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In the Wikipedia definition of principal curvature, it says that the principal curvatures are the eigenvalues of the symmetric matrix $\left[I\!I_{ij}\right] = \begin{bmatrix} I\!I(X_1,X_1)&I\!I(X_1,X_2)\\ I\!I(X_2,X_1)&I\!I(X_2,X_2) \end{bmatrix}.$

By $I\!I(X_1,X_2)$ does this mean that we evaluate $I\!I$ which contains differentials at the vector $X_1$ and $X_2$? So what I mean is, in second fundamental form $I\!I$, there only is differentials with scalar multiplication and their sums. So I am wondering that by $I\!I(X_1,X_2)$, this is similar to evaluating $dx(v)$ where $dx$ is differential and $v$ is vector.

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As far as I understand your question, you are asking if the components $$ II_{i j} = II(X_i,X_j) $$ are computed by evaluating a matrix $II$ on vectors $X_i$ and $X_j$.

Answer. Yes, they are.

Caveat. In the Wikipedia's article that you likely to have in mind $(X_1, X_2)$ is assumed to be an orthonormal basis. In this case one can say that the principal curvatures are "eigenvalues of the second fundamental form", but this is just because in the case of an orthonormal basis the second fundamental equals to the shape operator.

More precisely, the principal curvatures are eigenvalues of the shape operator which is the differential of the Gauss map.

So, indeed you differentiate some data, form a matrix, and then calculate the components of this matrix in a basis that you like.

I would recommend these lecture notes of Greg Galloway for those who want to get the right definitions and a number of worked examples.

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