Given a smooth scalar function $f(x_1, x_2, \ldots, x_n)$ defined in $\mathbb{R}^n$, we have at each point a gradient vector $g=\nabla f$ and a Hessian matrix $H$ with $H_{ij}=\partial _{x_i}\partial_{x_j}f$.
Suppose $C(t)$ $(t\ge0)$ is an integral curve of the gradient vector field $g$, with the property that at the starting point $t=0$ the tangent vector of $C$ is an eigenvector of the Hessian matrix at that point.
My question is, is it true that for $t>0$ the tangent vector of $C(t)$ keeps being an eigenvector of the Hessian matrix at the corresponding positions?
Best Answer
No. Take $f(x_1,x_2)=\frac12 x_1^2+F(x_2)$. Then $$\nabla f=\begin{pmatrix} x_1 \\\\ F'(x_2) \end{pmatrix},\qquad H=\begin{pmatrix} 1 & 0 \\\\ 0 & F''(x_2) \end{pmatrix},$$ thus $\nabla f$ is an eigenvector of $H$ if and only if either $x_1=0$, $F'(x_2)=0$ or $F''(x_2)=1$. For generic $F$'s, this defines three curves, among which the last one is not an integral curve of $\nabla f$, because $$\nabla f\nabla F''(x_2)=F'(x_2)F'''(x_2)\ne0$$ in general.