I was shown a proof in class, where it was said that all Runge-Kutta methods preserve the quadratic invariant $$I(y) = y^TCy \quad ,$$ where $C$ is a quadratic symmetric matrix, under the condition $$b_i a_{ij} + b_j a_{ji} - b_ib_j = 0 \quad .$$ In general I understood the form of the prove, however I am struggling with why $$\frac{d}{dt}I(y) = 2y^TCf(y) = 0 \quad, $$ holds and is zero as per my professor. If I try to perform the same step I arrive at the following form $$\frac{d}{dt}I(y) = (C+C^T)y \quad, $$ where I do not see why it would be zero for a symmetric matrix C.
I would appreciate if someone could show me where I went wrong and especially why the statement of my professor is true. If any more of the underlying proof is necessary to show I can add it.
The question seems not so much to be about Runge-Kutta methods but about the derivative of quadratic forms. There you get by the product rule $$ \frac{d}{dt}I(y(t))=\frac{d}{dt}y(t)^TCy(t)=\dot y(t)^TCy(t)+y(t)^TC\dot y(t) $$ Now the terms are scalars, thus invariant under transposition, apply that to the first term $$ =y(t)^T(C^T+C)\dot y(t) $$ and then apply the symmetry of $C$ to get $C^T+C=2C$.
That the derivative of a constant is zero is trivial, the last step then is to insert $\dot y(t)=f(y(t))$.