Consider the Lagrangian $$L(q_1, q_2, \dot{q_1}, \dot{q_2}) = \dot{q_1}^2 - \dot{q_2}^2 + q_1 ^2 - q_2 ^2$$ (Set aside any concerns about the possibility of the kinetic energy being negative.) Show that L is invariant under the transformation $$ \begin{pmatrix} Q_1 \\ Q_2 \\ \end{pmatrix} = \begin{pmatrix} \cosh\theta & \sinh\theta \\ \sinh\theta & \cosh\theta \\ \end{pmatrix} \begin{pmatrix} q_1 \\ q_2 \\ \end{pmatrix}$$
My question is how can one explicitly show that the transformation is invariant here? By using the Noether's theorem one can compute the conserved quantity under the transformation, however it doesn't explicitly show that $$ L(q_1, q_2, \dot{q_1}, \dot{q_2}) = L(Q_1, Q_2, \dot{Q_1}, \dot{Q_2}) $$ which is the definition of an transformation invariant Lagrangian.
We can use the inverse transfotmation (givig the $q_i$ as function of $Q_i$) and substitute in $L$ or start with $L(Q_1, Q_2, \dot{Q_1}, \dot{Q_2}) = \dot{Q_1}^2 - \dot{Q_2}^2 + Q_1 ^2 - Q_2 ^2$ and check the form $L'(q_1, q_2, \dot{q_1}, \dot{q_2}) = L(Q_1, Q_2, \dot{Q_1}, \dot{Q_2})$ has.
$Q_1^2=(q_1\cosh\theta+q_2\sinh\theta)^2=q_1^2\cosh^2\theta+2q_1q_2\cosh\theta\sinh\theta+q_2^2\sinh^2\theta$
$Q_2^2=(q_1\sinh\theta+q_2\cosh\theta)^2=q_1^2\sinh^2\theta+2q_1q_2\sinh\theta\cosh\theta+q_2^2\cosh^2\theta$
$Q_1^2-Q_2^2=q_1^2\cosh^2\theta-q_1^2\sinh^2\theta+2q_1q_2\cosh\theta\sinh\theta-2q_1q_2\cosh\theta\sinh\theta-(q_2^2\cosh^2\theta-q_2^2\sinh^2\theta)$
$Q_1^2-Q_2^2=q_1^2(\cosh^2\theta-\sinh^2\theta)-q_2^2(\cosh^2\theta-\sinh^2\theta)=q_1^2-q_2^2$
Similarly for the derivatives inasmuch $\dot Q_1=\dot q_1\cosh\theta+\dot q_2\sinh\theta$ and $\dot Q_2=\dot q_1\sinh\theta+\dot q_2\cosh\theta$. So,
$\dot{Q_1}^2 - \dot{Q_2}^2 + Q_1 ^2 - Q_2 ^2=\dot{q_1}^2 - \dot{q_2}^2 + q_1 ^2 - q_2 ^2$
Proving the invariance.