Carbon dioxide molecule model A simple model of the carbon dioxide molecule can be modeled by a system of three masses and two springs. The oxygen atoms have mass M and carbon has mass m. The springs (bonds) have spring constant $k$.
The corresponding mathematical model is
$$
Mx'' = k(y-x)$$
$$my'' = -k(y-x) - k(y-z)$$
$$Mz'' = -k(z-y)
$$
I want to model this such that there is a system of equations where $\vec{x} = Ax$. I set $x_1 = x - y$, $x_2 = y - z$, $x_3 = x_1'$, and $x_4 = x_2'$. But I don't know what to do next.
 A: Take $x'=a, y'=b, z'=c$, so $a'=k(y-z)/M, b'=k(z-x)/m, c'=k(y-z)/M$.  If you row reduce the matrix you might find that you can reduce the number of variables from 6.
I'm assuming you meant $x'=Ax$.  Otherwise I'm not sure what you meant by $\bar{x}=Ax$
A: $$
\left[ \begin{array}{rrr}
 m_{o} x'' \\
 m_{c} y'' \\
 m_{o} z''
\end{array} \right]
%
=
%
k
\left[ \begin{array}{rrr}
 -1 & 1 & 0 \\
 1 & -2 & 1 \\
  0 & 1 & -1
\end{array} \right]
%
\left[ \begin{array}{rrr}
 x \\
 y \\
 z
\end{array} \right]
$$
The key is to diagonalize
$$
\mathbf{A} = 
\left[ \begin{array}{rrr}
 -1 & 1 & 0 \\
 1 & -2 & 1 \\
  0 & 1 & -1
\end{array} \right]
$$ 
Eigenvalues:
$$ 
\lambda \left( \mathbf{A} \right) = \left( -3, -1, 0 \right)
$$
Matrix of eigenvectors:
$$
\mathbf{P} = 
\left[ \begin{array}{rrr}
 1 & -2 & 1 \\
 -1 & 0 & 1 \\
 1 & 1 & 1 \\
\end{array} \right]
$$
Diagonalization:
$$
\mathbf{P} \, \mathbf{A} \, \mathbf{P}^{-1} = 
\left[ \begin{array}{rrr}
 -3 & 0 & 0 \\
 0 & -1 & 0 \\
 0 & 0 & 0 \\
\end{array} \right]
$$
