Second Fundamental Form, Shape Operator, Curvatures and their relation Consider a parameterization $r:U\rightarrow M$ for a regular surface $M$.Denote by $n$ the unit normal to the surface. I was given the following defition for the matrix $B$ that defines the second fundamental form:
$$B=
    \begin{pmatrix}
    r''_{uu}\cdot n & r''_{uv}\cdot n \\
    r''_{vu}\cdot n & r''_{vv}\cdot n \\
    \end{pmatrix}
$$
Also, I was given the following definition for the matrix $g$ that defines the Riemannian metric:
$$g=
    \begin{pmatrix}
    r'_u\cdot r'_u & r'_u\cdot r'_v \\
    r'_v\cdot r'_u & r'_v\cdot r'_v \\
    \end{pmatrix}
$$
The shape operator $S$ is $-dG$ (minus the differential of the Gauss map). It can be shown that $-dG=Bg^{-1}$.
What can be said about the eigenvalues of $B$? What can be said about the eigenvalues of $S$? Are they the same? Which of these eigenvalues are defined to be the principal curvatures of $M$? Is $\det B$ considered to be the Gauss curvature, or either $\det S$?   
There is alot of information in Do Carmo's book and in other sources, but it is not consistent with the above definitions, which got me very cofused.

I find the best way to resolve this kind of confusion is to take the perspective of abstract linear algebra: think of $B$ and $S$ as more than matrices. Really $B$ is a bilinear form on the tangent space $T_pM$, while $S$ is a linear operator $T_p M \to T_p M$. Without extra structure, it only makes sense to talk about eigenvalues and determinant of a linear operator; so the principal curvatures are the eigenvalues of $S$, and likewise $K = \det S.$ 
In a context where we have a preferred inner product, we often also talk about eigenvalues (or determinant, or...) of a bilinear form $b$; but what we mean by this is simply eigenvalues of the linear operator $L$ obtained from $b$ via the usual correspondence $b(X,Y) = g(L(X), Y).$ Since the second fundamental form and shape operator are related in this way (up to a sign), the eigenvalues will agree (up to sign).
