= Solution
Use <mathematical induction> on the iteration. Initially $x_0=b=1\cdot b$. If $x_n=\lambda_nb$, then $Ax_n=\lambda_nb/2$, $x_n^Tb=\lambda_n$, and $x_n^TAx_n=\lambda_n^2/2$. Therefore
$$
\nabla h(x_n)=\tfrac32\lambda_n^2b.
$$
A <gradient descent> step with any scalar step length $\alpha_n$ consequently has the form
$$
x_{n+1}=x_n-\alpha_n\nabla h(x_n)
=\left(\lambda_n-\tfrac32\alpha_n\lambda_n^2\right)b.
$$
Thus every iterate remains on the same <eigenline>, with
$$
\boxed{\lambda_0=1,\qquad
\lambda_{n+1}=\lambda_n-\tfrac32\alpha_n\lambda_n^2.}
$$
This proves that <gradient descent preserves an eigenline for a cubic product>. The statement is invariance, not convergence to a minimum: along the negative direction of this line the <cubic polynomial> objective is unbounded below.
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