= Solution
The <kinetic energy> of the two equal masses is
$$
T=\frac m2(\dot x_1^2+\dot x_2^2).
$$
The extensions of the left, coupling, and right springs are respectively $x_1$, $x_2-x_1$, and $x_2$. Their <potential energy> is therefore
$$
U=\frac12k_1x_1^2+\frac12k_3(x_2-x_1)^2+\frac12k_2x_2^2.
$$
Thus the mechanical <Lagrangian> $L=T-U$ has the requested quadratic form with the symmetric <small-oscillation mass and stiffness matrices>
$$
\boxed{
(T_{ij})=
\begin{pmatrix}m&0\\0&m\end{pmatrix},
\qquad
(V_{ij})=
\begin{pmatrix}
k_1+k_3&-k_3\\
-k_3&k_2+k_3
\end{pmatrix}
}.
$$
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