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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.

For a finite-dimensional Hermitian matrix $A$, the Rayleigh quotient has precisely the range $[{\lambda }_{\min },{\lambda }_{\max }]$, where the endpoints are its extreme eigenvalues. The finite-dimensional spectral theorem writes a nonzero vector as $x={\sum }_j{c}_j{e}_j$ in an orthonormal eigenbasis, givingThus the quotient is a convex combination of eigenvalues. The bounds follow immediately, and equality at either endpoint means that $x$ lies in the corresponding extreme eigenspace, including any repeated eigenvalue. Every intermediate value occurs: for orthonormal extreme eigenvectors, $x=\sqrt{1-t}{e}_{\min }+\sqrt{t}{e}_{\max }$ gives $R_A(x)=(1-t){\lambda }_{\min }+t{\lambda }_{\max }$ for $0\le t\le 1$. If all eigenvalues are equal, the quotient is constant.

The Hermitian hypothesis matters. The real matrix $B=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has only the eigenvalue $1$, but $(1,1)B(1,1{)}^T/2=3$. Its quotient is governed by the symmetric part of a matrix $(B+B^T)/2$, not by a real orthonormal eigenbasis of $B$. A nonsymmetric matrix therefore need not obey the displayed eigenvalue bounds.

Let $x_1,\dots ,x_N$ be the interior nodes, and let ${\varphi }_1,\dots ,{\varphi }_N$ be the nodal basis of the common conforming finite element space, with ${\varphi }_j(x_i)={\delta }_{ij}$ and zero boundary values. These piecewise-linear hat functions give the expansionsTesting the first equation with ${\varphi }_i$ and then the second equation with ${\varphi }_i$ gives the linear systemHere $S$ is the diffusion stiffness matrix and $M$ the mass matrix. Their entries are symmetric in $i,j$, so both are symmetric matrices.

For $c\in {\mathbb{R}}^N$, put $v_c={\sum }_j{c}_j{\varphi }_j$. The two quadratic forms areIf $c\ne 0$, the linear independence of the basis ensures $v_c\ne 0$, so $c^{\mathsf{T}}Mc>0$. If $c^{\mathsf{T}}Sc=0$, the Poincaré inequality and zero Dirichlet boundary condition force $v_c=0$, contradicting $c\ne 0$. Thus both $S$ and $M$ are positive-definite matrices.

For $X=(p,q)\in {\mathbb{R}}^N\times {\mathbb{R}}^N$, symmetry of $M$ cancels the cross terms:ThereforeThis is the positivity of a quadratic form intended by the last sentence of the question. For $N>0$, $A$ is a nonsymmetric matrix, since $M\ne 0$. Under the convention that “positive-definite matrix” includes symmetry, that last sentence needs this qualification: the symmetric part of a matrix is positive definite. The quadratic lower bound for a nonsymmetric matrix still implies that $A$ is an invertible matrix: $AX=0$ would give $X^{\mathsf{T}}AX=0$, hence $X=0$.

For $A^{\prime }$, the centre and constant remain $\mathbf{a}=(-1,0,0)$ and $C=3$. Its eigenvalues are $3,1,-1$, so rotated coordinates put the surface in the form $3u^2+w^2-y^2=3$. This is a one-sheet hyperboloid. Since $3\Vert \mathbf{x}{\Vert }^2\ge {\mathbf{x}}^{\mathsf{T}}{A}^{\prime }\mathbf{x}=3$, the nearest centre-to-surface distance is at least one, attained in the eigen-direction with eigenvalue three. The coordinate $y$ is unbounded, because arbitrarily large $|y|$ can be balanced by $u$ or $w$. Thus the nearest distance is one, and there is no maximum value of $y$.

For the printed nonsymmetric $A^{\prime \prime }$, use the symmetric part of a matrix:The skew-symmetric part contributes zero to this scalar. The same centre solves $2S\mathbf{a}+\mathbf{b}=0$, and again $C=3$. Sylvester's positive principal minors are $2,2,5/2$, so this surface is an ellipsoid.

Let ${\lambda }_{\ast }$ be the largest eigenvalue of $S$. Its characteristic equation is $4{\lambda }^3-20{\lambda }^2+27\lambda -10=0$, and the Rayleigh quotient givesFor the maximum $y$, the support function of a centred ellipsoid gives the bound below. Equivalently, apply the Cauchy-Schwarz inequality in the $S$-weighted inner product: $y^2\le 3{\mathbf{e}}_y^{\mathsf{T}}{S}^{-1}{\mathbf{e}}_y$. The inverse entry is ${\mathbf{e}}_y^{\mathsf{T}}{S}^{-1}{\mathbf{e}}_y=6/5$, and equality is attained for a centred vector proportional to $S^{-1}{\mathbf{e}}_y$. Thus

Using the eigenvalues of the raw nonsymmetric $A^{\prime \prime }$ would be incorrect: it is its symmetric part that determines the surface and its Euclidean geometry.

A local Maxwellian can be written as $\mathcal{M}=\exp (a+b\cdot v+c|v{|}^2)$, with $c<0$ and coefficients depending on $(t,x)$. Its logarithm is a linear combination of collision invariants, so ${\mathcal{M}}^{\prime }{\mathcal{M}}_{\ast }^{\prime }=\mathcal{M}{\mathcal{M}}_{\ast }$ and $Q(\mathcal{M},\mathcal{M})=0$ at each spatial point.

Comparing powers of $v$ in $({\partial }_t+v\cdot {\nabla }_x)\log \mathcal{M}=0$ givesHere $\mathrm{sym}\nabla b$ is the symmetric part of a matrix formed by the coefficients of the quadratic polynomial. For constants $A,\alpha ,\beta >0$, choose $a=\log A-\alpha |x{|}^2$, $b=2\alpha tx$ and $c=-\beta -\alpha t^2$. All four coefficient equations hold. The resulting expanding Gaussian solution of the Boltzmann equation isDirect verification is particularly simple: $x-tv$ and $v$ are constant along free characteristic curves, so the transport derivative is zero. At fixed $(t,x)$ its logarithm is the displayed Maxwellian quadratic, so the collision term is also zero. Its spatial dependence is nonconstant, it is positive and rapidly decaying, and its Gaussian velocity integral is finite because $\beta +\alpha t^2>0$. It is not a compactly supported example, nor does this final part require one.

Differentiate the quadratic energy:For $x_1\ne 0$, put $r=x_2/x_1$. Immediate energy growth occurs precisely when ${\lambda }_2{r}^2+r+{\lambda }_1>0$. With both eigenvalues negative, such directions exist iff $1-4{\lambda }_1{\lambda }_2>0$. Writing $\delta =\sqrt{1-4{\lambda }_1{\lambda }_2}$, the growing initial conditions are the two opposite open wedgesIf $x_1=0$, energy decreases. No energy growth at any time, for any initial condition, is possible exactly whenIndeed, this makes the symmetric part of a matrix $H$ a negative semidefinite matrix, so $\dot{E}\le 0$ along every trajectory. If the product is smaller, the wedges already supply counterexamples at $t=0$. This is the instantaneous energy-growth criterion for a linear system.

Write $N=(n_1,n_2,n_3)$. Restriction to the plane $z=0$ givesThis is an equation of degree at most two in two Cartesian coordinates, the general algebraic form of a conic, with possible degenerate conics and an empty set of real points. No determinant hypothesis is needed for this substitution.

For an arbitrary plane, choose a point ${\mathbf{x}}_0$ and an orthonormal basis $\mathbf{e},\mathbf{f}$ of its direction linear subspace. Put $B=(\mathbf{e}\mathbf{f})$ and $\mathbf{x}={\mathbf{x}}_0+B\mathbf{\xi }$, where $\mathbf{\xi }=(s,t{)}^T$. With $S=(M+M^T)/2$, the symmetric part of a matrix, the restricted equation isThis proves the plane-section result whenever “conic” allows the appropriate degeneracies. A rigid motion taking the chosen plane to $z=0$ gives the same argument.

The source's assumption $\det M\ne 0$ does not guarantee a nondegenerate conic. Even with $M=I$ and $f=x^2+y^2+z^2-1$, the section $z=1$ is a single point, and $z=2$ has no real points. With the nonsingular symmetric matrix giving $f=2xz+y^2$, the section $z=0$ is the repeated straight line $y=0$.

There is a further literal omission: the source does not assume $M$ is symmetric. Forone has $\det M=2$ but $f=2xz$. Thus the section $z=0$ is the entire plane, rather than a conic. Only $S$ controls the quadratic form; invertibility of an arbitrary $M$ cannot remove this exception. If the intended hypothesis is that $M$ is symmetric and nonsingular, the quadratic restriction to a plane cannot vanish identically: otherwise $B^{T}MB=0$, so $M$ would map a two-dimensional linear subspace injectively into its one-dimensional orthogonal complement, which is impossible. Degenerate and empty sections still remain possible.

The symmetric part of a matrix isIt satisfies $S^{\mathsf{T}}=S$. The complementary term $(A-A^{\mathsf{T}})/2$ is a skew-symmetric matrix.

The strict positivity of a quadratic form on every nonzero vector. A real symmetric matrix has positive definiteness exactly when its associated quadratic form does, making it a positive-definite matrix. A Hermitian form is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.

Think of a positive-definite quadratic form as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real symmetric matrixis a positive-definite matrix because the two squares can vanish together only at $(0,0)$. The mixed term couples the coordinates, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive level sets are ellipses with axes along the eigenvectors, rather than along the original coordinates.

The real and complex settings require different symmetry assumptions. For a real symmetric matrix use $q(x)=x^{T}Ax$ with $x\in {\mathbb{R}}^n$; for a complex Hermitian matrix use $q(z)=z^{\ast }Az$ with $z\in {\mathbb{C}}^n$, where $z^{\ast }$ is the conjugate transpose. The latter expression is real, so strict positivity is meaningful. A complex quadratic form such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A Hermitian form uses complex conjugation, which changes this example to $|z_1{|}^2+|z_2{|}^2$.

For a real matrix $M$ without a symmetry hypothesis, its quadratic form depends only on its symmetric part of a matrix:Positive eigenvalues of $M$ alone do not establish positive definiteness of this quadratic form. For example, $M=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has both eigenvalues equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the symmetric part of a matrix, not the spectrum of an arbitrary matrix.^[ref]

Let $B$ be a real square matrix and suppose $x^{T}Bx\ge c|x{|}^2$ for every real vector $x$, with $c>0$. The Cauchy-Schwarz inequality gives $x^{T}Bx\le |x||Bx|$, hence $|Bx|\ge c|x|$ for all $x$. In particular the kernel is zero, so $B$ is an invertible matrix. Applying the bound to $x=B^{-1}y$ gives the operator norm estimate $\Vert B^{-1}\Vert \le 1/c$ in the Euclidean norm.

Symmetry of $B$ is unnecessary: $x^{T}Bx=x^{T}Hx$ for its symmetric part of a matrix $H=(B+B^T)/2$. Thus the hypothesis is exactly a positive lower bound on that symmetric matrix, even when the antisymmetric part is nonzero.

For a real orthogonal matrix $A$, $A^{\mathsf{T}}=A^{-1}$. If $Av=\lambda v$ for a nonzero complex eigenvector, then $A^{\mathsf{T}}v={\lambda }^{-1}v$. Therefore the same eigenvector has eigenvalue $(\lambda +{\lambda }^{-1})/2$ for the symmetric part of a matrix $(A+A^{\mathsf{T}})/2$. Orthogonality implies $|\lambda |=1$, so writing $\lambda =e^{i\theta }$ gives $\cos \theta$. The complete complex orthonormal eigenbasis of an orthogonal matrix shows that all the eigenvalues of its symmetric part lie in $[-1,1]$.