Lagrangian Oscilattor I want to know how to calculate the normal modes from a Lagrangian.
I make the T (kinetic energy matrix) and U (potential energy matrix), and then I calculate the determinant of $|T-\omega ^2 U|$ (with omega as the angular speed). This way I find the $\omega$'s but I dont know how to find the normal modes.
Are they just the eigenvectors of the $T-\omega ^2 U$?
 A: $\newcommand{\oh}[0]{\frac{1}{2}}$You're basically right. It is the eigenvector times $\cos(\omega t)$. I will put in more details,plaigarizing from a pervious answer of mine. 
I will adapt the discussion in landau mechanics 3rd edition starting at the bottom of page 65. In general the hamiltonian is $L = \oh \dot{x}_i m_{ij} \dot{x}_j - \oh x_i V_{ij} x_j$. The equation of motion (EOM) is $m_{ij} \ddot{x}_j = - V_{ij} x_j$. We seek a solution where each coordinate oscillates with the same frequency: $x_j(t) = A_j \sin(\omega t + \phi_j)$. Plugging this into the EOM we get $\omega^2 m_{ij} A_j - V_{ij} A_j = 0$. i.e. $A_j$ is a null vector of $\omega^2 m_{ij} - V_{ij}$. This is only possible if $\omega$ takes on such a value where the determinant of $\omega^2 m_{ij}  - V_{ij}$ is zero. The number of $\omega^2$'s which give zero determinant will be equal to the dimensionality of our system. We will index this set of $\omega$'s by the greek letter $\alpha$.
Now say we have found a $\omega_\alpha$ where the determinant of $\omega_\alpha^2 m_{ij}  - V_{ij}$ is zero. Then we must find the $A_{k\alpha}$ which is a null vector of $\omega_\alpha^2 m_{ij}  - V_{ij}$. Having done this for each $\alpha$, we find that the most general solution is $x_k(t) = A_{k \alpha} q_\alpha \sin(\omega_\alpha t - \phi_\alpha)$. The $q_\alpha$ are seen to be the normal mode coordinates. 
