Are normal modes the eigenvectors of the matrix $(\omega ^2 T- V)$ where $T$ is the matrix of kinetic energy and $V$ is the matrix of potential energy?
Is it the only way to express them?
How can I express them using the coordinates that I have choosen at the beginning of the exercise?
Lets look at this example
$$T=\frac{m_1}{2}\dot x_1^2+\frac{m_2}{2}\dot x_2^2\\ V=-\frac{k_1}{2}\,x_1^2-\frac{k_2}{2}\,x_2^2-\frac{k_3}{2}\,(x_1-x_2)^2$$
from here you obtain
$$M_{ij}=\frac{\partial}{\partial \dot x_i}\left(\frac{\partial T}{\partial \dot x_j}\right)\\ K_{ij}=-\frac{\partial}{\partial x_i}\left(\frac{\partial V}{\partial x_j}\right)$$
hence
$$\underbrace{\left[M\,\omega^2 -K\right]}_{A}\,\vec v=0$$
the solution
from the matrix A you obtain the eigenvalues $\omega_1~,\omega_2~$ and for each eigenvalue the eigen vector $~\Rightarrow\quad \vec v_1~,\vec v_2~$
$$\begin{bmatrix} x_1(t) \\ x_2(t) \\ \end{bmatrix}=\underbrace{c_1\,\vec v_1\,\cos(\omega_1\,t\,+\phi_1)}_{\vec\eta_1}+\underbrace{c_2\,\vec v_2\,\cos(\omega_2\,t\,+\phi_2)}_{\vec\eta_2}$$
where $c_i~,\phi_i~$ are the initial conditions and $~\vec\eta_i~$ are the normal modes
$$ M=\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\quad, K= -\begin{bmatrix} k_1+k_3 & -k_3 \\ -k_3 & k_2+k_3 \\ \end{bmatrix}$$
