Consider, for instance, the catenoid given by $x^2+y^2 = \cosh^2 z$, and suppose that we want to find the principal curvatures and directions at some point $p=(x,y,z)$ of the surface. Of course, we can do this by parametrizing the catenoid as $$(u,v) \mapsto (\cosh u \cos v, \cosh u \sin v, u),$$ and then computing the Gauss map etc.
However, I was wondering if this is possible to do without making reference to a parametrization. Since the catenoid is a level surface $f^{-1}(\{0\})$ of $$ f(x,y,z) = x^2+y^2 -\cosh^2 z, $$ we can find a Gauss map in Cartesian coordinates at the point $p=(x,y,z)$ by computing $$ N(p) = \frac{\nabla f(p)}{|\nabla f(p)|} = \frac{1}{\cosh^2 z}(x,y,-\cosh z\sinh z). $$ Furthermore, we can also (after a somewhat tedious, but not too difficult computation) see that the differential $\mathrm dN_p$ is given by $$ \mathrm dN_p = \frac{1}{\cosh^2 z} \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ -2x\tanh z & -2y\tanh z & -1 \end{pmatrix}. $$ Since this is a triangular matrix, we immediately spot the eigenvalues, namely $\lambda_1 = \cosh^{-2} z$ of multiplicity $2$, and $\lambda_2 = -\cosh^{-2} z$ of multiplicity $1$.
It should be expected that there are only two distinct eigenvalues, since there can only be two principal curvatures, and the result also agrees with the fact that the eigenvalues of the differential $\mathrm dN_p$ turn out to be $\pm \cosh^{-2} u$ when computed in the $(u,v)$ coordinates that I mentioned in the first paragraph.
The eigenvector corresponding to $\lambda_2$ is $(0,0,1),$ which is expected since the meridians of the catenoid are clearly lines of curvature. However, assuming that I did everything correctly, the eigenspace corresponding to $\lambda_1$ is two-dimensional, and spanned by $$ (1, 0, -x\tanh z) \quad \text{and} \quad (0,1,-y\tanh z), $$ which, as far as I can see, does not agree with the fact that there should only be two principal directions.
How am I to interpret this?
