If for $\vec{x}=(x_1,x_2,\ldots,x_n)\in S^{n-1}$ I have a linear ODE system $$\frac{d}{dt}\vec{x}=A\vec{x}$$ with $A$, say, diagonalizable (i.e. with $n$ simple eigenvalues) and an equilibrium $\vec{x}_0$ which is a saddle then in full dimension $n$ this equilibrium is unstable. But since everything lives on the sphere, stability then just means that I need those negative eigenvalues whose eigenspaces are tangent to the sphere at $\vec{x}_0$, right? There are, so to speak, too much dimensions and some can be neglected for the dynamics on the sphere. So I would say that the equilibrium can still be considered stable on the sphere under the condition that the unstable eigenvalues have eigenspaces which are not tangent to the sphere at $\vec{x}_0$ and there is at least one negative eigenvalue whose eigenspace is tangent to the sphere at $\vec{x}_0$. Is that correct?
Or to put it otherwise: The ODEs are only describing the flow tangentially to the sphere. Hence all transverse eigendirections can be neglected and are meaningless on the sphere.
If this is correct understanding:
How can I find, say, the largest negative eigenvalue whose eigenspace is tangent to the sphere at $\vec{x}_0$? And how can I show whether the eigenspace corresponding to another eigenvalue is transverse?
Am I right that the negative eigenvalues whose eigenspaces are not tangent to the sphere at $\vec{x}_0$ can be also neglected because they are irrelevant for the dynamics on the sphere?
To sum it up: How can I select those eigenvalues whose eigenspaces are tangent to the sphere at the equilibrium considered? And if I have selected these, how can I next find the negative eigenvalues of this type?
To be more concrete:
Lets say, $n=3$, and the eigenvalues are $\lambda_1=-1, \lambda_2=-4$ and $\lambda_3=7$. Then in dimension $n$ only $\lambda_{1,2}$ are stable eigenvalues. How can I know whether the corresponding eigenspaces are tangent to $\vec{x}_0$ and thus are relevant for the dynamics on the sphere? Is it possible that none/ both/ exactly only one of them is?
In full dimension $n$, the largest negative eigenvalue should be dominant, here $\lambda_2$. But can it be that its eigenspace is not tangent but that of $\lambda_1$ is so that only $\lambda_1$ is relevant for stability on the sphere?
Much text, but I hope you got my problem.
Maybe I can illustrate my problem with a concrete easy example.
Let $\vec{x}=(x_1,x_2)\in S^1=\{\vec{y}\in\mathbb{R}^2: \lVert\vec{y}\rVert=1\}$ and consider the ODE system $$ \begin{align*} x_1'&=x_1(-x_1+x_1^3+x_2^3-x_1x_2^2)\\ x_2'&=x_2(x_1-x_2+x_1^3+x_2^3-x_1x_2^2). \end{align*} $$ Consider the equilibrium $\vec{e}=\left(\sqrt{\frac{1}{5}},2\sqrt{\frac{1}{5}}\right)$. The linearization in this equilibrium gives $$ \vec{x}'=\begin{pmatrix}-\frac{6}{5}\sqrt{\frac{1}{5}} & \frac{8}{5}\sqrt{\frac{1}{5}}\\\frac{8}{5}\sqrt{\frac{1}{5}}&\frac{6}{5}\sqrt{\frac{1}{5}}\end{pmatrix}(\vec{x}-\vec{e}) $$ This matrix has eigenvalue $\lambda_1=2\sqrt{\frac{1}{5}}$ and $\lambda_2=-2\sqrt{\frac{1}{5}}$. So in dimension $n=2$ it is a saddle point and thus unstable. But is this equilibrium stable on $S^1$?
One eigenvector corresponding to $\lambda_1$ is $\vec{v}_1=\begin{pmatrix}1\\\frac{1}{2}\end{pmatrix}$.
One eigenvector corresponding to $\lambda_2$ is $\vec{v}_2=\begin{pmatrix}1\\-\frac{1}{2}\end{pmatrix}$.
It looks a bit as if the unstable direction (spanned by $\vec{v}_1$) is transverse to $S^1$ in $\vec{e}$, while the stable direction is tangent to $S^1$ in $\vec{e}$.
Indeed for the scalar products, $\langle\vec{e},\vec{v}_2\rangle=0$ while $\langle\vec{e},\vec{v}_1\rangle\neq 0$.
Doesn't that mean that for the question of stability on the sphere $S^1$, only the Eigenspace corresponding to $\lambda_2$ (the stable one) is relevant for the dynamics on $S^1$ so that $\vec{e}$ is locally stable on $S^1$?