I have a surface embedded in 3-space. It is parametrized using a set of two Gaussian coordinates, $x^1$ and $x^2$. From the parametrization, we may determine ${\bf a}_1$ and ${\bf a}_2$, the covariant basis vectors for the tangent space, generally a non-orthonormal basis at every point in the manifold. Using the duality property, ${\bf a}_i \cdot {\bf a}^j = \delta_i^j$, we can find a contravariant basis for the tangent space as well.
Using the basis vectors in the tangent space, we may perform a cross product to determine a unit normal, $\bf n$ to the surface that sticks out of the tangent plane and exists in the ambient 3-space.
Finally, the curvature tensor $\bf b$ may be defined. We obtain its components as $b_{\alpha \beta} = {\bf a}_{\alpha, \beta} \cdot {\bf n}$, where the comma is understood to indicate partial differentiation of ${\bf a}_{\alpha}$ with respect to the coordinate $x^\beta$. With these covariant components, $\bf b$ is naturally defined on the contravariant basis: ${\bf b} = b_{\alpha\beta} {\bf a}^{\alpha} \otimes {\bf a}^{\beta}$.
My question: I have found the curvature tensor for a surface. It is diagonal. However, the contravariant basis is non-orthogonal. Is this possible? Generally, is it possible for any symmetric linear operator to be diagonal in a non-orthogonal basis?
I believe that the eigenbasis and corresponding eigenvalues of $\bf b$ will given me the directions of the lines of principle curvature and the principle curvatures. The lines of principle curvature for any surface occur at right angles to each other.
I was under the impression that a diagonal tensor was in its orthogonal eigenbasis. I suppose this isn't true?