# Why eigen values and poles of a system are equivalent?

In control systems engineering, the stability of a system (modeled in the form of Transfer Function) is determined by the poles of the system in the right or left hand sides. When the model is represented using State Space approach, the eigen values of the (A) state matrix are equivalent to the poles in the Transfer Function approach.

why are both eigen values and poles equivalent? why do they give the same information?

• Is this indeed true? Can you cite some references where you found this? I would be personally greatly curious to know how... Welcome to DSP.SE – Dipan Mehta Jun 16 '15 at 4:33

Let's consider a discrete-time state space model (the derivation for a coninuous-time system is completely analogous):

\begin{align}\mathbf{q}[n+1]&=\mathbf{Aq}[n]+\mathbf{b}x[n]\\ y[n]&=\mathbf{c}^T\mathbf{q}[n]+dx[n]\tag{1} \end{align}

where $x[n]$ is the input, $y[n]$ is the output, and $\mathbf{q}[n]$ is the state vector. Taking the $\mathcal{Z}$-transform of (1) gives

\begin{align}z\mathbf{Q}(z)&=\mathbf{AQ}(z)+\mathbf{b}X(z)\\ Y(z)&=\mathbf{c}^T\mathbf{Q}(z)+dX(z)\tag{2} \end{align}

From the top equation of (2) we can express $\mathbf{Q}(z)$ in terms of $\mathbf{A}$, $\mathbf{b}$, and $X(z)$:

$$\mathbf{Q}(z)=(z\mathbf{I}-\mathbf{A})^{-1}\mathbf{b}X(z)\tag{3}$$

where $\mathbf{I}$ is the identity matrix of appropriate dimension. Plugging (3) into the second equation in (2) gives

$$Y(z)=\mathbf{c}^T(z\mathbf{I}-\mathbf{A})^{-1}\mathbf{b}X(z)+dX(z)\tag{4}$$

from which follows the expression for the transfer function

$$H(z)=\frac{Y(z)}{X(z)}=\mathbf{c}^T(z\mathbf{I}-\mathbf{A})^{-1}\mathbf{b}+d\tag{5}$$

The matrix $(z\mathbf{I}-\mathbf{A})^{-1}$ can be written as

$$(z\mathbf{I}-\mathbf{A})^{-1}=\frac{1}{\text{det}(z\mathbf{I}-\mathbf{A})}\mathbf{B}\tag{6}$$

where $\mathbf{B}$ is the adjugate matrix of $(z\mathbf{I}-\mathbf{A})$ whose elements are polynomials in $z$.

Plugging (6) into the expression for the transfer function (5) shows that the denominator of $H(z)$ is given by $\text{det}(z\mathbf{I}-\mathbf{A})$, which is the characteristic polynomial of $\mathbf{A}$ the roots of which are the eigenvalues of $\mathbf{A}$. Since the roots of the denominator of $H(z)$ are the system's poles, the eigenvalues of $\mathbf{A}$ and the poles of $H(z)$ are equivalent.