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Use the real Hilbert space $V=H_0^1(0,1)$, the zero-boundary Sobolev space, with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincaré inequality makes this equivalent to its usual Sobolev space norm. Define the symmetric bilinear form and continuous linear functionalThe Cauchy-Schwarz inequality and the sharp interval Poincaré inequality implyThus $a$ is a bounded bilinear form and a coercive bilinear form, and $\ell$ is bounded because $f$ is square-integrable. The Lax-Milgram theorem supplies a unique weak solution:This is obtained from the differential equation by integration by parts; the test functions have zero boundary trace. Conversely, the weak solution satisfies $y^{\prime \prime }=xy-f$ in the sense of distributions. Since $xy-f\in L^2$, it belongs to $H^2(0,1)\cap H_0^1(0,1)$ and solves the equation almost everywhere.

Equivalently, the Ritz method minimizesover $V$. Indeed $J(y+w)-J(y)=a(w,w)/2$, since the mixed term is $a(y,w)-\ell (w)=0$. This proves existence and uniqueness of the minimizer directly from the weak solution, as well as its equivalence to the variational problem.

A quadratic form on a finite-dimensional real vector space $V$ is a function $q(v)=B(v,v)$ for some symmetric bilinear form $B$. A positive-definite quadratic form satisfies $q(v)>0$ for every $v\ne 0$.

Completing squares givesSet $u=x+y$, $v=y+z$, $w=z$. The corresponding ordered basis in the original coordinates isand the diagonal form is $u^2+v^2+2w^2$. All diagonal coefficients are positive, so the form is positive definite.

The positive uniform ellipticity condition from part (i) is needed here. It is not explicitly imposed in the printed preamble. Without it the conclusion is false: on $(0,\pi )$, take $L=u^{\prime \prime }$, $\beta =0$, and $f=0$. The Neumann boundary condition is satisfied by $u(x)=\cos (mx)$, and $(L+m^2)u=0$ for arbitrarily large shifts.

There is also an actual error in the PDF's supplied trace inequality: a nonzero constant has nonzero trace and zero gradient. Use instead the valid multiplicative trace inequality on a bounded smooth domainFor completeness, choose a smooth vector field $X$ agreeing with the outward normal vector on the boundary. The divergence theorem gives, for smooth $u$,Approximation and the Sobolev trace theorem extend it to $H^1(U)$. Thus the intended existence result remains true after correcting the hint.

Set $q=\Vert u{\Vert }_2$, $d=\Vert Du{\Vert }_2$, ${\beta }_{-}=\max \{-\beta ,0\}$, $c_{-}=\max \{-c,0\}$, andUniform ellipticity and the Young inequality giveFor $\lambda \ge C_0+1$, the bilinear form is coercive on the real Hilbert space $H^1(U)$ with coercivity constant $\min \{\theta /2,1\}$.

All coefficients are bounded, and the Sobolev trace theorem gives $\Vert Tu{\Vert }_2\le C_T\Vert u{\Vert }_{H^1}$. The Cauchy-Schwarz inequality applied to each volume and boundary term therefore proves $|B_{\lambda }[u,v]|\le C_{\lambda }\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Also $\ell (v)={\int }_{U}fv$ is a continuous linear functional, with $|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Every hypothesis of the Lax-Milgram theorem is now verified, and it gives

On the zero-boundary Sobolev space $H_0^1(U)$, the associated bilinear form isThe homogeneous Dirichlet boundary condition removes the boundary term, so $(Lu,v{)}_{L^2}=B[u,v]$ for sufficiently regular $u$ and $v\in H_0^1$.

Formal self-adjointness means that $L$ equals its formal adjoint, or equivalently $(L\varphi ,\chi )=(\varphi ,L\chi )$ for compactly supported smooth functions. HereWith the symmetric real principal coefficients in the question, $L=L^{\ast }$ is equivalent to $b=0$. It makes $B$ a symmetric bilinear form.

For the assertions in part (b), positivity must mean strict Dirichlet positivity:This is the positive-definite operator convention. The weaker positive semidefinite operator convention $B[u,u]\ge 0$ does not suffice: $L=-\Delta -{\lambda }_1$ has zero energy on a first Dirichlet eigenfunction. Under strict positivity, the compactness argument in part (b)(i) gives a uniform lower bound as well.

Use the decomposition into Lagrangian subspaces $V=V_{+}\oplus V_{-}$ from part (a). The bilinear form $\Omega$ pairs $V_{+}$ and $V_{-}$ nondegenerately: a vector in either summand annihilating the other already annihilates its own summand and hence all of $V$, so is zero.

Choose a basis $e_1,\dots ,e_n$ of $V_{+}$. The linear mapis an linear isomorphism. Let $f_j$ correspond to the dual basis of the $e_i$. Then $\Omega (e_i,f_j)={\delta }_{ij}$, and the pairings within each summand vanish. Thus

A bilinear form $B$ on $V$ is nondegenerate whenIn finite dimensions this is equivalent to nondegeneracy in the first argument.

Define the linear mapNondegeneracy makes ${\beta }_1$ injective. Since $V$ and its dual space have the same finite dimension, ${\beta }_1$ is an isomorphism. Similarly define ${\beta }_2(w)=B_2(-,w)$ and setThen $\alpha$ is linear and

If $w\in \ker \alpha$, this identity gives $B_2(v,w)=0$ for every $v$. Conversely, if $B_2(v,w)=0$ for every $v$, then $B_1(v,\alpha w)=0$ for every $v$, so nondegeneracy gives $\alpha w=0$. HenceThis is the representation of a bilinear form relative to a nondegenerate bilinear form.

Define on $H$ the bilinear formand the linear functionalThe Hardy inequality on an interval, Cauchy-Schwarz inequality, and the one-sided Poincaré inequality show that $a$ is a bounded bilinear form and that

For $v\in H$, $v(x)={\int }_0^x{v}^{\prime }(t)dt$, so Cauchy--Schwarz and Fubini's theorem giveHenceThus $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique $u\in H$ satisfying $a(u,v)=\ell (v)$ for every $v\in H$. This is precisely the unique weak solution described by the weak boundary value problem with an inverse-square potential.

For a finite-dimensional Lie algebra representation $\rho$ on $V$, the Trace form of a Lie algebra representation isWrite again $A=\rho (a)$, $B=\rho (b)$, and $C=\rho (c)=[A,B]$. The operator $C$ commutes with both $A$ and $B$. Direct use of the cyclic property of the trace givesIn the last line, cyclicity and $AC=CA$ turn $\mathrm{tr}(ABC)$ into $\mathrm{tr}(BAC)$. Thus the nonzero vector $c\in \mathfrak{g}$ is orthogonal to the basis $a,b,c$, and hence to all of $\mathfrak{g}$. The bilinear form is therefore degenerate.

Let $V$ be the real vector space with basis $(e_i{)}_{i\in I}$. The Coxeter Gram matrix defines the symmetric bilinear formIts diagonal entries are $2$. The Geometric representation of a Coxeter group is generated by the reflectionsEach has square one, and on $\mathrm{span}\{e_i,e_j\}$ the product $\sigma (x_i)\sigma (x_j)$ has order $m_{ij}$. The reflections therefore satisfy the Coxeter relations and define a group representation $\sigma :W\to GL(V)$.

All edges in the Type E(p,q,k) Coxeter graph are unlabelled, so the factor $a^2$ in part c is one. Separate the arm of length $p$ from the central vertex. The remaining two arms form a type $A_{q+k+1}$ chain, while deleting the central vertex leaves the disjoint type $A_q$ and type $A_k$ chains. Using $\det G(A_r)=r+1$ in the formula from part c yieldsThe associated bilinear form is degenerate exactly whenThe positive-integer solutions of $1/a+1/b+1/c=1$, up to permutation, are $(3,3,3)$, $(2,4,4)$, and $(2,3,6)$. Consequently the degenerate arm-length triples are the permutations ofFor every other allowed $(p,q,k)$ the determinant is nonzero, so the form is nondegenerate.

Use part (a) to put the four projective points atA quadratic polynomial vanishing at the first three has no square terms, and vanishing at the fourth givesThese form a two-dimensional vector space. Choose the two membersTheir coefficient triples are independent and both satisfy the relation. Their symmetric bilinear form matrices are of the formFor $(A,B,C)=(1,1,-2)$ and $(1,-2,1)$, both determinants are $-1/2$, so both selected conics are nonsingular. Any coefficient triple with zero sum is uniquely a linear combination of these two. Hence every conic through the four points has matrixUndoing the projective transformation carries the entire family back to the original four points, preserving nonsingularity. This is a pencil of conics through four points.

The Petrov-Galerkin method chooses an approximate solution $u_h$ in a finite-dimensional trial vector space $V_h$ and tests a weak formulation against a possibly different test space $W_h$. For a bilinear form $a$ and a linear functional $\ell$, it imposes $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in W_h$. Taking $W_h=V_h$ recovers the usual Galerkin method. Different test functions can alter stability and accuracy; the Galerkin orthogonality is then against the test space and need not be an orthogonal projection in an energy norm.

For a closed oriented $d$-dimensional manifold over a field $F$, the cup product pairingis a perfect pairing between complementary degrees. The cap product with the fundamental class is an isomorphism by Poincare duality, and the universal coefficient theorem for cohomology over $F$ identifies the complementary cohomology with the full dual space of the resulting homology. Evaluation is therefore a perfect pairing. Taking the top-degree part of the same formula defines a nondegenerate bilinear form on the full graded cohomology. A boundary requires relative groups and Poincare-Lefschetz duality instead.

The displayed bilinear form is the polar form of a quadratic form. In odd characteristic of a field, $Q(x)=B_Q(x,x)/2$ recovers the quadratic form; in characteristic of a field two, different forms can have the same alternating bilinear form. Over ${\mathbb{F}}_2$, two such forms differ by a linear map to the field, giving the affine parametrization of binary quadratic refinements.

A real symmetric bilinear form is positive-definite when $B(v,v)>0$ for every nonzero vector. It defines an inner product and the associated norm $\sqrt{B(v,v)}$. Symmetry is part of this definition: positive diagonal values by themselves do not imply symmetry.^[ref]

The identity matrix $A=\mathrm{diag}(1,1)$ gives $B(v,w)=v_1{w}_1+v_2{w}_2$ and $B(v,v)=v_1^2+v_2^2$. For $v=(3,4)$ this is $25$, giving norm $5$. A weighted example, $A=\mathrm{diag}(4,1)$, gives norm $\sqrt{4x^2+y^2}$: the first coordinate costs twice as much in length. Such inner products let geometry assign different weights to different directions.

Every positive-definite bilinear form is nondegenerate: a vector in its radical of a bilinear form has $B(v,v)=0$, so positivity forces $v=0$. The converse fails, as $\mathrm{diag}(1,-1)$ shows. Even positivity of the associated quadratic form does not force a general bilinear form to be symmetric: $A=\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$ has $v^{T}Av=x^2+y^2$ but $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$.

The complex Killing form is not positive-definite on the entire complex Cartan subalgebra; its restriction to the Euclidean subspace of a Cartan subalgebra is a real positive-definite form.

For a vector bundle carrying a smooth nondegenerate symmetric bilinear form of constant metric signature, a pseudo-orthonormal frame is a smooth local frame of a vector bundle whose Gram matrix is the constant diagonal signature matrix $\eta$. It identifies the metric bundle locally with a fixed model ${\mathbb{R}}^{p,q}$. Two such frames differ by a smooth map into the corresponding orthogonal group. Positive-definite signature is included as the ordinary orthonormal case. A global pseudo-orthonormal frame is a section of a fiber bundle of the orthonormal frame bundle and trivializes that principal bundle.

A quadratic form on a vector space over a field $F$ is a function $Q:V\to F$ satisfying $Q(av)=a^{2}Q(v)$ whose polar bilinear form of a quadratic form $B_Q(v,w)=Q(v+w)-Q(v)-Q(w)$ is a bilinear form. In finite dimension, it is represented in coordinates by a homogeneous degree-two polynomial, or by $Q(v)=v^{T}Av$ with a suitable matrix $A$.

When the characteristic is not two, one may choose $A$ to be a symmetric matrix and recover it from $B_Q/2$. In characteristic two, the symmetric-matrix restriction would lose examples: $Q(x,y)=xy$ has an upper-triangular representing matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, while $v^{T}Av$ for a symmetric matrix has no $xy$ cross term. Its polar form is $B_Q((x,y),(u,v))=xv+yu$, an alternating bilinear form. The distinction matters when studying quadratic forms over finite fields.^[ref]

For a symmetric bilinear form or an alternating bilinear form, the radical is the linear subspaceThese vectors are invisible to all pairings, and the bilinear form is nondegenerate exactly when this linear subspace is $\{0\}$. In coordinates, $\mathrm{rad}B=\ker A$ for both kinds of bilinear form, since $A^T=A$ or $A^T=-A$. Thus $\mathrm{diag}(1,0)$ has radical of a bilinear form $\mathrm{span}\{(0,1)\}$.

Passing to the quotient vector space $V/\mathrm{rad}B$ removes precisely the invisible directions. The rule $\overline{B}([v],[w])=B(v,w)$ is well-defined because adding a vector in the radical of a bilinear form to either input changes no pairing, and the resulting bilinear form is nondegenerate. For $B((x,y),(s,t))=xs$, this quotient vector space retains just the $x$ direction and its form is ordinary multiplication.^[ref]

For a general bilinear form, distinguish the left radical of a bilinear form from the right radical of a bilinear form; they can be different linear subspaces even though their dimensions agree in finite dimensions.

For $\mathcal{L}={\varphi }_{1,t}{\varphi }_{2,t}-c^2{\varphi }_{1,x}{\varphi }_{2,x}$, the real linear internal transformations preserving the density are $\mathrm{diag}(r,r^{-1})$ and $\left(\begin{array}{cc}0& r\\ r^{-1}& 0\end{array}\right)$ with $r\ne 0$. There are four connected components; the identity component is the displayed reciprocal rescaling with $r=e^s>0$. In the sum-and-difference basis it preserves a bilinear form of signature $(1,1)$, with the identity component isomorphic to that of the One-dimensional Lorentz group. Its Noether current in coordinates $(t,x)$ is $j^0={\varphi }_1{\varphi }_{2,t}-{\varphi }_2{\varphi }_{1,t}$ and $j^x=c^2({\varphi }_2{\varphi }_{1,x}-{\varphi }_1{\varphi }_{2,x})$. The two wave equations give ${\partial }_t{j}^0+{\partial }_x{j}^x=0$. The energy quadratic form is indefinite: the fields ${\varphi }_{\pm }=({\varphi }_1\pm {\varphi }_2)/\sqrt{2}$ have equal and opposite kinetic signs.

For continuous $P\ge 0,Q$, a scalar second variation $q=\int (P{\phi }^{\prime 2}+Q{\phi }^2)$ is strictly positive on nonzero fixed-endpoint Lipschitz variations if and only if every regularized form $q_{\eta }$ for $\eta >0$ has no conjugate point in $(a,b]$, and the original weak Jacobi boundary problem has no nonzero Lipschitz solution. The Jacobi positivity criterion applies to each regularized coefficient $P+\eta >0$. Letting $\eta$ decrease to zero gives $q\ge 0$. For a nonnegative quadratic form, $q(\phi )=0$ forces its mixed bilinear form with every test function to vanish, by considering $q(\phi +t\psi )$ for both signs of $t$. This is the weak Jacobi equation, so the kernel condition gives strict positivity. Necessity follows by the same argument in reverse. It distinguishes strict positivity from a uniform coercive estimate when the derivative coefficient degenerates.