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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.

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]

On a rectangular grid with zero Dirichlet boundary conditions, the nine-point Poisson equation matrix has diagonal $10/3$, axial off-diagonal entries $-2/3$, and diagonal-neighbor entries $-1/6$. Its quadratic form is the sum of squared differences over internal links, weighted by $2/3$ or $1/6$, plus the weighted squares of values linked to the fixed zero boundary. The connected axial grid forces a zero quadratic form to have all entries zero. This proves positive-definite matrix in every ordering. The Jacobi method converges because its nonnegative iteration matrix has row sums at most one, with a strict deficit at boundary-adjacent vertices; a hypothetical unit-modulus eigenvector would propagate its maximum modulus to such a deficit row, which is impossible.

Vary the coupling in the opening example. The eigenvectors $e_{+}=(1,1{)}^T/\sqrt{2}$ and $e_{-}=(1,-1{)}^T/\sqrt{2}$ form a fixed orthonormal eigenbasis, with eigenvalues $3+t$ and $3-t$. Writing a vector as $u{e}_{+}+v{e}_{-}$ givesFor $|t|<3$, both directions cost energy and $A_t$ is a positive-definite matrix. At $t=3$, the quadratic form is $6u^2$: the entire line spanned by $e_{-}$ has zero energy, and $A_t$ is a singular positive semidefinite matrix. At $t=-3$, the flat line is instead spanned by $e_{+}$. For $|t|>3$, the quadratic form is an indefinite quadratic form; one direction lowers the energy without bound as its displacement grows. At $t=4$, for instance, $q_t(e_{+})=7$ while $q_t(e_{-})=-1$.

The geometric transition is precise. For $|t|<3$, $q_t=1$ is an ellipse. At $t=3$, it consists of the two parallel lines $u=\pm 1/\sqrt{6}$, extending freely in the zero-energy direction. For $t>3$, it is a hyperbola. The boundary case retains nonnegative energy but loses both the bounded positive level sets and the unique zero-energy vector.

A self-adjoint operator $A$ is positive definite when $⟨Av,v⟩>0$ for every nonzero vector $v$ in its domain. In finite dimension it is invertible and has positive eigenvalues. In infinite dimension, strict positivity of individual quadratic-form values need not give a bounded inverse; the stronger uniform bound $⟨Av,v⟩\ge c\Vert v{\Vert }^2$ with $c>0$ supplies coercivity.

For maps $\beta :U\to V$ and $\gamma :V\to W$ between finite-dimensional inner product spaces, with $\mathrm{im}\beta =\ker \gamma$, the operator $\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma$ is positive definite. Its quadratic form is $\Vert {\beta }^{\ast }v{\Vert }^2+\Vert \gamma v{\Vert }^2$, whose zero set is $\mathrm{im}\beta \cap (\mathrm{im}\beta {)}^{\perp }=\{0\}$. This uses the adjoint operator relation $\ker {\beta }^{\ast }=(\mathrm{im}\beta {)}^{\perp }$.

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

Let $H$ be a symmetric positive-definite matrix, $a\ne 0$, and $b\in \mathbb{R}$. Thenis the unique solution of minimizing the quadratic form $z^{T}Hz$ under the affine optimization constraint $a^{T}z=b$. The minimum value is $b^2/(a^T{H}^{-1}a)$.

The denominator is positive because $H^{-1}$ is a positive-definite matrix. The Lagrange multiplier equations $2Hz=\lambda a$ give the displayed candidate and verify the constraint. For a global proof, any feasible perturbation $u=z-z_{\ast }$ satisfies $a^{T}u=0$. HenceThis is positive for every nonzero $u$. The result also handles several independent quadratic blocks by taking $H$ to be their block diagonal matrix and concatenating their variables.

The sums $P+N$ of a real positive semidefinite matrix $P$ and a symmetric nonnegative matrix $N$ form a convex cone inside the copositive cone. Both terms have nonnegative quadratic forms on the nonnegative orthant. The Horn copositive matrix shows that the inclusion is strict in dimension five; the sum of squares criterion for a biquadratic form explains this cone's relation to semidefinite programming.

Let $B$ be a finite-dimensional Hermitian matrix and $a<b$ real. The spectral theorem gives $v={\sum }_j{c}_j{x}_j$ in an orthonormal eigenbasis, soIf the spectrum avoids the closed interval, every factor product is positive. Consequently the quadratic form is positive for $v\ne 0$ and zero for $v=0$. More precisely it is bounded below by ${\min }_j[({\lambda }_j-a)({\lambda }_j-b)]\Vert v{\Vert }^2$. This is a finite-dimensional instance of testing a spectral gap with a polynomial in the operator.

For a vector parameter with finite second moments and posterior mean $m_y$, the posterior covariance matrix records its uncertainty and linear dependence after conditioning on the observations. Its diagonal entries are posterior variances. In a regular locally flat-prior normal distribution approximation it is the inverse observed information. A quadratic form has posterior expectation equal to the form at the posterior mean plus the trace of its matrix times this covariance.

The principal axes of an ellipse are the two perpendicular symmetry axes through its center of an ellipse. For a symmetric positive-definite matrix $M$, its orthonormal eigenvectors give those directions and its positive eigenvalues ${\lambda }_1,{\lambda }_2$ give semiaxis lengths ${\lambda }_1^{-1/2},{\lambda }_2^{-1/2}$. This follows by orthogonal diagonalization of the defining quadratic form. In ${\xi }^2/A^2+{\upsilon }^2/B^2=1$, the axes are already horizontal and vertical, with semiaxes $|A|,|B|$ and full lengths $2|A|,2|B|$. Equal semiaxes give a circle, for which every perpendicular pair through the center can serve.

For a positive-definite real quadratic form $x^{T}Mx=c>0$, an orthonormal eigenvector of $M$ with eigenvalue $\lambda$ gives a principal semi-axis of length $\sqrt{c/\lambda }$. If an eigenvalue is repeated, any orthonormal basis of its eigenspace gives equally valid principal axes. Thus an angle attached to “the longest axis” need not be unique when the smallest eigenvalue is repeated. For example, the form with eigenvalues $6,1,1$ and short direction $(2,0,1)$ has every direction in $2x_1+x_3=0$ as a longest-axis direction. Its unit directions $(0,1,0)$ and $(1,0,-2)/\sqrt{5}$ make different angles, $0$ and $\arcsin (1/\sqrt{5})$, to the plane $x_1=0$.

Write a real quadratic polynomial as $x^{T}Hx+2q^{T}x+c$, with $H$ symmetric. If $H$ is invertible, translation by $x_0=-H^{-1}q$ removes the linear term. An orthonormal eigenbasis then makes the remaining quadratic form diagonal. Eigenvalue signs and the translated constant determine the real geometry, such as an ellipsoid or a one-sheet hyperboloid. A singular $H$ requires separate treatment of linear terms in its null directions; a center need not exist.

A projective conic is the zero locus in a projective plane of a nonzero homogeneous degree-two polynomial. The equation is considered up to multiplication by a nonzero scalar. Over the complex numbers, a symmetric matrix for the quadratic form has rank three, two, or one: these give a nonsingular conic, two distinct projective lines, or one line counted twice. Diagonalization of a quadratic form and rescaling its nonzero diagonal entries prove this classification. The smooth model $XZ-Y^2=0$ is parametrized by the degree-two Veronese map $[s:t]\mapsto [s^2:st:t^2]$.

The quadratic form of a matrix only sees its symmetric part. Thus $\lambda |\xi {|}^2\le {\xi }^{T}A\xi \le \Lambda |\xi {|}^2$ need not bound the skew part of a measurable coefficient matrix. For $A=\left(\begin{array}{cc}1& k\\ -k& 1\end{array}\right)$ the quadratic form is exactly $|\xi {|}^2$, regardless of $k$. On a square, take $k(x)=1/|x_1|$ off the line $x_1=0$, with a finite arbitrary value on that line. Two compactly supported smooth functions that equal $x_2$ and $x_1$ near the origin have $(A\nabla u)\cdot \nabla v=k$ there; the purported energy pairing is not integrable. A Lax-Milgram theorem proof therefore requires bounded full coefficients or another hypothesis that makes the bilinear form bounded. Symmetry supplies the missing matrix bound from the quadratic upper bound.

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]

Let $F$ generate a nonsingular smooth flow on a closed manifold. Suppose a continuous nondegenerate indefinite quadratic form on the transverse vector bundle $TN/\mathbb{R}F$ has constant signature, is differentiable along the linearized flow, and has continuous positive-definite derivative along that flow. Then the flow is an Anosov flow. Compactness makes the positivity uniform; the positive and negative cones give the unstable and stable transverse directions. The flow direction must be quotiented out: its derivative cannot satisfy strict positivity.

Proof. For a skew-symmetric matrix, the scalar $x^{\top }Ax$ equals its transpose $x^{\top }{A}^{\top }x=-x^{\top }Ax$, so it vanishes. Conversely, substituting standard basis vectors $e_i$ into the vanishing quadratic form gives $a_{ii}=0$. Substituting $e_i+e_j$ gives $a_{ij}+a_{ji}=0$. Thus $A^{\top }=-A$. In particular, a real quadratic form depends only on a matrix's symmetric part.

A quadratic functional on a real vector space combines a quadratic form $B(v,v)$, a linear functional $\ell (v)$ and a constant. If $B$ is a symmetric bounded bilinear form, its first variation is $DI(v)[w]=2B(v,w)-2\ell (w)$. For a strictly positive $B$, an existing stationary point is the unique minimizer by the quadratic variational principle for a symmetric positive operator.

Optimization of a quadratic form or quadratic-plus-linear objective over specified constraints. Convexity depends on the matrix and feasible domain; discrete restrictions lead to binary quadratic optimization.