Why eigen values and poles of a system are equivalent?
Let's consider a discrete-time state space model (the derivation for a coninuous-time system is completely analogous):

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

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

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

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.