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Apply symmetric Gaussian elimination without row exchanges. At step $k$, let $d_k$ be the leading diagonal entry of the remaining symmetric Schur complement. If $d_k\le 0$, stop and report that $A$ is not positive definite. If $d_k>0$, use it to eliminate the rest of its row and column. If all steps succeed, this constructs an LDL decompositionwhere $L$ is unit lower triangular and $D=\mathrm{diag}(d_1,\dots ,d_n)$ has positive diagonal.

The test is correct from first principles. If all $d_k>0$, then for every nonzero $x$,because $L$ is invertible. Conversely, if $A$ is positive definite, its first pivot is $a_{11}>0$, and completing the square givesChoosing $s=-b^{T}y/a$ shows that the Schur complement is positive definite. Induction forces every pivot to be positive. This is also Sylvester's criterion.

At step $k$, updating the remaining matrix costs $O((n-k{)}^2)$ arithmetic operations. Hence the total iswhich proves the existence of the required algorithm.

Introduce a Lagrange multiplier $\lambda$ for the normalization and varyFor a smooth variation $u+\epsilon \eta$ with $\eta =0$ on $\partial D$, integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equation

Multiplying by $u$ and integrating, while usingthe divergence theorem and $u=0$ on the boundary giveThe normalization is one, so the multiplier equals the stationary value:

The integrand has an order-two pole at $0$ and a simple pole at $2$. Write it near zero as $h(z)/z^2$, whereIts residue at zero isAt $z=2$ the residue isThe residue theorem now gives

The condition $D{f}_p\ne 0$ says that $0$ is a regular value of $f$. The regular level set theorem therefore makes $f^{-1}(0)$ a smooth submanifold of codimension one in ${\mathbb{R}}^3$, hence a smooth surface.

Forthe gradient isOn $F^{-1}(0)$, $x^2+y^2=\cosh (z^2)\ge 1$, so $x$ and $y$ cannot both vanish. Thus $\nabla F\ne 0$ there, and the given set is a smooth surface.

Not every smooth surface in ${\mathbb{R}}^3$ is a global zero set. Every set $f^{-1}(0)$ is closed because $f$ is continuous, whereas the open unit diskis a smooth surface but is not closed in ${\mathbb{R}}^3$.

The mapis a nonzero linear functional: otherwise $v$ would lie in the radical of a bilinear form, contradicting nondegeneracy. Its kernel is $v^{\perp }$, so the rank-nullity theorem gives $\dim v^{\perp }=n-1$. Antisymmetry gives $⟨v,v⟩=-⟨v,v⟩=0$, hence $v\in v^{\perp }$.

Suppose $w\in W$ is orthogonal to every vector of $W$. Since $W\subseteq v^{\perp }$, it is also orthogonal to $v$, and therefore toFor a nondegenerate bilinear form, $(v^{\perp }{)}^{\perp }=\mathbb{R}v$: both sides have dimension one and the latter is contained in the former. Thus $w\in W\cap \mathbb{R}v=\{0\}$, proving that the restriction to $W$ is nondegenerate.

The space $W$ has dimension $n-2$ and again carries a nondegenerate antisymmetric form. Induction, starting from the zero-dimensional space, shows that $\dim W$ is even. Therefore $n=\dim W+2$ is even. Equivalently, every finite-dimensional symplectic vector space has even dimension.