \tilde{A}\, \underline{{\tilde x}}(n) + {\tilde B}\, \underline{u}(n)$$

$$\underline{y}(n) = {\tilde C}\, \underline{{\tilde x}}(n) + {\tilde D}\, \underline{u}(n)$$ (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):

$$\begin{eqnarray*} {\tilde H}(z) &=& {\tilde D}+ {\tilde C}(zI - \tilde{A})^{-1}{\tilde B}\\ &=& D + (C E) \left(zI - E^{-1}AE\right)^{-1}(E^{-1}B)\\ &=& D + C \left[E\left(zI - E^{-1}AE\right)E^{-1}\right]^{-1} B\\ &=& D + C \left(zI - A\right)^{-1} B\\ &=& H(z) \end{eqnarray*}$$

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

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

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:

$$\begin{eqnarray*} {\tilde h}(n) &=& {\tilde C}\tilde{A}^n{\tilde B}= (CE)(E^{-1}AE)(E^{-1}AE)\cdots(E^{-1}AE)(E^{-1}B)\\ &=& {\tilde C}\tilde{A}^n{\tilde B}= C(EE^{-1}) A(EE^{-1}) \cdots A(EE^{-1}) B)\\ &=& CA^nB \end{eqnarray*}$$