Attractors of nonlinear dynamical systems on the sphere This is related to my other recent question, which involved linear flows on the unit sphere. Here, we are going to consider nonlinear flows.

Let $\mathbb S^{d-1}=\{x\in\mathbb R^d\ :\ x^Tx=1\}$ denote the unit sphere. Let moreover $A$ be a real $d\times d$ matrix.
Following this MathOverflow question we consider the unique solution $x(t, x_0)$ to the nonlinear initial value problem
$$
\dot{x}=(I-x x^T)Ax, \quad x(0)=x_0\in \mathbb S^{d-1}. $$
For $x\in \mathbb S^{d-1}$, that is $\lvert x \rvert^2=1$, we see that
$$\tfrac{d}{dt}(x^T x)=x^TA^Tx-x^TA^Tx\lvert x\rvert^2+x^TAx - \lvert x \rvert^2 x^TAx=0,$$
so $x(t, x_0)$ remains on $\mathbb S^{d-1}$ for all $t>0$.
Now the linked MathOverflow question states, without proof, that if $A$ is negative semi-definite, then

$x(t, x_0)$ converges to a stable equilibrium.

Can you prove an appropriate version of this statement?

Here's some of my thoughts.
I can think of two versions of the statement to prove. But I cannot prove either of them. First of all, it is easy to see that the normalized eigenvectors of $A$ correspond to equilibria; precisely, if $Av=\lambda v$ and $v\in \mathbb S^{d-1}$ then
$$
\left.\tfrac{d}{dt} x(t, v)\right|_{t=0}= (I-vv^T)\lambda v=0,$$
which implies that $x(t, v)=v$ for all $t\ge 0$. This leads me to think that the "stable equilibrium" mentioned in the statement above is an eigenvector. The two conjectures follow.
Conjecture 1. For each $x_0\in \mathbb S^{d-1}$ there is an eigenvector $v\in\mathbb S^{d-1}$ of $A$ such that $x(t, x_0)\to v$ as $t\to \infty$.
Conjecture 2. (stronger). Let $\lambda_j$ denote the eigenvalues of $A$ and suppose that $0>\lambda_1>\lambda_j$ for all $j>1$, and that $\lambda_1$ is non-degenerate. Let $v\in \mathbb S^{d-1}$ be a $\lambda_1$-eigenvector of $A$. Then
$$
x(t, x_0)\to v,\quad \text{or}\quad x(t, x_0)\to -v$$
as $t\to \infty$, unless $v^Tx_0=0$. (In the latter case the system never leaves the $(d-2)$ dimensional sphere $\{x\in\mathbb R^d\ :\ v^Tx=0,\ \lvert x\rvert^2=1\}$).
 A: This is based on the substantial input given by Evgeny in comments. Thank you Evgeny!

Since $A$ is definite negative, up to an orthogonal change of coordinates we can rewrite the system
$$\tag{1}\dot{x}=(I-xx^T)Ax, \quad x\in\mathbb R^n,$$
as
$$\tag{2}
\dot{x}_i=x_i(-\lambda_i +\sum_{j=1}^n\lambda_j x_j^2),\quad i=1, \ldots, n,$$
where
$$0<\lambda_1< \lambda_2\le \lambda_3\ldots$$
and
$$
Ae_j=-\lambda_j e_j, \qquad \text{ where }e_i=(0, \ldots, 1, \ldots, 0).$$
(Searching the literature, I see that $\lambda_1$ is often called spectral gap, and $-\lambda_1$ is the  leading eigenvalue).
Assumption. I am assuming that $\lambda_1$ is simple, meaning that the only normalized eigenvectors corresponding to $-\lambda_1$ are $\pm e_1$.
Side remark. The eigenvectors $e_i$ are equilibria of (1). The change of variable $x=y+e_i$ produces the equivalent system
$$\tag{3}
\dot{y}=(A+\lambda_i I + 2\lambda_i e_i e_i^T)y+ 2\lambda_iy_i y - y^TAy e_i -y^TAy y.$$
The matrix in round brackets is the linearization of (1) around the equilibrium point $e_i$. However, this does not seem to play a role in the present analysis.

CLAIM . For each $x_0\in\mathbb S^{n-1}$, either
$$\lvert x(t)-e_1\rvert \to 0, \quad \text{or} \quad \lvert x(t)+e_1\rvert\to 0,$$ or $e_1^Tx(t)=0$ for all times. (In this latter case, (1) reduces to an ODE on $\mathbb S^{n-2}$, and if $\lambda_2$ is simple, this claim applies to it).
Proof. Letting $y_i=x_i^2$ we see that these functions satisfy a differential system;
$$\tag{4}
\frac{d y_i}{dt}=2y_i(-\lambda_i + \sum_{j=1}^n \lambda_j y_j), \qquad i=1,\ldots,n.
$$
Also, obviously, $y_i\in[0,1]$ and $\sum y_i=1$.
Now, $\sum \lambda_j y_j\ge \lambda_1 y_1$, with strict inequality unless $y_1=1, y_2=0, \ldots, y_n=0$. So we see that $\dot{y}_1>0$, unless $y_1=0$ or $y_1=1$. In the first case, we have $e_1^Tx(t)=0$ for all times. In the latter case, we have $x(t)=e_1$ or $x(t)=-e_1$ for all times.
It remains to consider the case $y_1\in (0, 1)$ and $\dot{y}_1>0$. Since $y_1$ is strictly increasing, it has a limit as $t\to \infty$, and by uniform continuity it must be $\dot{y}_1(t)\to 0$. Thus, (4) shows that $(-\lambda_i + \sum_{j=1}^n \lambda_j y_j)$ must vanish as $t\to \infty$, and this can only happen if $y_1^2\to 1$. This concludes the proof. $\Box$
Side remark. The system (4) can be written in matrix form by letting $\mathbf 1^T=(1, 1,\ldots, 1)$;
$$
\frac{dy}{dt}=(2I -2y \mathbf 1^T)Ay.$$
A: First a remark: It seems rather odd to ask for $A$ to be negative definite, since the sphere becomes repelling and the problem is unstable when running forward in time! One has:
$$ \frac{d}{dt} (1-x^t x)^2 = - 2 (1-x^tx)^2 (x^t A x).$$
So for $x\neq 0$, $|1-x^tx|$ is decreasing when $A$ is positive definite  and the equation is globally stable only in that case.
Suppose $v\in {\Bbb R}^d\setminus\{0\}$ is a fixed point. Then you find $Av=\lambda v$ and $v^t v=1$.
Linearizing  around the fixed point, $x=v+z$ ($z$ small), yields:
$$ \dot{z}= (A-\lambda) z - 2 \lambda v (v^t z)+o(z)= (A-\lambda-2\lambda vv^t) z + o(z).$$
In the $v$-direction you need $\lambda>0$ for global stability as already mentioned. In an orthogonal direction, $v^tz=0$, i.e. tangent to the sphere, we have $\dot{z}=(A-\lambda)z$ which gives stability precisely when $\lambda$ is a largest isolated eigenvalue of $A$.
