Eigenfrequencies of an Hamiltonian dynamical system in different bases Consider the Hamiltonian 
$$
   H=H(x,y,p_x,p_y)
$$
which generates the dynamical system
$$ 
  \dot{x}=+\frac{\partial H}{\partial p_x}
$$
$$ 
  \dot{y}=+\frac{\partial H}{\partial p_y}
$$
$$ 
  \dot{p_x}=-\frac{\partial H}{\partial x}
$$
$$ 
  \dot{p_y}=-\frac{\partial H}{\partial y}
$$
I then discover that this system admits a certain fixed point
$$
    \vec{z}_0:=(x_0,\,y_0,\,p_{x,0},\,p_{y,0})
$$
To determine the stability properties of $\vec{z}_0$, I compute the Jacobian matrix $J$ associated to the dynamical system, I evaluate it at fixed point $\vec{z}_0$, and I compute the eigenvalues, which are of the type:
$$
    \lambda_1=+i\omega_a
$$
$$
    \lambda_2=-i\omega_a
$$
$$
    \lambda_3=+i\omega_b
$$
$$
    \lambda_4=-i\omega_b
$$
Therefore, I can recognize the presence of two characteristic frequencies, namely $\omega_a$ and $\omega_b$.
So far, so good.
At this point, I try to do the same computation with a different set of dynamical variables, for example making use of polar coordinates instead of cartesian coordinates. So, I start from Hamiltonian 
$$
   H^\prime=H^\prime(r,\theta,p_r,p_\theta),
$$
I find the same fixed point found before, but now written in polar coordinates, i.e. $\vec{z}_0^\prime=(r_0,\theta_0,p_{r,0},p_{\theta,0})$. I then compute the Jacobian matrix $J^\prime$ associated to $H^\prime$, evaluate it at fixed point $\vec{z}_0^\prime$ and compute its eigenvalues. The latter have the following structure: 
$$
    \lambda_1^\prime=+i\omega_a^\prime
$$
$$
    \lambda_2^\prime=-i\omega_a^\prime
$$
$$
    \lambda_3^\prime=+i\omega_b^\prime
$$
$$
    \lambda_4^\prime=-i\omega_b^\prime
$$
So, i can recognize two characteristic frequencies: $\omega_a^\prime$ and $\omega_b^\prime$.
What I find really counterintuitive is that
$$
    \omega_a\neq\omega_a^\prime
$$
$$
    \omega_b\neq\omega_b^\prime
$$
that means that, the two eigenfrequencies are different in the two schemes (the cartesian one and the polar one) that I developed. I would have expected them to be the same, i.e. that $\omega_a=\omega_a^\prime$ and $\omega_b=\omega_b^\prime$. 
So, here my question comes: is it possible that different set of dynamical variables lead to different characteristic frequencies for a certain fixed point?
To be noted: actually the two sets of eigenfrequencies, i.e. $\{\omega_a, \omega_b\}$ and $\{\omega_a^\prime, \omega_b^\prime\}$ include similar terms and one can obtain the frequencies in the first set by properly combining the ones in the second set, and viceversa. But, as I said, I expected the two sets to be really identical.
 A: If you change the generalized coordinate the eigenvalues of your new equation of motion must be the same otherwise the dynamic of your new system is changed. to obtain the eigenvalues you must linearize your equation of motions. 
Example:
$$\underbrace{\begin{bmatrix}
  \ddot{x} \\
  \ddot{y} \\
 \end{bmatrix}}_{\vec{\ddot{q}}}+\underbrace{\left[ \begin {array}{cc} 2\,c&-c\\ -c&2\,c
\end {array} \right]}_{C}\,\underbrace{\left[ \begin {array}{c} x\\ y\end {array} \right]}_{\vec{q}}  =0\tag 1 $$
where C is the stiffness matrix. to obtain the eigenvalues we transform equation (1) to first order differential equation.
$$\vec{\dot{Y}}=\underbrace{\begin{bmatrix}
   0_{2\times 2} & -C \\
   I_{2\times 2} &  0_{2\times 2}\\
 \end{bmatrix}}_{A}\,\vec{Y}$$
the eigenvalues of the matrix A are:
$$\vec{\lambda}=\left[ \begin {array}{c} i\sqrt {c}\\ -i\sqrt {c}
\\ i\sqrt {3}\sqrt {c}\\-i\sqrt {
3}\sqrt {c}\end {array} \right] 
\tag 2$$
if you choose new generalized coordinates ($r\,,\varphi$) ,polar coordinate ,  for example 
$$ \begin{bmatrix}
  x \\
  y \\
 \end{bmatrix}= \left[ \begin {array}{c} r\cos \left( \varphi  \right) 
\\ r\sin \left( \varphi  \right) \end {array}
 \right] 
 $$
equation (1) $\mapsto$:
$$J^T\,J\,\underbrace{\begin{bmatrix}
  \ddot{r} \\
  \ddot{\varphi} \\
 \end{bmatrix}}_{\vec{\ddot{w}}}+J^T\,C\,J\,\underbrace{\begin{bmatrix}
  r \\
  \varphi \\
 \end{bmatrix}}_{\vec{w}}=0\tag 3$$
where J is the Jacobean :
$$J=\left[ \begin {array}{cc} \cos \left( \varphi  \right) &-r\sin
 \left( \varphi  \right) \\ \sin \left( \varphi 
 \right) &r\cos \left( \varphi  \right) \end {array} \right] 
$$ 
we can solve equation (3) for $\vec{\ddot{w}}$ and get:
$$\vec{\ddot{w}}=-\underbrace{J^{-1}\,C\,J}_{C_w}\vec{w}\tag 4$$
to calculate the eigenvalues we linearize the matrix $C_w\bigg|_{\varphi=0}:=C_L$
$ \Rightarrow$
$$C_{L}= \left[ \begin {array}{cc} 2\,c&-rc\\ -{\frac {c}{r}
}&2\,c\end {array} \right] 
$$
equation (4) $\mapsto$:
$$\vec{\ddot{w}}_L=-C_{L}\vec{w}_L\tag 5$$
again transformed to first order differential equation and obtain the eigenvalues give you the same eigenvalues equation (2).   (determinate of matrix C is equal to the determinate of matrix $C_L$)
