Matrix of shape operator of the sphere Problem: Let $M$ be a sphere of radius $a$ in $\mathbb{R}^3$, defined by $$ x^2 + y^2 + z^2 = a^2. $$ Parametrize the sphere using spherical coordinates $$x = a \sin \phi \cos \theta \\ y = a \sin \phi \sin \theta \\ z = a \cos \phi $$ where $0 \leq \phi < \pi, 0 \leq \theta < 2 \pi$. 
Then for each $p \in M, e_1 = \partial/\partial\phi$ and $e_2 = \partial/ \partial \theta$ is a basis for the tangent space $T_p M$ for $p \in M$. Let $N_p$ be the unit outward normal vector at $p$ on the sphere.
Find the matrix of the shape operator of the sphere with respect to the basis $e_1, e_2$. 
Attempt: I need to calculate $L(e_j) = - D_{e_j} N$ and then find the matrix $[a^{i}_{j}]$ such that $L(e_j) = \sum a^{i}_{j} e_i. $ Here $L$ denotes the shape operator. 
I know that in cartesian coordinates, a normal unit vector is $N = \frac{1}{a} (x, y, z)$. But do I have to write this first in the $(\theta, \phi)$ basis?
I would calculate e.g. $$ D_{\partial_{\phi}} \frac{1}{a} (a \sin \phi \cos \theta, a \sin \phi \sin \theta, a \cos \phi) \\ = ( \cos \phi \cos \theta, \cos \phi \sin \theta, - \sin \phi). $$ But I need to write this as a linear combination of the basis vectors $\partial_{\phi}$ and $\partial_{\theta}$ to find the matrix representation? 
 A: I guess that this exercise is from the book Differential Geometry of Loring Tu. It is misleading that in the body of the exercise appear the map of the change from polar to Cartesian coordinates, or at least I didn't use this map to solve the exercise.
This is what I did: define the map $f(x,y,z):=\sqrt{x^2+y^2+z^2}-a$, then $f^{-1}(0)=a\mathbb{S}^2$, and as zero is a regular value of $f$ then we find that $\nabla f(x,y,z)=\tfrac1{a}(x,y,z)$ is the outward pointing unit normal vector to every $(x,y,z)\in a\mathbb{S}^2$, therefore
$$
N=\frac1{a}\left(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}\right)
$$
and from the definition of the shape operator you get
$$
L(e_1)=-\frac1{a}\left(\frac{\partial x}{\partial \phi }\frac{\partial}{\partial x}+\frac{\partial y}{\partial \phi }\frac{\partial}{\partial y}+\frac{\partial z}{\partial \phi }\frac{\partial}{\partial z}\right)=-\frac1{a}\frac{\partial}{\partial \phi }
$$
where the last identity follow easily applying the coordinate functions $x,y,z$ to the right of both expressions for the vector field. Therefore in basis $e_1,e_2$ we find that $L=-\tfrac1a\left[\begin{smallmatrix}1&0\\0& 1\end{smallmatrix}\right]$.∎
