Differential linear system limit Suppose that $u\in \mathcal{L}(\mathbb{R}^n)$ and 
$$\forall x\not\in \ker u,\quad (x,u(x))<0,$$
where $(.,.)$ is a scalar product on $\mathbb{R}^n$. Then consider 
$$ \left\{ \begin{array}{rcl}x'(t)&=&u(x(t))\\
x(0)&=& x_0 \end{array} \right.$$
I know that $\mathbb{R}^n$ is the direct sum of the kernel of $u$ and the range of $u$. I've to show that $x(t)$ converges when $t\to +\infty$.
My idea : if $x_0\in \ker u$ then $x(t)=x_0$ for all $t$. 
We know that $t\mapsto\left\| x(t)\right\|^2$ is non-increasing. 
I think that if $x_0$ is in the range of $u$ then $x(t)\to 0$. To prove that I want to say (using compacity) that $(x(t),u(x(t)))\le \alpha <0$ to conclude.
If $\ker u = \{ 0 \}$, I can conclude since $t\mapsto\left\| x(t)\right\|^2$ admits a limit when $t\to +\infty$. Then I suppose that this limit is $>0$. Then $x(t)$ takes its value in the compact $\left\{ Y\in \mathbb{R}^n,\ 0<l\le \left\| Y \right\| \le \left\| x_0 \right\| \right\}$ and in this compact we know that $(Y,u(Y))<0$ so $(x(t),u(x(t))\le \alpha <0$. So $0\le \left\| x(t) \right\|^2-\left\| x_0 \right\|^2\le \alpha t$ which is impossible.  
But what can I do if $\ker u\neq \{0 \}$ ? 
Any ideas?
 A: As has been stated in the text of the question, $\Bbb R^n$ is the direct sum of $\ker u$ and $\text{Im} u$, where $\text{Im} u$ is the image or range of $u$, that is $\text{Im} u = u(\Bbb R^n)$.  For any vector $x \in \Bbb R^n$, we may thus uniquely decompose $x$ as $x = y + z$, 
where $u(y) = 0$ and $z = u(w)$ for some $w \in \Bbb R^n$, and we observe that application of the operator $u$ respects this decomposition; i.e. $\ker u$ and $\text{Im} u$ are invariant subspaces of the operator $u$.  Indeed, we see that $u(y) = 0 \in \ker u$ and  obviously $u(z) = u(u(w)) \in \text{Im} u$; thus $u(\ker u) \subseteq \ker u$ and $u(\text{Im} u) \subseteq \text{Im} u$.  For any solution $x(t)$ to the differential equation
$x'(t) = u(x(t)), \; x(0) = x_0, \tag{1}$
we may thus uniquely write
$x(t) = y(t) + z(t) \tag{2}$
with $y(t) \in \ker(u)$ and $z(t) \in \text{Im} u$ for all $t \in \Bbb R$.  Furthermore $y(t)$ and $z(t)$ are both differentiable functions of $t$; this follows from the fact that $\ker u$ and $\text{Im} u$ are closed subspaces of $\Bbb R^n$ and the projections $P_{\ker u}$, $P_{\text{Im u}}$ are constant, continuous linear maps (all vector spaces involved being of finite dimension), so that we indeed have
$y'(t) = (P_{\ker u} (x(t)))' = P_{\ker u}(x'(t)), \tag{3}$
$z'(t) = (P_{\text{Im} u} (x(t)))' = P_{\text{Im} u}(x'(t)). \tag{4}$
(3) and (4) also show that
$y'(t) \in \ker u \tag{5}$
and
$z'(t) \in \text{Im} u. \tag{6}$
It follows from these considerations that, using (2), (1) yields
$y'(t) + z'(t) = x'(t) = u(x(t)) = u(y(t)) + u(z(t)) = u(z(t));\, \tag{7}$
and using (5) and (6), along with the invariance of $\ker u$ and $\text{Im} u$ under $u$ now allows us to conclude that
$y'(t) = 0 \tag{8}$
and
$z'(t) = u(z(t)) \tag{9}$
by virtue of the direct sum decomposition $\Bbb R^n = \ker u \oplus \text{Im} u$.  (8) implies that $y(t) = y_0$ is constant; as for $z(t)$, we see from (9) that
$(z(t), z(t))' = 2(z(t), z'(t)) = 2(z(t), u(z(t)) < 0 \tag{10}$
by virtue of the hypothesis $(x, u(x)) < 0$ for all $x \in \Bbb R^n$, $x \notin \ker u$.  This hypothesis in fact implies a slightly sharper statement which holds on $\text{Im} u$ and is better adapted to the present purposes.  Since the unit sphere $S$ in $\text{Im} u$ is compact, and $(e, u(e)) < 0$ for all unit vectors $e \in S$, there is an $\alpha < 0$ with $(e, u(e)) < \alpha$ for $e \in S$.  But then any nonzero $z \in \text{Im} u$ satisfies
$(z, u(z)) = (\Vert z \Vert e_z, u(\Vert z \Vert e_z)) = \Vert z \Vert^2 (e_z, u(e_z)) < \alpha \Vert z \Vert^2, \tag{11}$
where $e_z = z / \Vert z \Vert$ is the unit vector in the direction of $z$; here of course $\Vert z \Vert = (z, z)^{\frac{1}{2}}$.  Applying (11) to (10) we find
$(z(t), z(t))' = 2(z(t), z'(t)) = 2(z(t), u(z(t)) < 2\alpha (z(t), z(t)) \tag{12}$
for $z(t) \ne 0$.  But then
$(\ln (z(t), z(t)))' < 2\alpha \tag{13}$
which may be integrated:
$\ln (\dfrac{(z(t), z(t))}{(z(0), z(0))} = \ln ((z(t), z(t))) - ((z(0), z(0)))$
$= \int_0^t (\ln (z(s), z(s)))' ds < 2\int_0^t \alpha ds = 2\alpha t, \tag{14}$
that is, after some standard algebraic and logarithmic maneuvering, 
$\ln (\dfrac{\Vert z(t) \Vert}{\Vert z(0) \Vert}) < \alpha t, \tag{15}$
which of course implies, in the usual manner, and finally
$\Vert z(t) \Vert < \Vert z(0) \Vert e^{\alpha t}; \tag{16}$
since $\alpha < 0$, (16) shows that $z(t) \to 0$ as $t \to \infty$.  Thus $x(t) = y(t) + z(t) \to y(t) = y_0$.
Hope this helps!  Cheerio,
and as always, 
Fiat Lux!!!
