Show that $T_p\nu$ is symmetric with eigenvalues $0$ and $1$ Im stuck in this exercise on page 278 of Analysis II of Amann and Escher

Suppose $M:=\{(x,y,z)\in\Bbb R^3: x^2+y^2=1\}$ and $$\nu:M\to\mathrm S^2,\quad (x,y,z)\mapsto (x,y,0)$$ Show that $\nu\in C^\infty(M,\mathrm S^2)$ and that $T_p\nu$ is symmetric with eingenvalues $0$ and $1$.

(Here $T_p\nu$ is the tangential of $\nu$ around $p$, and $i_M$ is the canonical injection $i_M:M\to\Bbb R^n$ for some $M\subset\Bbb R^n$.)
My work so far: clearly $\nu$ is continuous, we want to show now that it is also smooth. Let 
$$
\psi_\pm:M\setminus F_{\pm}\to(-\pi,\pi)\times\Bbb R,\quad (x,y,z)\mapsto(\arg(\mp x\mp iy),z)
$$
That is $\{\psi_-,\psi_+\}$ is an atlas of $M$. Now define
$$
\varphi_-:\mathrm S^2\setminus H_2\to (-\pi,\pi)\times (0,\pi),\quad (x,y,z)\mapsto \left(\arg\left(\frac{x+iy}{\sin(\arccos z)}\right),\arccos z\right)
$$
where $H_2:=(-\infty,0]\times\{0\}\times \Bbb R$ and $\arg$ is the principal argument, defined in $(-\pi,\pi)$. Its easy to see that $\varphi_-$ is a chart and together with $\varphi^*:=\varphi_-\circ\sigma$, where $\sigma:=(123)$ is a permutation, they are an atlas of $\mathrm S^2$.
Now observe that, under the action of $\nu$, $\psi_-$ and $\varphi_-$ map the same territories and 
$$
\nu_{\psi_-,\varphi_-}(\theta,r)=(\varphi_-\circ\nu)(\cos\theta,\sin\theta,r)=\varphi_-(\cos\theta,\sin\theta,0)=(\theta,\pi/2)\\\implies[\partial(\nu_{\psi_-,\varphi_-})(\theta,r)]=\begin{bmatrix}1&0\\0&0\end{bmatrix}
$$
A similar result can be shown for $\psi_+$ and $\varphi_+(x,y,z):=\varphi_-(-x,-y,z)$. Thus $\nu\in C^\infty(M,\mathrm S^2)$ as expected. 
Now to define $T_p\nu$, for some $p\in M$, we need to use the functional equation 
$$
T_{\psi(p)}\nu_{\psi,\varphi}\circ T_p\psi=T_{\nu(p)}\varphi\circ T_p\nu
$$
where $\psi$ is a chart around $p\in M$ and $\varphi$ is a chart around $\nu(p)\in\mathrm S^2$. Also we knows that $T_{x_0}g=(g(x_0),\partial g(x_0))$ for a parametrization $g$ and it must hold that $(T_{\psi(p)}g)(T_p\psi)=i_{T_pM}$ for $g=i_M\circ \psi^{-1}$. Hence the tangential part $A_\psi$ of the chart $\psi$ is defined by the equation $\partial \psi^{-1}(x_0)A_\psi=I_n$, for $M\subset\Bbb R^n$.
Observe that
$$
\psi_\pm^{-1}(\theta,r)=(\mp \cos\theta,\mp \sin\theta,r)\implies[\partial\psi^{-1}_\pm(\theta,r)]=\begin{bmatrix}\pm\sin\theta&0\\\mp\cos\theta&0\\0&1\end{bmatrix}\implies A_\psi=\begin{bmatrix}???\end{bmatrix}
$$
But Im unable to find an $A_\psi$ that holds my hypothesis when $\cos\theta=0$ or $\sin\theta=0$. Probably Im doing an overkill or something but I dont know how to continue. Some help will be appreciated, thank you.
 A: Ok, I solved it. It was not as hard as I thought, I just needed a little more of thinking.
Define $g_\pm:=i_M\circ \psi_\pm^{-1}$ and $V_\pm:=\psi_\pm(M)$, then
$$
T_{\psi(p)}g:T_{\psi(p)}V\to T_p\Bbb R^2\quad\text{and}\quad T_p\psi:T_pM\to T_{\psi(p)}V\tag1
$$
for some $p:=(x,y,z)\in M$. This together with $T_{\psi(p)}g\circ T_p\psi=i_{T_pM}$ means that the tangential part of $T_p\psi$ is just the transpose of the tangential part of $T_{\psi(p)}g$, that is, $A_{\psi_\pm}=[\partial\psi^{-1}_\pm(\theta,r)]^T$. Also observe that
$$
T_{\psi(p)}\nu_{\psi,\varphi}=\big((\varphi\circ \nu)(p),\partial\nu_{\psi,\varphi}(\psi(p))\big)\tag2
$$
because $\nu_{\psi,\varphi}$ is a function between open sets in euclidean spaces. Also
$$
\varphi^{-1}_\pm(\theta,\eta)=(\mp\cos\theta\sin\eta,\mp\sin\theta\sin\eta,\cos\eta)\\\implies[\partial\varphi^{-1}_\pm(\theta,\eta)]=\begin{bmatrix}\pm\sin\theta\sin\eta&\mp\cos\theta\cos\eta\\\mp\cos\theta\sin\eta&\mp\sin\theta\cos\eta\\0&-\sin\eta\end{bmatrix}\tag3
$$
and for such $p:=(x,y,z)$
$$
(\varphi_\pm\circ\nu)(p)=\varphi_\pm(x,y,0)=(\arg(\mp x\mp iy),\pi/2)\quad\text{and}\quad \psi_\pm(x,y,z)=(\arg(\mp x\mp iy),z)\tag4
$$
Thus the tangent part of $T_p\nu$ have the form
$$
\begin{bmatrix}\sin\theta&0\\\mp\cos\theta&0\\0&-1\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}\sin\theta&\mp\cos\theta&0\\0&0&1\end{bmatrix}=\begin{bmatrix}\sin\theta&0\\\mp\cos\theta&0\\0&0\end{bmatrix}\begin{bmatrix}\sin\theta&\mp\cos\theta&0\\0&0&1\end{bmatrix}\\=\begin{bmatrix}\sin^2\theta& \mp\sin\theta\cos\theta&0\\\mp \sin\theta\cos\theta&\cos^2\theta&0\\0&0&0\end{bmatrix}\tag5
$$
And according to the expected, the tangential part of $T_p\nu$ is symmetric, with characteristic polynomial
$$
q(x):=(x-\sin^2\theta)(x-\cos^2\theta)x-\sin^2\theta\cos^2\theta x=x(x^2-x)=x^2(x-1)\tag6
$$
so the eigenvalues of the tangential part of $T_p\nu$ are $0$ and $1$.
