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At a critical point $p$ of a twice continuously differentiable real function, $\nabla f(p)=0$ and the Taylor expansion isThe real symmetric Hessian matrix has an orthonormal eigenbasis, so writing $h={\sum }_i{a}_i{u}_i$ gives $h^{T}Hh={\sum }_i{\lambda }_i{a}_i^2$. If every eigenvalue is positive, this quadratic form is at least ${\lambda }_{\min }|h{|}^2$, and the remainder cannot change its sign for sufficiently small nonzero $h$. Thus $p$ is a strict local minimum. If every eigenvalue is negative, the analogous argument gives a strict local maximum. If positive and negative eigenvalues both occur, displacement along the corresponding eigenvectors gives values above and below $f(p)$, so $p$ is a saddle point of a scalar function.

The second-derivative test can be inconclusive when the Hessian matrix is singular and semidefinite: higher-order terms along directions in the kernel of a linear map matter. For example $x^2+y^4$ and $x^2-y^4$ have the same Hessian $\mathrm{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle point of a scalar function. A zero eigenvalue does not prevent a saddle conclusion if other eigenvalues already have both signs.

For the given function write $r^2=x^2+y^2$. Its first derivatives areThe origin is a critical point for every real $\alpha$, withIts eigenvalues are $1,-1$, so it is always a saddle point of a scalar function. If one coordinate at a critical point is zero, the other must also be zero. If both are nonzero, the derivative equations force $x^2=y^2=1/(2\alpha )$, possible only for $\alpha >0$. Hence for $\alpha \le 0$ the origin is the only critical point.

For $\alpha >0$ there are four additional critical points,At such a point,Thus the two equal-sign points are strict maxima, and the two opposite-sign points are strict minima. They are global: $2|xy|\le r^2$ implieswhere the last maximum occurs at $r^2=1/\alpha$ and equality also requires $x^2=y^2$.

As $\alpha \downarrow 0$, the four nonzero critical points have distance ${\alpha }^{-1/2}$ from the origin and escape to infinity, rather than merging with the origin. Their values diverge as $\pm 1/(2e\alpha )$. As $\alpha \uparrow 0$ from below there are no extra critical points, while the origin stays a saddle throughout. This is critical points escaping to infinity in a Gaussian-weighted bilinear function.