Matrix representation of shape operator Let $f$ be a parametrized surface $f: \Omega \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ and $N : \Omega \rightarrow Tf$ the Gauß map. Then the shape operator is defined as $L = -DN \circ Df^{-1}.$ Now the thing is that $Df$ is a $3 \times 2$ matrix, so I cannot invert this matrix easily. So how do I get a matrix representation for my shape operator?
If anything is unclear, please let me know. 
 A: The Gauss' map is $\Omega\to\Bbb R^3$ given by
$p\longmapsto \dfrac{\partial_1\times\partial_2}{||\partial_1\times\partial_2||}$,
where $\partial_1=\frac{\partial f}{\partial v}$ and $\partial_2=\frac{\partial f}{\partial w}$.
We can see this map as a vector field $N:\Sigma\to\Bbb R^3$, i.e. an assignation
$
\left(\begin{array}{c}
x\\
y\\
z
\end{array}\right)
\longmapsto 
\left(\begin{array}{c}
N_1\\
N_2\\
N_3
\end{array}\right)$. So its Jacobian is
$DN=
\left(\begin{array}{ccc}
\dfrac{\partial N_1}{\partial x}&\dfrac{\partial N_1}{\partial y}&\dfrac{\partial N_1}{\partial z}\\
\\
\dfrac{\partial N_2}{\partial x}&\dfrac{\partial N_2}{\partial y}&\dfrac{\partial N_2}{\partial z}\\
\\
\dfrac{\partial N_3}{\partial x}&\dfrac{\partial N_3}{\partial y}&\dfrac{\partial N_3}{\partial z}
\end{array}\right)$.
Now observe that -by the use of the Chain's Rule- one gets
$$DN(\partial_1)=
\left(\begin{array}{ccc}
\dfrac{\partial N_1}{\partial x}&\dfrac{\partial N_1}{\partial y}&\dfrac{\partial N_1}{\partial z}\\
\\
\dfrac{\partial N_2}{\partial x}&\dfrac{\partial N_2}{\partial y}&\dfrac{\partial N_2}{\partial z}\\
\\
\dfrac{\partial N_3}{\partial x}&\dfrac{\partial N_3}{\partial y}&\dfrac{\partial N_3}{\partial z}
\end{array}\right)
\left(\begin{array}{c}
\dfrac{\partial x}{\partial v}\\
\\
\dfrac{\partial y}{\partial v}\\
\\
\dfrac{\partial z}{\partial v}
\end{array}\right)
=
\left(\begin{array}{c}
\dfrac{\partial N_1}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial N_1}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial N_1}{\partial z}\dfrac{\partial z}{\partial v}\\
\\
\dfrac{\partial N_2}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial N_2}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial N_2}{\partial z}\dfrac{\partial z}{\partial v}\\
\\
\dfrac{\partial N_3}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial N_3}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial N_3}{\partial z}\dfrac{\partial z}{\partial v}\
\end{array}\right)
=
\left(\begin{array}{c}
\dfrac{\partial N_1}{\partial v}\\
\\
\dfrac{\partial N_2}{\partial v}\\
\\
\dfrac{\partial N_3}{\partial v}
\end{array}\right)
=
\dfrac{\partial N}{\partial v}
$$
and similarly
$$DN(\partial_2)
=
\left(\begin{array}{c}
\dfrac{\partial N_1}{\partial w}\\
\\
\dfrac{\partial N_2}{\partial w}\\
\\
\dfrac{\partial N_3}{\partial w}
\end{array}\right)
=
\dfrac{\partial N}{\partial w}
$$
In another hand we have $||N||=1$, then $DN\bullet N=0$, that is, $DN$ is orthogonal to $T_p\Sigma$.
Hence we can construct
$$L:T_p\Sigma\longrightarrow T_p\Sigma$$
$$\partial_1\longmapsto L(\partial_1)=\frac{\partial N}{\partial v}$$
$$\partial_2\longmapsto L(\partial_2)= \frac{\partial N}{\partial w}$$
which is a basis change in case that $\frac{\partial N}{\partial v}$ and $\frac{\partial N}{\partial w}$ 
are linearly independent. 
