Finding the eigenvalues of differential Normal mapping of the Hyperbolic Paraboloid Consider the hyperbolic paraboloid $S$, $S=\{(x,y,z)\in\mathbb{R}^3 |z=x^2 -y^2\}$
Study the second fundamental form of this surface at $(0,0,0)$ and show that it is not a semi-definite bilinear form, In other words, show that its Gauss Curvature is negative at this point.
My solution:
First I define $f$ as $f(x,y,z)=-x^2+y^2+z$ and see that $\bigtriangledown f(x,y,z)=(-2x,2y,1)$ for all $(x,y,z)$ to define the Gauss map $N(p):S\rightarrow T_{p}S$ by 
$N(p)=\displaystyle\frac{\bigtriangledown f(x,y,z)}{|\bigtriangledown f(x,y,z)|}$ $\forall p=(x,y,z)\in S$
where $T_{p}S$ is the tangent plane of $S$ at $p$
so, the Gauss curvature by definition is the product of two eigenvalues of  the differential $dN(p)$ that is given by the matrix
$DN_{(x,y,z)}=\frac{1}{|\bigtriangledown f(x,y,z)|}\left[\begin{array}{ccc}
2 &  0  & 0 \\
0 & -2  & 0 \\
0 &  0  & 0
\end{array}
\right]$
and so $dN_{(x,y,z)}.(v_{1},v_{2},v_{3})=(-2v_{1},2v_{2},0)$, where $v\in T_{p}. S$
I checked that the eigenvalues of $DN$ are $\{2,-2,0\}$ but by the fact that the operator $dN(p)$ is self-adjoint it must have only two eigenvalues, so how can I choose these two? By the exercise, I assume that the curvature is $-2.2=-4$ but I don't know the argumentation for choosing these values.
I found an example  asking this same thing, but $S$ is a cylinder, in this case 
$N(x,y,z)=\frac{1}{r}(x,y,0)$, $r$ is the radius of $S$
and 
$dN_{(x,y,z)}(v_{1},v_{2},v_{3})=\frac{1}{r}(v_{1},v_{2},0)$
the matrix that defines $dN$ is 
$DN=(1/r)\left[\begin{array}{ccc}
1 &  0  & 0 \\
0 & 1  & 0 \\
0 &  0  & 0
\end{array} \right]$
and the exercise concludes that the eigenvalues $k_{1}=1/r$ and $k_{2}=0$, in that way we can see there are only two options, but in my exercise there are 3.
I know this is kind of linear algebra exercise but i would appreciate any help of geometers and algebraists.
 A: You can use the Monge chart $X:\Bbb R^2 \to S$ given by $$X(u,v) = (u,v,u^2 - v^2).$$One has $X_u(u,v) = (1,0,2u)$ and $X_v(u,v) = (0,1,-2v)$, and so the First Fundamental Form is given by $$E(u,v) = 1+4u^2, \quad F(u,v) = -4uv, \quad G(u,v) = 1+4v^2.$$The Gauss map is then given by $$\begin{align}N(X(u,v)) &= \frac{X_u(u,v) \times X_v(u,v)}{\|X_u(u,v) \times X_v(u,v)\|} \\ &= \frac{1}{\sqrt{1+4u^2+4v^2}} \begin{vmatrix}e_1 & e_2 & e_3 \\ 1 & 0 & 2u \\ 0 & 1 & -2v \end{vmatrix} \\ &= \frac{(-2u, 2v, 1)}{\sqrt{1+4u^2+4v^2}}.\end{align}$$Computing $X_{uu}(u,v) = (0,0,2)$, $X_{uv}(u,v) = 0$ and $X_{vv}(u,v) = (0,0,-2)$, we can get the coefficients of the Second Fundamental Form: $$e(u,v) = \frac{2}{\sqrt{1+4u^2+4v^2}}, \quad f(u,v) = 0, \quad g(u,v) = \frac{-2}{\sqrt{1+4u^2+4v^2}}.$$With this, the Gaussian Curvature of $S$ at the point $X(u,v)$ is given by $$K(X(u,v)) = \frac{e(u,v)g(u,v)-f(u,v)^2}{E(u,v)G(u,v)-F(u,v)^2} = \frac{-4}{(1+4u^2+4v^2)^2}.$$So $$K(0,0,0) = K(X(0,0)) = -4 < 0,$$and this tells us that the Second Fundamental Form is indefinite (assumes positive and negative values). As a bonus, one can compute the Mean Curvature as $$\begin{align} H(X(u,v)) &= \frac{1}{2}\frac{E(u,v)g(u,v) - 2F(u,v)f(u,v) + G(u,v)e(u,v)}{E(u,v)G(u,v)-F(u,v)^2} \\ &= \frac{-2(1+4u^2) + 2(1+4v^2)}{2(1+4u^2+4v^2)^{3/2}} \\ &= \frac{-4u^2 +4v^2}{(1+4u^2+4v^2)^{3/2}}.\end{align}$$For the proof of these coordinate expressions for $K$ and $H$ see do Carmo, for example.
A: I will just leave this alternative answer here for completeness. This is based on the comments to Ivo's answer.
We know that the hyperbolic paraboloid is the graph of a function $f(x, y) = x^2 - y^2$, hence an orientable surface with unit normal field
$$
N = \frac{1}{\sqrt{1+|\nabla f|^2}}(-f_x, -f_y, 1)
$$
where the subindex denotes partial differentiation. Then, for our function, 
$$
N(x, y, z) = \frac{1}{\sqrt{1+4(x^2 + y^2)}}(-2x, 2y, 1).
$$
A parametrization of this surface is given by $X: \Bbb{R}^2 \longrightarrow \Bbb{R}^3$ where 
$$
X(u, v) = (u, v, f(u, v)).
$$
Then a basis for the tangent space $T_pS$ at $p$ is given by $\{X_u(u,v), X_v(u, v)\}$. But
$$
X_u(u, v) = (1, 0, f_u(u, v)), \quad X_v(u, v) = (0, 1, f_v(u,v)).
$$
Then, at the point $(0, 0, 0)$ the tangent space has basis $\{(1, 0, 0), (0, 1, 0)\}$. Hence the matrix of $(dN)_{(0, 0, 0)}$ when seen as a map from $T_{(0, 0, 0)}S$ to $T_{(0, 0, 0)}S$ is
$$
\begin{bmatrix}
-2 & 0 \\
0 & 2
\end{bmatrix}.
$$
Now, 
$$
K((0, 0, 0)) = -4, \quad H((0, 0, 0)) = 0.
$$
Hence, at the origin the hyperbolic paraboloid has negative Gauss curvature and zero mean curvature.
