more details :
The equations of motion in your case are:
$$\ddot{\vec{q}}+\frac{\partial U}{\partial \vec{q}}=0\tag 1$$
where $\vec{q}$ is the vector of generalized coordinates and $U$ is the potential energy
if $\frac{\partial U}{\partial \vec{q}}$ is a linear function of $\vec{q}$
you can write equation (1) as:
$$\ddot{\vec{q}}+C\,\vec{q}=0\tag 2$$
where $C$ is a quadratic constant matrix :
$C=\frac{\partial}{\partial \vec{q}}\left(\frac{\partial U}{\partial \vec{q}}\right)$
to transformed equation (2) to diagonal shape with the transformation matrix $T$, we calculate the eigenvalues and the eigenvectors of the matrix $C$.
with $\vec{q}=T\,\vec{Q}$ and
$T=\left[\vec{E}_{v1}(\lambda_1)\,,\vec{E}_{v2}(\lambda_2)\,,\ldots\right]$ the transformation matrix.
,where $\lambda_i$ are the eigenvalues of the matrix $C$ and $\vec{E}_{vi}$ are the eigenvectors. we obtain for equation (2)
$$T\ddot{\vec{Q}}+C\,T\,\vec{Q}=0\tag 3$$
multiply equation (3) with $T^T$ we get:
$$T^T\,T\ddot{\vec{Q}}+T^T\,C\,T\,\vec{Q}=0\tag 4$$
because $T^T\,T=I$ unity matrix and $T^T\,C\,T=\text{Diag}\left[\lambda_1\,,\lambda_2\,,\ldots\,,\lambda_n\right]$ diagonal shape , we get
$$\ddot{{Q}_i}+\lambda_i\,Q_i=\boxed{\ddot{{Q}_i}+\omega_i^2\,Q_i=0}$$
Example:
$U=1/2\,{\it k1}\, \left( {x_{{1}}}^{2}+{x_{{3}}}^{2} \right) +1/2\,{\it
k2}\,{x_{{2}}}^{2}+1/2\,{\it k3}\, \left( x_{{1}}x_{{2}}+x_{{2}}x_{{3}
} \right)
$
$\Rightarrow$
$C=\left[ \begin {array}{ccc} {\it k1}&1/2\,{\it k3}&0
\\ 1/2\,{\it k3}&{\it k2}&1/2\,{\it k3}
\\ 0&1/2\,{\it k3}&{\it k1}\end {array} \right]
$
eigenvalues:
$\vec{\lambda}=\left[ \begin {array}{c} {\it k1}\\ 1/2\,{\it k1}+1
/2\,{\it k2}+1/2\,\sqrt {{{\it k1}}^{2}-2\,{\it k2}\,{\it k1}+{{\it k2
}}^{2}+2\,{{\it k3}}^{2}}\\ 1/2\,{\it k1}+1/2\,{\it
k2}-1/2\,\sqrt {{{\it k1}}^{2}-2\,{\it k2}\,{\it k1}+{{\it k2}}^{2}+2
\,{{\it k3}}^{2}}\end {array} \right]
$
