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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?

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1 Answer 1

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.

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When $x_1=0$, $F'(x_2)$ does not have to equal zero, right? The curve $F''(X_2)=1$ may not be an integral curve of the gradient vector field, so it should not be considered here, right? –  D.F.J. Jan 20 '12 at 14:24
    
@D.F.J. Yes, this is what I mean. Among the three curves, two are integral curves of $\nabla f$, namely $x_1=0$ and $F'(x_2)=0$, but the third one is not. –  Denis Serre Jan 20 '12 at 14:58

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