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$f(x,y,z)=x^3+y^3+z^3-3xy-3yz$

Is $(0,0,0)$ saddle point?

$\nabla f(0,0,0)=0$, so $(0,0,0)$ is one of the stationary point.

Also, because of my posture The reason why $f(0,0,0)$ is not a extreme value. , $f(0,0,0)$ is not a extreme value.

Therefore, I think $(0,0,0)$ is a saddle point.

However, according to wolfram alpha, saddle points of $f(x,y,z)=x^3+y^3+z^3-3xy-3yz$ don't exist.

daㅤ
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1 Answers1

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The Hessian at $(0,0,0)$ is: \begin{pmatrix} 0 & -3 & 0 \\ -3 & 0 &-3 \\ 0 & -3 & 0 \end{pmatrix} Its npn-zero eigenvalues are $\pm3\sqrt2$. So by definition it is indefinite (not positive or negative definite) and so $(0,0,0)$ is a saddle point.

am301
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