I want to compute the shape operator $A$ of the unit sphere $\mathbb S^2$ which is given by $$A=-I^{-1 }II$$ where $I^{-1}$ is the inverse of the first fundamental form $I$ and $II$ being the second fundamental form. From the parametrization $$X(\theta,\phi)=\bigl(\sin(\theta)\cos(\phi), \sin(\theta)\sin(\phi, \cos(\theta)\bigr)^T$$ one obtains the first fundamental form and its inverse: $$I=\begin{pmatrix}1 & 0 \\ 0 & \sin^2(\theta)\end{pmatrix}\quad\quad I^{-1}=\frac{1}{sin^2{\theta}} \, \begin{pmatrix}\sin^2(\theta) & 0 \\ 0 & 1 \end{pmatrix}$$
and by the computation $N=X_{\theta} \times X_{\phi}$ the normal vector $N=\begin{pmatrix}\sin^2(\theta)\cos(\phi)\\ \sin^2(\theta)\sin(\phi)\\\cos(\theta)sin(\theta)\end{pmatrix}$.
The coefficients of $II$ are given by $II_{ij}=X_{ij} \cdot N$, thus , if I didn't miscalculate$$II=\begin{pmatrix}-\sin(\theta) & 0 \\0 &-sin^3(\theta) \end{pmatrix}.$$
Then finally the shape operator should be $$A=-I^{-1}II=\begin{pmatrix}\sin(\theta) & 0 \\ 0 & \sin(\theta)\end{pmatrix}.$$
The mean curvature $H$ of the sphere can then be calculated by $$H = tr(A)=2\sin(\theta).$$
But shouldn't the mean curvature of the unit sphere be $1$? Where did I miscalculate?
You need to take a normal vector with unit length. So the correct choice is
$$ N = \frac{X_\theta \times X_\phi}{|X_\theta \times X_\phi|},$$
then you should get
$$N = \bigl(\sin(\theta)\cos(\phi), \sin(\theta)\sin(\phi, \cos(\theta)\bigr)^T$$
and from there you find $H = 2$.