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  "body": "\n## Proof\n\nWe compare $f$ near $a$ with its second-order Taylor polynomial.\n\n1. Let $U$, $f$, $a$ and the [[Hessian matrix|0SCSQWXRA4]] $H = H_f(a)$ be as in the statement. Since $f$ is of [[class C^2|0TCBZMEYGN]], [[Schwarz's theorem|0NNQY8WCW6]] shows that $H$ is symmetric. Its quadratic form $Q(h) = h^{\\mathsf{T}} H h$ is as in [[the definition of definite forms|01RBSSGMY5]].\n2. Since $\\nabla f(a) = 0$, the [[second-order Taylor expansion|0P30YYZN85]] gives $f(a + h) - f(a) = \\tfrac{1}{2} Q(h) + R(h)$ for $a + h \\in U$, with $R(h) / \\norm{h}^2 \\to 0$ as $h \\to 0$.\n3. Part 1. Assume that $H$ is positive definite. Let $S = \\set{h \\in \\R^n : \\norm{h} = 1}$ be the unit sphere. It is closed and bounded, so it is compact by the [[Heine–Borel theorem|07PFYV65MB]].\n4. The function $Q$ is a polynomial in the coordinates of $h$, so it is continuous. By the [[extreme value theorem|0V4DET59NW]], there is $h_0 \\in S$ with $Q(h) \\ge Q(h_0)$ for all $h \\in S$. Let $m = Q(h_0)$. Then $m \u003e 0$, since $h_0 \\ne 0$ and $Q$ is positive definite.\n5. For $h \\ne 0$, the vector $h / \\norm{h}$ lies in $S$, and $Q(h) = \\norm{h}^2 Q(h / \\norm{h})$. So $Q(h) \\ge m \\norm{h}^2$ for every $h \\in \\R^n$.\n6. By step 2, there is $\\delta \u003e 0$ such that $a + h \\in U$ and $\\abs{R(h)} \u003c \\tfrac{m}{4} \\norm{h}^2$ whenever $0 \u003c \\norm{h} \u003c \\delta$.\n7. For such $h$, steps 2, 5 and 6 give $f(a + h) - f(a) \\ge \\tfrac{m}{2} \\norm{h}^2 - \\tfrac{m}{4} \\norm{h}^2 = \\tfrac{m}{4} \\norm{h}^2 \u003e 0$.\n8. So $f(x) \u003e f(a)$ for all $x \\in U$ with $0 \u003c \\norm{x - a} \u003c \\delta$. This means that $f$ has a strict [[local minimum|04CPG9K1C4]] at $a$.\n9. Part 2. Assume that $H$ is negative definite. The function $-f$ is of class $C^2$, has gradient $0$ at $a$, and has Hessian $-H$, which is positive definite.\n10. By Part 1, $-f$ has a strict local minimum at $a$. So $f$ has a strict local maximum at $a$.\n11. Part 3. Let $\\lambda \u003e 0$ and $\\mu \u003c 0$ be eigenvalues of $H$. The real matrices $H - \\lambda I$ and $H - \\mu I$ are singular. So there are real [[eigenvectors|0B5H4208Z9]] $v, w \\in \\R^n$ with $H v = \\lambda v$ and $H w = \\mu w$. Then $Q(v) = \\lambda \\norm{v}^2 \u003e 0$ and $Q(w) = \\mu \\norm{w}^2 \u003c 0$.\n12. For real $t \\ne 0$ small enough that $a + t v \\in U$, step 2 gives $\\frac{f(a + t v) - f(a)}{t^2 \\norm{v}^2} = \\frac{\\lambda}{2} + \\frac{R(t v)}{\\norm{t v}^2}$.\n13. As $t \\to 0$, the right side tends to $\\lambda / 2 \u003e 0$. So there is $\\eta \u003e 0$ with $f(a + t v) \u003e f(a)$ for $0 \u003c \\abs{t} \u003c \\eta$.\n14. In the same way, using $w$ and $\\mu \u003c 0$, there is $\\eta' \u003e 0$ with $f(a + t w) \u003c f(a)$ for $0 \u003c \\abs{t} \u003c \\eta'$.\n15. Let $\\varepsilon \u003e 0$. By steps 13 and 14, the ball of radius $\\varepsilon$ around $a$ contains a point $x \\in U$ with $f(x) \u003e f(a)$. It also contains a point $y \\in U$ with $f(y) \u003c f(a)$.\n16. So $a$ is neither a local maximum nor a local minimum of $f$. Since $\\nabla f(a) = 0$, the point $a$ is a [[saddle point|00N3T18QNA]] of $f$.\n\n## Idea\n\nAt a critical point the linear term of the Taylor expansion is zero, so $f(a + h) - f(a)$ is $\\tfrac{1}{2} h^{\\mathsf{T}} H h$ plus a remainder that is small compared to $\\norm{h}^2$. A positive definite form is at least $m \\norm{h}^2$, by compactness of the sphere, so it wins against the remainder. For an indefinite Hessian, moving along eigenvectors makes $f$ go up in one direction and down in another.\n",
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