Question about an O.D.E I have this theorem:

Suppose that $U$ is a neighborhood of $\theta$ in a Hilbert space $H$
  and that $f\in C^2(U,\mathbb{R}^1)$. Assume that $\theta$ is the only
  critical point of $f$ and that $A=d^2f(\theta)$ ($A$ is a self adjoint operator) with kernel $B$. If
  $0$ is either an isolated point of the spectrum $\sigma(A)$ or not in
  $\sigma(A)$, then there exist a ball $B_\delta$, and a $C^1$ mapping
  $h:B_\delta\cap B\to B^\perp$ such that $$
> f\circ\phi(z+y)=\frac12(Az,z)+f(h(y)+y),\quad\forall x\in B_\delta, $$
  where $y=P_Nx$, $z=P_{B^\perp}x$, and $P_B$ is the orthogonal
  projection onto the subspace $B$.

Let $\eta$ be the flot defined by the following o.d.e
$$
\begin{cases}
\displaystyle\eta '(s)=- \frac{A\eta(s)}{||A\eta (s)||}\\
\eta(0)=u
\end{cases}
$$
we have that A is a continuous operator, and that $A_{|_{N}}$ is invertible
if $u \in N$ why $\eta(t,u)\in N$ ?
Thank you.
 A: From the quoted text provided by Vrouvrou, we have that $A = d^2f(\theta)$ for some isolated critical point $\theta$ of $f \in C^2(U, \Bbb R^1)$, and (here we have a "critical" fact) $A$ is self-adjoint, that is, $A = A^T$. 
Once it is granted that $A$ is continuous and self-adjoint, the desired result follows immediately; the origins of $A$ as the second derivative of $d^2f(\theta)$ at a critical point $\theta$ may be left behind.
First, we recall the following well-known fact:  with $A$ as above, and $B = \ker A$, we have $A(B^\bot) \subset B^\bot$, since $y \in B^\bot$ if and only if $\langle x, y \rangle = 0$ for any $x \in B$ .  Then for such $x$ and $y$, $\langle x, Ay \rangle = \langle A^Tx, y \rangle = \langle Ax, y \rangle = 0$, since $Ax = 0$, showing $A(B^\bot) \in B^\bot$.  $B$ is also closed since it is the kernel of the continuous operator $A$.
We can use this result to address the problem at hand.  First, note that for $0 \ne \eta \notin B$, $A\eta \ne 0$, for $A\eta = 0$ would imply $\eta \in B = \ker A$, contradicting $\eta \notin B$.  Thus $\Vert A \eta \Vert \ne 0$, so the vector field $-(1 / \Vert A \eta \Vert) A \eta$ is well defined on the open set $H - B$.  If $\eta(s, u)$ satisfies the equation
$\eta'(s, u) = -\dfrac{1}{\Vert A \eta \Vert}A\eta(s, u) \tag{1}$
with
$0 \ne \eta(0, u) = u \in B^\bot, \tag{2}$
then for any constant $x \in B$ we have
$\dfrac{d}{ds}\langle x, \eta(s, u) \rangle = \langle x, \eta'(s, u) \rangle =  \dfrac{-1}{\Vert A \eta \Vert}\langle x, A\eta(s, u) \rangle = \dfrac{-1}{\Vert A \eta \Vert}  \langle Ax, \eta(s, u) \rangle = 0, \tag{3}$
since $Ax$ vanishes by virtue of $x \in B = \ker A$.  This shows that $\langle x, \eta(s, u) \rangle$ is constant along the integral curves $\eta(s, u)$ of (1).  Since $u \in B^\bot$, $\langle x, u \rangle = 0$ for all $x \in B$.  Hence $\langle x, \eta(u, s) \rangle = 0$ as well, showing $\eta(s, u) \in B^\bot$ for all $s$ for which it is defined.  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
