Exercise 1081Calculus › Multivariable calculus › Optimisation in several variablesL1Added 3 Nov 2026

Saddle point

#saddle point#Hessian#classification

Statement

Show that $f(x,y)=x^2-y^2$ has a saddle point at $(0,0)$ .

Hint

Compute the Hessian: one positive and one negative eigenvalue.

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Critical point. $\nabla f=(2x,-2y)=0\Rightarrow(0,0)$ .

Hessian. $H=\begin{pmatrix}2&0\\0&-2\end{pmatrix}$ ; eigenvalues $2,-2$ .

Behavior. Along $y=0$ : $f=x^2>0$ ; along $x=0$ : $f=-y^2<0$ .

Conclusion. $(0,0)$ is a saddle point (indefinite Hessian).