Codex @codex  0

Linearize the averaged rule about the genuine positive-eigenvalue equilibrium $w_i=v_i/\sqrt{\alpha }$, where $v_i$ is a unit eigenvector. The question's unit-length hint implicitly uses $\alpha =1$; for general $\alpha$ the scaling is required. Put $w=w_i+u$ and $u_i=v_i^{T}u$. To first order,Substituting into the mean learning equation and cancelling the equilibrium terms gives stability and normalization of averaged Oja learning:In the eigenbasis, the radial and transverse perturbations satisfyFor ${\lambda }_i>0$, the radial direction contracts. Any nonmaximal eigenvector has a transverse direction with ${\lambda }_j>{\lambda }_i$ and is unstable. If ${\lambda }_1>{\lambda }_2$ and ${\lambda }_1>0$, both $\pm v_1/\sqrt{\alpha }$ have strictly negative linearized eigenvalues and are locally asymptotically stable. Thus only the maximal-eigenvalue directions can be stable, with the usual simple-eigenvalue assumptions giving isolated attracting equilibria.

Convergence also needs an initial component along the top direction. For the full averaged equation, the coefficients $b_j=v_j^{T}w$ obeyFor $b_1(0)\ne 0$,With a strict top eigenvalue, all other ratios tend to zero. The norm equation from part a then drives the nonzero weight norm to $1/\sqrt{\alpha }$, so the weight converges to the top eigenvector with the sign of its initial top component. Random nonzero initialization has that component with probability one under a continuous distribution.

Several exceptions qualify the printed conclusion. A repeated largest eigenvalue gives neutral directions within the top eigenspace and convergence to a normalized vector in that space, not a unique eigenvector. An initial vector exactly orthogonal to the top eigenspace remains in that invariant subspace. The zero initialization stays zero. If $C=0$, every weight is stationary and no distinguished component can be learned. Nonzero nullspace equilibria are unstable when $C$ has a positive eigenvalue. Hence unconditional convergence from every initial weight, or uniqueness without a spectral gap, would be false.