Why are the Eigenvalues of the shape operator the principle curvatures? For now I would be content with understanding why the eigenvalues of the shape operator of a surface are the principle curvatures, let's call them $k_1,k_2$. 
Let $f: M \rightarrow S^2$ be the Gauss map of an oriented surface $M$ into the sphere. This map simply sends the unit normal vector at any point of our surface to it's point on the sphere, I like to think of this map sort of like a trippy compass.
The differential of this map is called the Shape Operator.
Given a point $x \in M$, the tangent plane at $x$ is denoted $T_xM$ is an inner product space. The shape operator can be defined as a linear operator on $T_xM$ by the equation:
$$ (S_x(v),w)=(df_x(v),w) \quad \text{for any $v,w \in T_x M.$} $$
Apparently, the equation is above is symmetric in $v$ and $w$, and thus the shape operator is a self-adjoint operator. Hm... So it's symmetric in $v$ and $w$, so $(df_x(v),w)=(df_x(w),v)$? Can somebody explain to me why that makes sense??
And then yeah, if somebody could help me understand why the eigen-values of this operator are the principle curvatures, i.e. the maximum and minimum values of possible curvatures as you depart from our given point, $x$, I'd really appreciate it.
Thanks!
 A: Let $\mathbf{x}(u,v)$ be a parametrisation of $M$ around $x$, then $\{\mathbf{x}_u,\mathbf{x}_v\}$ is a basis of $T_x M$. If $N$ is the normal vector on $M$, then $df_x(\mathbf{x_u}) = N_u$ and $df_x(\mathbf{x}_v)=N_v$. Derive $(N,\mathbf{x}_u) = 0$ w.r.t $v$ and $(N, \mathbf{x}_v)=0$ w.r.t. $u$:
$$
\begin{align*}
 (N_v, \mathbf{x}_u) + (N_, \mathbf{x}_{uv}) &= 0  \\
 (N_u, \mathbf{x}_v) + (N_, \mathbf{x}_{vu}) &= 0
\end{align*}
$$
Hence
$$
   (N_u, \mathbf{x}_v) = -(N,\mathbf{x}_{uv}) = (N_v,\mathbf{x}_{u}). 
$$
By linearity, it follows that $(df_x(v),w)= (df_x(w),v)$ for all $v$, $w \in T_x M$.
For your second question, let $\mathbf{e}_1$,  $\mathbf{e}_2 \in T_x M$ be the eigenvectors of $df_x$, i.e. the principal directions, and let $k_1$, $k_2$ be the corresponding eigenvalues. Let us say that $k_1 \geq k_2$. The normal curvature $k(v)$
in a direction $v=\cos \theta\, \mathbf{e}_1+\sin \theta\, \mathbf{e}_2$ is by definition $(df_x(v),v)$. We get
$$
\begin{align*}
 k(v)= (df_x(v),v) &= (k_1 \cos \theta \,\mathbf{e}_1+ k _2 \sin \theta \,\mathbf{e}_2,\cos \theta \mathbf{e}_1+\sin \theta \mathbf{e}_2)\\
&= k_1\cos^2\theta + k_2\sin^2 \theta.
\end{align*}
$$
In some references this formula is called Eulers formula for the normal curvatures. Finally we note that.
$$
  k_1 = k_1 \cos^2\theta + k_1\sin^2\theta \geq k_1\cos^2\theta + k_2\sin^2\theta \geq k_2\cos^2\theta +k_2\sin^2\theta = k_2,
$$
so $k_1\geq k(v) \geq k_2$. This shows that the eigenvalues of $df_x$ are the maximal en minimal normal curvatures.
