Linear ODE induces new ODE on the sphere: stability questions. Assume we have a time-invariant linear ODE that arose from a non-linear ODE by linearizing at some point $U_1=(u_1,v_1,w_1)$ and that has the following form:
$$
\dot{\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}}=\underbrace{\begin{pmatrix}0 & 1 & 0\\-f'(u_1) & -c & 1\\0 & 0 & 0\end{pmatrix}}_{=:A}\cdot\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix},
$$
where $f'(u_1)<0, c>0$. Then the matrix has eigenvalues
$$
\lambda_1=-\frac{c}{2}-\sqrt{\frac{c^2}{4}-f'(u_1)},~~~~~\lambda_2=-\frac{c}{2}+\sqrt{\frac{c^2}{4}-f'(u_1)},~~~~~\lambda_3=0.
$$
Since $f'(u_1)<0$, $\lambda_1<0$ and $\lambda_2>0$. 
In terms of stability theory, this means that unstable subspace, stable subspace and centre subspace are non-empty. Moreover, as far as I know, the centre subspace here is stable (in the Lyapunov sense), too. So we can say that we have one unstable subspace and two stable ones. Let these be spans of the eigenvectors $X_1$ (unstable), $X_2$ and $X_3$.
Now, project this equation down to the unit sphere $S^2$ by defining
$$
s(t)=\frac{x(t)}{\lvert x(t)\rvert}.
$$
Then, by the chain rule, we get
$$
\dot{s}(t)=h(s(t)),~~~\text{where }h(s)=(A-s^TAs\cdot I)s.
$$
Now, there are some statements that I do not understand.
(1.) By homogeneity, the original equation induces a differential equation on $S^2$ obtained by identifying opposite points on $S^2$.
I do not understand this claim, in particular how the new ODE looks like and what it does on the sphere.
(2.) The new ODE on $S^2$ has two saddle points, two attracting points and two repelling points. These come from the eigenspaces. Let $X_2$ be the eigenvector that gives the saddle. Set $C=\text{span}\left\{X_2,X_3\right\}\cap S^2$. 
I do not understand where the two saddle points, two attracting points and two saddle points come from and how they are connected with the unstable subspace, stable subspace and centre subspace of the original ODE.
(3.) For the new ODE on $S^2$, the two points contained in $\text{span}\left\{X_1\right\}\cap S^2$ are attractors.
I do not understand why these two points are attractors for the new ODE on the sphere. That might be connected with my problem that I do not understand what the ODE on the sphere does and how it works on the sphere.
Maybe you can explain me this three statements. 
 A: After a discussion with the author of the OP, I write some explanations in the case where the "projection on the unit sphere" means intersection subspaces with it and renormalizion of vectors.
$(1)$ Let $s(t):=\frac{x(t)}{|x(t)|}$. Indeed, $|s(t)|=1$ (thus $s:\mathbb{R}\rightarrow\mathbb{S}^2$) and we have
\begin{align*}
\frac{\mathrm{d}s}{\mathrm{d}t} & = \frac{1}{|x(t)|}\frac{\mathrm{d}x}{\mathrm{d}t}-\frac{x(t)}{|x(t)|^3}\left(x(t)\cdot\frac{\mathrm{d}x}{\mathrm{d}t}\right) \\
 & = \frac{1}{|x(t)|}Ax(t)-\frac{x(t)}{|x(t)|^3}\left(x(t)\cdot Ax(t)\right) \\ 
 & = A\frac{x(t)}{|x(t)|}-\frac{x(t)}{|x(t)|}\left(\frac{x(t)^T}{|x(t)|}A\frac{x(t)}{|x(t)|}\right) \\
 & = As(t)-(s(t)^TAs(t))s(t) \\
 & = \left(1-s(t)^TAs(t)\right)As(t) \\
 & = B(s(t))s(t)
\end{align*}
with
$$B(s(t))=\left(1-s(t)^TAs(t)\right)A=\left(1-s_2\left(s_1(1-f'(u_2))-s_2+s_3\right)\right)A$$
if we write $s(t)=(s_1,s_2,s_3)(t)$. This is the ODE induces by the original one on the unit sphere. Not sure why they want to identify opposite points (as we discussed, identifying opposite points on the sphere is quotienting it by a relation of equivalence, and then they must define norm and ODE in the quotient...).
$(2)$ Let $(\lambda_k,X_k)_{1\leq k\leq3}$ the eigen pairs of the matrix $A$, that is
$$AX_k=\lambda_kX_k\quad\quad\quad\forall1\leq k\leq3.$$
Then, for all $1\leq k\leq3$, we have
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{X_k(t)}{|X_k(t)|}\right) & = B\left(\frac{X_k(t)}{|X_k(t)|}\right)\frac{X_k(t)}{|X_k(t)|} \\
 & = \left(1-\frac{X_k(t)^T}{|X_k(t)|}A\frac{X_k(t)}{|X_k(t)|}\right)A\frac{X_k(t)}{|X_k(t)|} \\
 & = \left(1-\frac{X_k(t)^T}{|X_k(t)|}\lambda_k\frac{X_k(t)}{|X_k(t)|}\right)\lambda_k\frac{X_k(t)}{|X_k(t)|} \\
 & = \left(1-\lambda_k\frac{|X_k(t)|^2}{|X_k(t)|^2}\right)\lambda_k\frac{X_k(t)}{|X_k(t)|} \\
  & = \left(1-\lambda_k\right)\lambda_k\frac{X_k(t)}{|X_k(t)|} \\
 & = \Lambda_k\frac{X_k(t)}{|X_k(t)|}
\end{align*}
with $\Lambda_k=\left(1-\lambda_k\right)\lambda_k$.
This shows that there are three new eigen pairs, namely $(\Lambda_k,\frac{X_k}{|X_k|})_{1\leq k\leq3}$. Now, following your notations, $\Lambda_3=0$ because $\lambda_3=0$, so $\frac{X_3}{|X_3|}$ is the eigenvector that gives the saddle (they call it $X_3$). Next, as $\lambda_1<0$, $\Lambda_1<0$ also. Finally, $\Lambda_2$ can have any sign, according to $\lambda_2\leq\frac{1}{2}$ or $\lambda_2>\frac{1}{2}$.
$(3)$ To see where are eigenspaces on the sphere (provided that I understand the question), we identify the eigenspaces in $\mathbb{R}^3$ spanned by the eigenvectors $\left(\frac{X_k}{|X_k|}\right)_{1\leq k\leq3}$ (these are the same that those spanned by $\left(X_k\right)_{1\leq k\leq3}$), and then we intersect them with the sphere. Lines and planes passing through the center of the sphere then becomes respectively points and great circles (or longitudes). For example, $C:=\mathrm{span}{X_1}\cap\mathbb{S}^2$ is the union of the two intersection points of $\mathbb{S}^2$ and the line spanned by $X_1$. Since $\Lambda_1<0$, they are attractors. Similaraly, the intersection $\mathrm{span}{X_2,X_3}\cap\mathbb{S}^2$ is the intersection of $\mathbb{S}^2$ with the vector plane spanned by $X_2$ and $X_3$, so it is a great circle passing through the projections of $X_2$ and $X_3$ on the sphere (it is a geodesic of the sphere). 
