= Second-derivative test
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Let $f$ have continuous second <partial derivatives> near an interior <critical point> $x_*$ of an <open subset> of $\mathbb R^n$, with $\nabla f(x_*)=0$. The <Hessian matrix> gives the following sufficient conditions:
* All <eigenvalues> positive: a strict <local minimum>.
* All <eigenvalues> negative: a strict <local maximum>.
* Both positive and negative <eigenvalues>: a <saddle point of a scalar function>.
A <positive semidefinite> or <negative semidefinite> <Hessian matrix> with a zero <eigenvalue> does not settle the classification. The <higher-order test for separated extrema> gives two quartic examples with exactly the same zero <Hessian>. These conditions apply to an interior <critical point>; a boundary optimum or a <constrained optimization> problem requires attention to the allowed directions. https://ocw.mit.edu/ans7870/18/18.013a/textbook/chapter11/section03.html[MIT 18.013A Calculus with Applications, §11.3: Hessian eigenvalues and the critical-point criterion]{ref}
The proof outline is the <Taylor expansion>
$$
f(x_*+h)-f(x_*)=\frac12h^TH_f(x_*)h+o(\|h\|^2).
$$
For a <positive-definite matrix>, the smallest <eigenvalue> supplies a positive bound proportional to $\|h\|^2$, which dominates the remainder for small enough $h$. Negate the <function> for the <local maximum> case. For opposite-sign <eigenvalues>, choose the two corresponding <eigenvectors>: the leading terms have opposite signs along their lines.
There are also necessary conditions. At an interior <local minimum>, the <Hessian matrix> is <positive semidefinite>; at an interior <local maximum>, it is <negative semidefinite>. Restriction to each <straight line> reduces this to the one-variable <second derivative> condition. Since the <Laplacian> is the <trace> of the <Hessian>, it is nonnegative at such a <local minimum> and nonpositive at such a <local maximum>. The sign of the <Laplacian> alone is insufficient: $x^2-2y^2$ has negative <Laplacian> but a <saddle point of a scalar function> at the origin.
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