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In my differential geometry course the professor has defined Gaussian curvature to be the determinant of the shape operator, $$K=\det(S(p))$$ However most books that I have been following along with define it to be the product of the principal curvatures, $$K=\kappa_1\kappa_2$$ I believe that these are equivalent definitions but have been having a hard time proving this fact, and would greatly appreciate if somebody could provide a proof of this.

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This is just a linear algebra fact.

The shape operator is known to be symmetric, so by the spectral theorem it has real eigenvalues, so there is an orthogonal basis in which this operator has a diagonal matrix, and the determinant of a diagonal matrix is easy to compute.

Of course, the validity of my "proof" depends on how you define the principal curvatures, but I rely here on the treatment given in the book of Theodore Shifrin "Differential Geometry. A first Course in Curves and Surfaces", where one can find a very accessible exposition.

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