Range of $a$ such that $w \leftarrow w-a x \langle w, x \rangle$ converges almost surely? Edit Sep 19 this answer on Mathoverflow matches simulation results
Suppose $x_i$ come from 2-d standard Normal centered at 0. What is the range of $a$ for which the following iteration converges almost surely?
$$w_{i+1} = w_i-a x_i \langle w_i, x_i \rangle$$
Mean norm of $w$ diverges for $a>2/3$ but this is vacuous because variance also diverges. The answer is somewhere between 0.9 and 1.1.
How do I get a more accurate estimate?

Here's a way to verify that $a=2$ converges for 1D, which doesn't generalize to 2D
Note that our process defines the following recursion on $w^2$
$$\log w_{k+1}^2 = \log (1-\alpha x^2)^2 + \log w_k^2$$
Here we see that $\log w_{k+1}^2$ is a sum of $k$ independent random variables, tending towards $-\infty$ as $k$ grows whenever $E[\log (1-\alpha x^2)^2]<0$, from Central Limit Theorem.
Since $E[\log (1-\alpha x^2)^2]\approx 0$ for $\alpha \approx
2.42129$, this also forms the critical rate, and can be verified in simulations by looking at median value of trajectories for $\alpha$ slightly above, and slightly below critical

Edit Aug 24
Expectation in answer has a closed form solution in terms of exponential integral, can be obtained in Mathematica. Plotting it for $d=2$ there are two critical points, for $a=0.121312$ and for $a=0.918611$


$\newcommand{\E}{\operatorname{E}}$
This is a new answer for the updated question, where we have a sequence of Gaussians instead of a single $x$.
We have that $w_n = \left[\prod_{i=1}^n (I - ax_ix_i^\top)\right]w_0$, and want to examine the behavior as $n\rightarrow \infty$. We're going to use the trick that an infinite product converges to zero iff the product of squares converges to zero:
\begin{align*}
\lim_{n\rightarrow\infty}\prod_{i=1}^n(I - ax_ix_i^\top)^2 
&= \lim_{n\rightarrow\infty}\exp\left(n\cdot \frac{1}{n}\sum_{i=1}^n\log(I - ax_ix_i^\top)^2\right) \\
&\overset{a.s.}{\longrightarrow}\lim_{n\rightarrow\infty}\exp\left(n\cdot \E\left[\log(I - ax_ix_i^\top)^2\right]\right)
\end{align*}
Above, we use the matrix exponential and matrix logarithm. Using the spectral decomposition
\begin{equation*}
(I - ax_ix_i^\top)^2 = \begin{bmatrix}
x_{i,2}/x_{i,1} & x_{i, 1}/x_{i,2} \\ 1 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\ 0 & (1-a||x_i||^2)^2
\end{bmatrix}
\begin{bmatrix}
-\frac{x_{1,i}x_{2,i}}{||x_i||^2} & \frac{x_{1,i}^2}{||x_i||^2} \\ \frac{x_{1,i}x_{2,i}}{||x_i||^2} & \frac{x_{2,i}^2}{||x_i||^2} 
\end{bmatrix}
\end{equation*}
we can calculate that
\begin{equation*}
\E\left[\log(I - ax_ix_i^\top)^2\right] = \begin{bmatrix}
\E\left[\frac{x_{1,i}^2}{||x_i||^2}\cdot\log(1-a||x_i||^2)^2\right] & 0 \\ 0 & \E\left[\frac{x_{2,i}^2}{||x_i||^2}\cdot\log(1-a||x_i||^2)^2\right]
\end{bmatrix}
\end{equation*}
Thus, we are concerned with when the following matrix has limit zero:
\begin{equation*}
\lim_{n\rightarrow\infty}\begin{bmatrix}
\exp\left(\E\left[\frac{x_{1,i}^2}{||x_i||^2}\cdot\log(1-a||x_i||^2)^2\right]\right) & 0 \\ 0 & \exp\left(\E\left[\frac{x_{2,i}^2}{||x_i||^2}\cdot\log(1-a||x_i||^2)^2\right]\right)
\end{bmatrix}^n
\end{equation*}
and this of course happens iff it has norm less than 1. That is,
\begin{equation*}
\sqrt{2\cdot \left[\exp\left(\E\left[\frac{x_{1,i}^2}{||x_i||^2}\cdot\log(1-a||x_i||^2)^2\right]\right)\right]^2} < 1
\end{equation*}
Some numerical integration suggests that the critical value is roughly $a \approx 0.92$.
