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For any real vector $u$, a Cholesky decomposition gives the quadratic form identitySince $L$ is an invertible matrix, this squared Euclidean norm is positive whenever $u\ne 0$. The factorization thus certifies positive definiteness and gives coordinates in which the quadratic form is a sum of ordinary squares. For the matrix above,The level set $u^{T}Au=1$ is an ellipse; the linear map $z=L^{T}u$ sends it to the unit circle. The points $u=(1/2,0{)}^T$ and $u=(-1/4,1/2{)}^T$ go to $(1,0{)}^T$ and $(0,1{)}^T$, respectively.

The same factor works in the reverse direction for Gaussian random vectors. If $Z$ has independent unit-variance, zero-mean Gaussian random variables as coordinates, $X=LZ$ has covariance matrix $L{L}^T=A$. In this example $X_1=2Z_1$ and $X_2=Z_1+2Z_2$: their variances are $4$ and $5$, while their covariance is $2$. This is a concrete way to generate specified correlations from independent draws.

A triangular factor is not generally a square root of a matrix in the sense $B^2=A$. Here $L^2=\left(\begin{array}{cc}4& 0\\ 4& 4\end{array}\right)\ne A$. The Cholesky decomposition uses the product with the transpose, which is also what makes the quadratic form a squared Euclidean norm.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

On real symmetric matrices of size $n\ge 2$, the displayed pairing is symmetric because $\det (AC)=\det A\det C$. It is not bilinear: $B(I,2I)=2^n$, whereas linearity would require $2B(I,I)=2$. Its self-pairing is always nonnegative, but a nonzero singular matrix has $B(A,A)=0$, so positive definiteness fails as well. For $n=1$ it reduces to the ordinary real inner product. The Frobenius inner product $\mathrm{tr}(A^{T}C)$ supplies the appropriate bilinear, positive-definite geometry in every size.

Let $P_n(\mathbb{R})$ be the vector space of real polynomials of degree of a polynomial at most $n$. Choose $n+1$ distinct real sample points $t_0,\dots ,t_n$ and positive real weights $w_0,\dots ,w_n$. Thenis an inner product. Symmetry and bilinearity follow from linearity of polynomial evaluation. Its value at $(f,f)$ is nonnegative, and equality forces all $f(t_j)=0$. By the root bound for a polynomial, a nonzero polynomial of degree of a polynomial at most $n$ cannot vanish at $n+1$ distinct points. Thus the form has positive definiteness.

Equivalently, the polynomial evaluation map on a finite point set identifies $P_n(\mathbb{R})$ with ${\mathbb{R}}^{n+1}$, equipped with a weighted Euclidean inner product. Too few sample points destroy positive definiteness: if the distinct points number at most $n$, the nonzero polynomial ${\prod }_j(x-t_j)$ lies in $P_n(\mathbb{R})$ and has zero sampled squared norm.

For three equally weighted samples at $-1,0,1$, an orthonormal basis isThis construction gives a finite sampling version of orthogonal polynomials and permits exact Gram-Schmidt orthogonalization using only finitely many evaluations.

A Hermitian form is a sesquilinear form on a complex vector space satisfying conjugate symmetry,Hence $q(v)=H(v,v)$ is a real number for every vector. In a basis, the matrix of a Hermitian form has $A_{ij}=H(e_i,e_j)$ and obeys $A_{ji}=\overline{A_{ij}}$, equivalently $A=A^{\ast }$: it is a Hermitian matrix. The coordinate rule remains $H(x,y)=x^{T}A\overline{y}$. Conjugate symmetry does not imply positivity.^[ref]

The diagonal contains more information than might be expected. For any complex sesquilinear form $B$, put $q(v)=B(v,v)$. Expanding four expressions gives the complex polarization identity in the linear-first convention:This is an algebraic identity, with no positivity assumption. If every $q(v)$ is real, interchanging $u,v$ preserves the first difference and negates the second: use $q(\alpha v)=|\alpha {|}^{2}q(v)$ for $\alpha =-1,i$. The identity then gives $B(v,u)=\overline{B(u,v)}$. Thus a complex sesquilinear form is a Hermitian form if and only if all its diagonal values are real. Testing only the basis vectors is insufficient: the off-diagonal example in sesquilinear form has zero diagonal entries in its matrix but a nonreal value at $(1,i)$.

Consider the indefinite Hermitian formIts values at $e_1,e_2$ are $1,-1$. Moreover $u=(1,1)$ and $v=(1,-1)$ are nonzero isotropic vectors: $q_{-}(u)=q_{-}(v)=0$, yet $H_{-}(u,v)=2$. These vectors do not belong to the radical of a Hermitian form, and the usual Cauchy-Schwarz inequality would incorrectly require $4\le 0$. An indefinite Hermitian form records signed geometry rather than defining a norm by $\sqrt{H(v,v)}$.

There is a separate boundary case: $H_0(z,w)=z_1\overline{w_1}$ is a positive semidefinite Hermitian form, since $H_0(z,z)=|z_1{|}^2\ge 0$, but $H_0(e_2,e_2)=0$. Its radical of a Hermitian form is the entire second coordinate line. Positive definiteness is exactly the additional condition in a Hermitian inner product. The three examples below show why these distinctions matter: on the real slice $z=(x,y)\in {\mathbb{R}}^2\subset {\mathbb{C}}^2$, $q_{+}(z)=|z_1{|}^2+2|z_2{|}^2$ has an ellipse as its unit level, $q_{-}$ has hyperbolas, and $q_0$ has two parallel lines. These are slices of the complex examples, not drawings of their full four-dimensional domains.

A Hermitian inner product is a Hermitian form $H$ on a complex vector space which has positive definiteness: $H(v,v)>0$ for every nonzero vector $v$. It is precisely a complex inner product. Only with this definiteness condition doesdefine a norm rather than a seminorm or an expression that may not be real. The standard Hermitian inner product on ${\mathbb{C}}^n$ is $H(z,w)={\sum }_j{z}_j\overline{w_j}$; positive weights $c_j$ give ${\sum }_j{c}_j{z}_j\overline{w_j}$, with squared norm ${\sum }_j{c}_j|z_j{|}^2$.^[ref]

A Hermitian inner product need not have a real matrix. For instance, with the linear-first coordinate convention,The last expression proves positive definiteness without a lengthy calculation of eigenvalues. For $z=(1,i)$ it gives $6$, while for $z=(1,-i)$ it gives $2$. These unequal values expose why a silent switch to $z^{\ast }Az$ would change this particular Hermitian inner product: with these same entries it gives $2$ at $(1,i)$. All entries must keep their roles from the matrix of a Hermitian form.

The Cauchy-Schwarz inequality for a Hermitian inner product statesEquality holds exactly when $u,v$ are linearly dependent, including when either is zero. For a nonzero $u$, outline the proof by setting $a=H(v,u)/H(u,u)$ and $r=v-au$. Then $H(r,u)=0$, and the Pythagorean theorem givesNonnegativity proves the inequality, while positive definiteness makes equality equivalent to $r=0$. The scalar $a$ has no extra conjugate because the first argument is linear.^[ref]

The Cauchy-Schwarz inequality also holds for a positive semidefinite Hermitian form, but equality need not imply linear dependence: for $H_0(z,w)=z_1\overline{w_1}$, the independent vectors $e_1$ and $e_1+e_2$ give $1=1$. Equality then describes dependence after passing to the quotient vector space by the radical of a Hermitian form. The indefinite Hermitian form in Hermitian form shows that conjugate symmetry alone does not even guarantee the inequality.

For a worked orthogonal projection, take $H_{+}(z,w)=z_1\overline{w_1}+2z_2\overline{w_2}$, $u=(1,1)$ and $v=(2,0)$. Then $a=2/3$, $au=(2/3,2/3)$ and $r=(4/3,-2/3)$, with $H_{+}(r,u)=0$ and $4=4/3+8/3$ for the squared norms. The residual is perpendicular in this Hermitian inner product even though its ordinary real dot product with $u$ is $2/3$. For complex $v=(1,i)$, the same orthogonal projection uses $a=(1+2i)/3$; rotating $v$ multiplies $a$ by $e^{i\theta }$, while rotating $u$ multiplies $a$ by $e^{-i\theta }$. This combines geometric decomposition with sensitivity to the relative complex arguments of the multipliers.

A Hermitian inner product therefore supplies more than a positive diagonal: orthogonality, orthogonal projection onto a span, and the triangle inequality for its norm make approximation and distance meaningful. For the function space of complex continuous functions on $[0,1]$, $H(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ has positive definiteness, because a nonzero continuous function has positive modulus on a small interval. The same construction turns overlap under integration into lengths and best approximation in finite-dimensional vector subspaces.

The Hessian matrix describes how the slope of a function changes in different directions. For a real-valued function $f$ with continuous second partial derivatives on an open subset of ${\mathbb{R}}^n$, it isThe symmetry of second derivatives makes $H_f$ a symmetric matrix under these hypotheses. It is also the Jacobian matrix of the gradient: near $x$, a small displacement $h$ changes $\nabla f$ by $H_f(x)h+o(\Vert h\Vert )$. Thus the gradient measures the current slope, while the Hessian measures its local variation. ^[ref]

For example, take the quadratic formIts gradient and Hessian matrix areThe only critical point is the origin. Completing the square gives $q=(x+y{)}^2+2y^2$, so the origin is a strict global minimum. The eigenvalues $4-2\sqrt{2}$ and $4+2\sqrt{2}$ are both positive, recovering the same conclusion from the positive definiteness of the Hessian. Its eigenvectors identify the rotated directions in which the quadratic form has its smallest and largest bending.

For a fixed vector $v$, restriction to a straight line gives the directional interpretationThis is the quadratic form associated with the Hessian matrix. For a unit eigenvector $v$, it equals the corresponding eigenvalue. Opposite signs therefore mean that some directions bend upwards and others downwards. The second-derivative test turns this information into a classification when the gradient vanishes.

The same object is useful for computation. Newton's method in optimization chooses a displacement $h$ by solving $H_f(x)h=-\nabla f(x)$ when the Hessian matrix is an invertible matrix; this makes the gradient of the local quadratic model vanish. For the displayed $q$, the equation gives $h=-(x,y{)}^T$, so one exact step reaches the minimum from any starting point. For a general function, a Hessian with opposite-sign eigenvalues or a large condition number need not give a useful minimizing step, and the local minimum need not be a global minimum.

The name honors Otto Hesse. ^[ref] An older geometric use is the Hessian criterion for flexes of a plane cubic, connecting a determinant of second partial derivatives with inflection points of an algebraic curve. Hesse's 1844 paper Über die Wendepuncte der Curven dritter Ordnung is an early source on that geometric problem. ^[ref]

A matrix decomposition expresses a matrix as a matrix product of factors with simpler structure. An LU decomposition uses triangular factors, a QR decomposition uses an orthogonal matrix and a triangular factor, and a singular value decomposition separates input directions, stretching factors and output directions. The factorization is an identity; the chosen structure determines what it helps compute.

For an original two-by-two example,The outer factors are a lower triangular matrix and an upper triangular matrix, each with unit diagonal as in a unit triangular matrix; the middle factor is a diagonal matrix. Their matrix product is $M$, and the final diagonal entry $5/2$ is the Schur complement of the first entry $2$. This is the scalar version of the factorization under block elimination.

To solve $Mz=(3,4{)}^T$, solve with the three factors in their displayed order from left to right: first the lower triangular factor, then the diagonal factor, then the upper triangular factor. The intermediate vectors are $(3,5/2{)}^T$ and $(3/2,1{)}^T$, giving the concise answer $z=(1,1{)}^T$. A new right-hand side uses the same factors. This is why Reusing LU factors for multiple right-hand sides is useful: factorization records work that otherwise would be repeated, while forward substitution and back substitution apply it cheaply.^[ref]

Different decompositions reveal different features. A Cholesky decomposition uses positive definiteness of a symmetric matrix, while a singular value decomposition also applies to rectangular matrices and exposes matrix rank. The existence and numerical behaviour of the factors depend on their own hypotheses; for example, an unpivoted LU decomposition can fail when a chosen pivot element is zero even though the matrix is an invertible matrix.

In a homogeneous overlapping two-component Bose-Einstein condensate, the long-wavelength energy change at fixed species numbers is one half the integral of the density perturbation quadratic form with matrix $\left(\begin{array}{cc}{U}_{11}& U_{12}\\ U_{12}& U_{22}\end{array}\right)$. Strict positive definiteness gives the displayed criterion. Strong positive cross-repulsion favors separation; sufficiently strong cross-attraction instead produces an unstable joint density mode. Equality is marginal at this order, not strict stability.

A bilinear form $B$ is nondegenerate when its left radical and right radical are both zero: every nonzero vector can be detected by some pairing, whichever input slot it occupies. On a finite-dimensional vector space, either condition implies the other, and the following tests are equivalent:Here $A$ is the matrix of a bilinear form, and $V^{\ast }$ is the dual space. The linear map $w\mapsto B(-,w)$ has matrix $A$ in the chosen basis and its dual basis, so its kernel is the right radical. The map $v\mapsto B(v,-)$ has matrix $A^T$. An invertible matrix has zero kernel on both sides, proving the finite-dimensional equivalence. In infinite dimensions, zero left radical and right radical imply these maps are injective, but do not generally make them surjective.^[ref]

For the real symmetric bilinear form $B((x,y),(s,t))=xs-yt$, the determinant is $-1$. Its isotropic vector $(1,1)$ has zero self-pairing but pairs to $1$ with $(1,0)$, so it is detected. The alternating bilinear form with matrix $J$ has determinant $1$ and detects every nonzero vector even though all self-pairings vanish. Neither zero self-pairings nor the absence of positive definiteness implies degeneracy.

Nondegeneracy is useful because it turns a linear functional into a unique vector. For $A=\mathrm{diag}(1,-1)$, the linear functional $\ell (s,t)=2s+3t$ is $B((2,-3),(s,t))$; solving for that representing vector amounts to solving an invertible system of linear equations. The analogous identification underlies bilinear orthogonal complements and the dual isomorphisms induced by a nondegenerate pairing.

For the inner product obtained by integrating with weight $|x|$ on $[-1,1]$, odd moments vanish and the moment of $x^{2m}$ is $1/(m+1)$. The first monic orthogonal polynomials are $1$, $x$ and $x^2-1/2$. Positive definiteness follows because the weight is positive almost everywhere and a nonzero polynomial cannot vanish almost everywhere.

The restriction of a positive definite form to each $V_m$ is positive definite. By the spectral theorem for real symmetric matrices, choose an orthogonal matrix $P$ such that $P^T{A}_{m}P=\mathrm{diag}(d_1,\dots ,d_m)$.

Each $d_i=\phi (f_i,f_i)>0$, since a basis vector is nonzero. Taking determinants in part (i) givesbecause an orthogonal matrix has determinant $\pm 1$. ConsequentlyThese determinants are the leading principal minors in Sylvester's criterion.

Writing $u={\sum }_i{u}_i{e}_i$ and $v={\sum }_i{v}_i{e}_i$, evaluation against the dual basis givesThis has codomain $\mathbb{C}$, not the printed $\mathbb{R}$. For a nonzero vector space, take $u=i{e}_1$ and $v=e_1$ to obtain $\varphi (u,v)=i$. Only the diagonal values $\varphi (v,v)$ must be real. The displayed formula defines the intended complex inner product.

It is linear in the first argument and conjugate-linear in the second:It has conjugate symmetry, since $\varphi (v,u)=\overline{\varphi (u,v)}$. Finally,with equality exactly when every coordinate vanishes, equivalently $v=0$. This proves positive definiteness. Since $\varphi (e_i,e_j)={\delta }_{ij}$, the original basis is orthonormal. This is the inner product determined by a basis.

Use the convention of part (a): the inner product is linear in its first argument. If $Tv=\lambda v$ with $v\ne 0$, the adjoint operator identity and $T^{\ast }=-T$ givePositive definiteness makes $⟨v,v⟩>0$, soThus every eigenvalue of a skew-adjoint operator is purely imaginary, including zero.

For $n\ge 2$, take the matrix whose leading $2\times 2$ block is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ and whose other entries are zero. Its square has leading block $-I_2$, soThis contradicts the positive definiteness required of an inner product. Even the nonzero matrix unit $E_{12}$ has $\phi (E_{12},E_{12})=0$.

The printed assertion has an exception: when $n=1$, $\phi (a,b)=ab$ is an inner product on $\mathbb{R}$. Thus the requested failure requires $n\ge 2$, a restriction absent from the printed question.

A real inner product must be symmetric. Here ${\beta }_a(e_1,e_2)=-2$ and ${\beta }_a(e_2,e_1)=-a$, forcing $a=2$. For this value,This is strictly positive for every nonzero vector. Bilinearity is already given, so symmetry and positive definiteness prove

The adjoint operator $T^{\ast }$ is characterized byThis identity works with either convention for which argument of the complex inner product is linear, provided the convention is used consistently.

For uniqueness, subtract the two given identities to get $⟨v,(S-T^{\ast })w⟩=0$ for every $v,w$. Fix $w$ and choose $v=(S-T^{\ast })w$. Then positive definiteness givesso $(S-T^{\ast })w=0$. As $w$ was arbitrary, $\boxed{S=T^{\ast }}$. The decisive fact is that a vector orthogonal to every vector, including itself, must be zero.

The adjoint on an inner product space ${\alpha }^{\ast }:V\to V$ is characterized byHere is a finite-dimensional construction that supplies every required basis fact. If $v_1,\dots ,v_n$ is any basis, its coordinate linear functionals form a basis of $V^{\ast }$: a linear functional is uniquely specified by its values on these $n$ vectors. Hence $\dim V^{\ast }=n$. The linear mapis injective, because $J(w)=0$ implies $⟨w,w⟩=0$, and positive definiteness implies $w=0$. Equal dimensions make $J$ an linear isomorphism. If ${\alpha }^{\prime }$ is the dual map, thenThis establishes existence and linearity. If two possible vectors $u,w$ represent the same linear functional through the inner product, then $J(u-w)=0$, so $u=w$. Applied to every argument, this proves uniqueness of the adjoint on an inner product space.

For linear maps $S,T$, two applications of the defining inner product identity giveUniqueness therefore gives $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$. Likewise $(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$, $(cT{)}^{\ast }=c{T}^{\ast }$ for real $c$, and $I^{\ast }=I$. An induction gives $({\alpha }^n{)}^{\ast }=({\alpha }^{\ast }{)}^n$, and applying additivity to a polynomial with real coefficients givesSymmetry of the real inner product also gives $({\alpha }^{\ast }{)}^{\ast }=\alpha$. If $m$ is the minimal polynomial of $\alpha$, the displayed identity gives $m({\alpha }^{\ast })=0$, so the minimal polynomial of ${\alpha }^{\ast }$ divides $m$. Applying the same argument to ${\alpha }^{\ast }$ gives reverse divisibility. Both polynomials are monic, hence their minimal polynomials are equal.

The distinct eigenvalues are exactly the roots of the minimal polynomial in the chosen field. An eigenvector $v$ with eigenvalue $\lambda$ satisfies $m(\alpha )v=m(\lambda )v$, so $m(\lambda )=0$. Conversely, if $m(X)=(X-\lambda )q(X)$ and $\alpha -\lambda I$ were a linear isomorphism, then $m(\alpha )=0$ would imply $q(\alpha )=0$, contradicting minimal degree. Thus $\alpha -\lambda I$ is not injective, giving an eigenvector. It follows that $\alpha$ and ${\alpha }^{\ast }$ have the same distinct eigenvalues. This includes the complex eigenvalues after complexification of a real vector space; over the real field it states equality of their real eigenvalues.

The adjoint operator $T^{\ast }$ is characterized byThis identity works with either convention for which argument of the complex inner product is linear, provided the convention is used consistently.

For uniqueness, subtract the two given identities to get $⟨v,(S-T^{\ast })w⟩=0$ for every $v,w$. Fix $w$ and choose $v=(S-T^{\ast })w$. Then positive definiteness givesso $(S-T^{\ast })w=0$. As $w$ was arbitrary, $\boxed{S=T^{\ast }}$. The decisive fact is that a vector orthogonal to every vector, including itself, must be zero.

For fixed-endpoint variations, the second variation iswith $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$ evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$. The Jacobi equation, written without differentiating $r$ separately, isWhen $r$ is differentiable this is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$.

Positive definiteness means $q(\phi )>0$ for every nonzero admissible variation. Uniform or coercive positivity of a second variation is the stronger assertion that $q(\phi )\ge c\int {\phi }^{\prime 2}$ for some fixed $c>0$. On a fixed-endpoint interval this is equivalent to a positive lower bound in the $W^{1,2}$ norm.

The strict Legendre implication uses the uniform interpretation of positivity. If $q\ge c\int {\phi }^{\prime 2}$, localized rapidly oscillating variations show $p(x)\ge c$: otherwise support them in an interval where $p-c<0$, so the derivative term of $q-c\int {\phi }^{\prime 2}$ is negative of order frequency squared, overwhelming its mixed and zero-order terms. Continuity extends the bound to the endpoints, giving $p>0$ on $[a,b]$. Under mere positive definiteness this strict conclusion is false: $f=u^2$ at $\bar{u}=0$ gives $q=2\int {\phi }^2>0$ but $p=0$. In that weaker convention only the ordinary Legendre necessary condition $p\ge 0$ follows. This distinction makes the implication precise.

Under the stated regular condition $p>0$, a conjugate point $c\in (a,b]$ to $a$ is an endpoint admitting a nonzero Jacobi solution with $z(a)=z(c)=0$. The disconjugacy form of the Jacobi positivity criterion says that no such point in $(a,b]$ exists if and only if there is a Jacobi solution strictly positive on all of $[a,b]$.

For the required identity, let that positive solution be $z$ and write $\phi =zh$. Set $A=pz{z}^{\prime }+r{z}^2$. The Jacobi equation givesExpanding the second-variation integrand therefore yieldsFor compactly supported $\phi$, the boundary term vanishes, provingThis is the Jacobi identity for the second variation. The divergence form is also appropriate under the original $C^1$ regularity; with $p>0$, the Euler-Lagrange equation gives the usual further regularity along the extremal.