TITLE: Principal, Gauss and Average Curvature
QUESTION [1 upvotes]: On my lecture note it is written as follow (sorry for the mistaken terms, it is translate):

Principal curvatures are the eigenvalues of $L^{i}_{j}$ which are $k_1,k_2$
The gauess curvature $K=det(L^{i}_{j})=k_1\cdot k_2$
The average curvature is $H=\frac{k_1+k_2}{2}=\frac{1}{2}trace(L^{i}_{j})$

now all that is fine when my shape operator is diagonal, but in the case I get
$$(u^2+1)^2 \begin{pmatrix} 2cos(2v)&-2usin(2v)\\ -\frac{sin(2v)}{u}&-2cos(2v) \end{pmatrix} $$
I first need to diagonal the shape operator?

REPLY [1 votes]: No. The trace and the determinant don't depend on your matrix representation. If the matrix is diagonal, you can read off the principal curvatures $k_1$ and $k_2$ as the entries, but the sum of the eigenvalues is always the trace of the matrix and the product of the eigenvalues is always the determinant of the matrix.
{"set_name": "stack_exchange", "score": 1, "question_id": 2542946}