Prove that, for a normal surface $\Sigma \subsetneq \mathbb{R}^3$ $(L \circ L)(u) - 2H(p) L(u) + K(p) u = 0$ for the Weingarten operator

Question

As per the title:

I need help proving that $(L \circ L)(u) - 2H(p) L(u) + K(p) u = 0$, where $H(p)$ is the mean curvature and $K(p)$ is the Gauss curvature.

I realy have zero clue even on how to start.

Answer 1

Hint: Think about the fact that $L(u)$ is a linear map from a 2-dimensional vector space to itself and that $K(u)$ and $H(u)$ are related to the determinant and the trace of $L(u)$. Then recall what the characteristic polynomial of a $2\times 2$-matrix looks like and use the Cayley-Hamilton Theorem.