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A vector $v$ can be included in a diagonal basis for a real quadratic form precisely when $v\ne 0$ and either its quadratic value is nonzero or it lies in the radical of a bilinear form. A nonzero quadratic value gives the direct splitting $\mathbb{R}v\oplus v^{\perp }$ and induction. A nonzero radical vector can be included in a radical basis. Conversely, an isotropic vector in a diagonal basis is orthogonal to every basis vector, and must lie in the radical.

A bilinear form is degenerate if some nonzero vector pairs to zero with every vector on at least one side. For a symmetric bilinear form or an alternating bilinear form, this says its radical of a bilinear form contains a nonzero vector. On a finite-dimensional vector space, it is equivalent to a singular matrix of a bilinear form, or $\det A=0$.

For $A=\mathrm{diag}(1,0)$, $B((x,y),(s,t))=xs$. The nonzero vector $(0,1)$ is invisible to every pairing, and changing $y$ never changes $B((x,y),w)$. The quadratic form $q=x^2$ therefore has two parallel lines $x=\pm 1$ as its unit level set. This is a lost direction, a stronger condition than being an isotropic vector: with $A=\mathrm{diag}(1,-1)$, $B((1,1),(1,1))=0$ but $B((1,1),(1,0))=1$.

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.

In a two-dimensional nondegenerate Hermitian form over a quadratic finite field, choose an orthonormal basis $v,w$ and $a$ with $a\bar{a}=-1$. Then $e=v+aw$ is an isotropic vector. Choose $g$ with $B(e,g)=1$, and choose $c$ with $c+\bar{c}=B(g,g)$. The vector $f=g-ce$ satisfies $B(e,f)=1$ and $B(f,f)=0$. Thus $e,f$ give the displayed matrix, including in characteristic two. Iterating on nondegenerate orthogonal complement for a sesquilinear form leaves only a unit-norm line in odd dimension.

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.

The group of invertible linear maps preserving a specified quadratic form over a finite field. In characteristic two one must preserve the quadratic form itself, not merely its polar alternating bilinear form. For a nonsingular form, any two nonzero isotropic vectors can be incorporated into hyperbolic pairs and standard adapted bases, producing an isometry between them.

No. On the real vector space $V={\mathbb{R}}^2$, use the symmetric bilinear formIts matrix of a bilinear form is $\mathrm{diag}(1,-1)$, whose determinant is $-1\ne 0$, so it is a nondegenerate bilinear form. LetA vector $(s,t)$ belongs to the bilinear orthogonal complement $U^{\perp }$ exactly when $s-t=0$. HenceIn particular, the claimed direct sum fails. The isotropic vector $(1,1)$ has zero self-pairing, although it pairs nontrivially with other vectors of $V$.

The dimension identity in part (i) counts both vector subspaces but does not force their intersection of vector subspaces to be zero. For a symmetric bilinear form, $U\cap U^{\perp }$ is the radical of a bilinear form restricted to $U$. Thus the precise criterion isIndeed, trivial intersection of vector subspaces and part (i) give a sum of vector subspaces of dimension $\dim V$. A positive-definite bilinear form would guarantee this for every $U$; symmetry alone does not.

We construct a Hermitian hyperbolic plane over a finite field inside each successive two-dimensional piece. By the field norm surjectivity proved above, choose $a\in {\mathbb{F}}_{q^2}$ with $a\bar{a}=-1$. From two vectors in an orthonormal basis, form $e=v_1+a{v}_2$. Then $B(e,e)=1+a\bar{a}=0$, so $e$ is a nonzero isotropic vector. Nondegeneracy supplies $g$ with $B(e,g)=1$; rescaling the second argument achieves this normalization. In fact $g=v_1$ already works in this two-dimensional piece. By quadratic finite-field norm and trace surjectivity, choose $c$ with $c+\bar{c}=B(g,g)$ and put $f=g-ce$. Direct expansion givesThus $e,f$ are linearly independent, and their Gram matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, which is invertible even in characteristic two. Their span is nondegenerate, so its orthogonal complement for a sesquilinear form is again nondegenerate. Repeat the construction there. In even dimension this exhausts $V$; in odd dimension the final one-dimensional complement has a nonzero diagonal value, which can be normalized to one using the field norm. Calling its vector $d$, we obtainwith the $d$ terms present only in odd dimension. This proves all the requested pairings and the basis assertion.

The printed hint needs a qualification. If $X^2-X-1$ is irreducible over ${\mathbb{F}}_q$, its two roots are conjugate, so $\zeta +\bar{\zeta }=1$ and $\zeta \bar{\zeta }=-1$; the hinted vectors then do work. They need not work when it splits. For example, over ${\mathbb{F}}_{11}$ its roots are $4,8$, fixed by conjugation in ${\mathbb{F}}_{121}$. For either choice the proposed $e$ has norm $1+{\zeta }^2$, respectively $6$ or $10$, rather than zero. The norm-and-trace construction above proves the assertion without that restriction.

A finite two-transitive group action is transitive on ordered pairs of distinct points. Equivalently, the action is transitive and a point stabilizer subgroup is transitive on the remaining points. Its rank of a transitive permutation group is two, and the orbit-stabilizer theorem gives $n(n-1)\mid |G|$ for degree $n$. Such an action is primitive: a block containing a point and another point must, by the transitivity of the stabilizer subgroup, contain every point.

These groups illustrate two quite different sources of symmetry. The structural theorem for finite two-transitive groups says that they have either a regular elementary-abelian normal subgroup, giving the affine case, or a nonabelian simple normal subgroup, giving the almost simple case. The full classification of the second case depends on the classification of finite simple groups; the useful point here is to describe the constructions, actions, and representative families rather than treat two-transitivity as an abstract property of a group without a specified action.

In the affine case let $V$ be a finite vector space over ${\mathbb{F}}_p$ and let $H\le GL(V)$ act transitively on $V\setminus \{0\}$. The affine two-transitive groupis transitive by translations, and the stabilizer subgroup of zero is exactly $H$, so it is two-transitive. Conversely, if a two-transitive group has an abelian minimal normal subgroup $V$, normal-subgroup orbits and primitivity make $V$ transitive. An abelian transitive faithful permutation subgroup is regular: an element fixing one point commutes with every element moving that point and hence fixes them all. Minimal normality makes $V$ elementary abelian, since its prime-primary subgroups and its subgroup of pth powers are characteristic. Conjugation by the stabilizer subgroup acts faithfully as linear automorphisms of $V$, and regularity gives $G=V⋊G_0$. This recovers precisely the construction above.

Many examples follow at once. The group $AG{L}_d(q)$ acts on the $q^d$ vectors of ${\mathbb{F}}_q^d$; $G{L}_d(q)$ sends any nonzero vector to any other by a basis map. For $d\ge 2$, the subgroup $AS{L}_d(q)$ also gives a two-transitive action, since the determinant of a basis map can be adjusted on a second basis vector while retaining the prescribed first vector. The affine group with linear part $S{p}_{2m}(q)$ gives another family: the nonzero-vector transitivity was proved using symplectic transvections in part 4(ii). Here the additive group of ${\mathbb{F}}_q^d$, for $q=p^f$, is an elementary abelian p-group of rank $fd$.

In particular $AG{L}_1(q)$ consists of $x\mapsto ax+b$ with $a\ne 0$. Given distinct $x,y$ and distinct $u,v$, the equations $ax+b=u$, $ay+b=v$ have the unique solution $a=(v-u)/(y-x)$, $b=u-ax$. Thus this action is sharply two-transitive, of degree $q$ and order $q(q-1)$. Adding field automorphisms gives $A\Gamma L_1(q)$, still two-transitive, with elements $x\mapsto a{x}^{p^i}+b$.

An almost simple group lies between a nonabelian simple group $T$ and its automorphism group. One must still specify an action: a group can have several permutation representations with different degrees or different transitivity properties. The natural actions of $S_n$ for $n\ge 2$ and $A_n$ for $n\ge 4$ are the most familiar examples of a two-transitive group action. For $n\ge 5$, the simple group $A_n$, a normal subgroup, places these groups in the almost simple case; the smaller two-transitive examples belong to the affine case. The symmetric group sends any ordered distinct pair to any other; for the alternating group one can adjust an odd map by swapping two letters outside the target pair. Thus both are two-transitive in the indicated ranges.

The projective linear examples are just as concrete. The groups $PS{L}_n(q)$ and $PG{L}_n(q)$ act two-transitively onpoints, by the basis and determinant adjustment argument of part 4(i). Their field-automorphism extensions are also two-transitive. On the projective line, $PG{L}_2(q)$ is sharply three-transitive: three distinct points determine a unique Möbius transformation. Its degree is $q+1$ and its order is $q(q^2-1)$. For even $q$, $PS{L}_2(q)=PG{L}_2(q)$; for odd $q$ the special projective group has half that order and remains two-transitive, but cannot be three-transitive. In projective dimension at least two, triples of distinct points may be collinear or noncollinear, so the natural action is not three-transitive.

The projective special unitary group over a finite field $PS{U}_3(q)$, for $q\ge 3$, acts on the isotropic points of a space with a nondegenerate Hermitian form of dimension three over ${\mathbb{F}}_{q^2}$. Its degree and order areFor two distinct isotropic lines the Hermitian form pairing is nonzero, since a three-dimensional nondegenerate Hermitian space has maximal totally isotropic dimension one. Scale their representatives to form a hyperbolic pair and extend to a basis adapted to the Hermitian form. The corresponding isometry sends one ordered pair to another, and a norm-one scalar on the orthogonal one-dimensional complement adjusts its determinant without moving the two lines. Thus the special projective unitary action is two-transitive.

Two further families of simple groups have analogous rank-two actions:The Suzuki group of Lie type and the Ree group of type G2 have stabilizer subgroups containing subgroups acting regularly on all the other points in these actions; this gives two-transitivity. For example $Sz(8)$ has degree 65 and order 29120. These are exceptional Lie-type families, distinct from the preceding linear and unitary constructions.

The symplectic group over a finite field in characteristic two has a particularly illuminating pair of two-transitive actions which are not their projective-vector actions. For $m\ge 3$, $S{p}_{2m}(2)$ acts on the two classes of quadratic forms whose polar form is the given alternating bilinear form, distinguished by the Arf invariant of a quadratic form. Their degrees areTransitivity on each class follows by choosing symplectic bases giving the standard quadratic-form normal forms. Fix a form $Q$. Every other form with the same polar form is $Q_a(x)=Q(x)+B(a,x)$, and has the same Arf invariant of a quadratic form exactly when $Q(a)=0$. The stabilizer subgroup of $Q$, an orthogonal group over a finite field, is transitive on its nonzero isotropic vectors: extend such a vector to a hyperbolic pair and map adapted quadratic-form bases. It is therefore transitive on the other forms of the same type. This proves two-transitivity of both actions. In particular $S{p}_6(2)$ has actions of degrees 28 and 36. The degree-six action of $S{p}_4(2)\cong S_6$ in part 4(ii) is the smaller-dimensional precursor.

Finally the Mathieu groups supply exceptional examples with even higher transitivity:In particular all five natural actions are two-transitive. Together these examples exhibit affine spaces, projective and Hermitian geometries, symplectic quadratic forms, and exceptional finite simple groups as different sources of the same pair symmetry.

The organizing distinction is affine versus almost simple; the action, its degree, and its stabilizer subgroup are essential data. The explicitly verified affine, symmetric, alternating, projective, unitary, and symplectic actions already give many infinite families, while the exceptional families show why no single matrix construction covers every example.

Associate to $f$ the symmetric bilinear form $B(u,v)=\frac{1}{2}(f(u+v)-f(u)-f(v))$. If $f$ is identically zero, every basis is diagonal. Otherwise choose $v$ with $f(v)\ne 0$. Every $w$ has the decompositionwhose second term lies in $v^{\perp }$. Thus $V=\mathbb{R}v\oplus v^{\perp }$, with no cross term in $f$. Induction on dimension gives a diagonal basis. Rescaling its nonzero entries reduces the form towith $k$ additional zero coordinates. Define the rank of a quadratic form $r=p+q$ and its signature $s=p-q$. Rank is intrinsic, since it is the rank of $v\mapsto B(v,\cdot )$, equivalently $n-\dim \mathrm{rad}B$. Also $p$ is the greatest dimension of a subspace on which the restriction of $f$ is a positive-definite quadratic form: any such subspace intersects the negative-plus-zero coordinate space only at zero, so its dimension is at most $p$, while the positive coordinate space attains $p$. The corresponding negative statement identifies $q$. This proves the basis independence of $r$ and $s$, the Sylvester's law of inertia conclusion.

For a totally isotropic subspace $W$ on which $f$ vanishes, polarization identity also makes $B$ vanish on $W\times W$. Quotient by the radical of a bilinear form. In the resulting nondegenerate form, projection of an isotropic subspace to either the positive or negative coordinate space is injective, so its dimension is at most $\min (p,q)$. Consequently $\dim W\le k+\min (p,q)$. The radical of a bilinear form together with the vectors $e_i^{+}+e_i^{-}$, $1\le i\le \min (p,q)$, attains this bound. The maximum dimension of a totally isotropic subspace is thereforeThe criterion for membership in a diagonal basis follows similarly: a nonzero vector can belong to a diagonal basis if it has nonzero quadratic value, by the same orthogonal splitting, or if it lies in the radical of a bilinear form, by choosing a radical basis containing it. An isotropic vector in a diagonal basis must be orthogonal to every basis vector, and hence must lie in that radical. For the specified form, its radical is the $z$-axis, giving exactlyThe zero vector never belongs to a basis.

Let $f_1,f_2\in V^{\ast }$ satisfy $f_1(v)f_2(v)=0$ for every vector $v$. If $f_1=0$ there is nothing to prove. Otherwise choose $u$ with $f_1(u)\ne 0$. For every $v\in \ker f_1$ and every real number $t$,so the hypothesis gives $f_2(u+tv)=0$. Taking two values of $t$ shows that $f_2(u)=f_2(v)=0$. Sincewe obtain $f_2=0$. Thus one of the two linear functionals is zero.

Sylvester's law of inertia says that, in a suitable basis, every real quadratic form isThe rank is $r=p+m$, and the signature is $\sigma =p-m$; both are independent of the chosen basis.

If $q=f_1{f}_2$, its polar form has image contained in $\mathrm{span}\{f_1,f_2\}$, so $r\le 2$. If $r=2$, the two functionals must be independent and, after taking their sum and difference, $q$ is a difference of two squares. Hence $\sigma =0$. If $r=1$, then $|\sigma |=1$. In every case,Conversely, this inequality leaves onlyup to a linear change of coordinates. These factor respectively as $0$, $(\pm x_1)x_1$, and $(x_1-x_2)(x_1+x_2)$. Therefore

Finally suppose that $q$ takes both positive and negative values. Sylvester's law supplies normalized basis vectorswith $q(e_i)=1$, $q(f_j)=-1$, and the $z_l$ spanning the radical. The $p+m+k$ vectorsare linearly independent and all satisfy $q(v)=0$. They form the required basis of isotropic vectors.

The symmetric bilinear form associated with a real quadratic form $Q$ is obtained by polarization identity:In coordinates $Q(x)=x^{T}Ax$, replacing $A$ by its symmetric part does not change $Q$, and the formula gives $phi(u,v)=u^{T}Av$; this proves existence. A symmetric form is positive semidefinite when $phi(v,v)\ge 0$ for every $v$, and positive definite when the inequality is strict for every $v\ne 0$.

The diagonalization theorem for real quadratic forms says that some basis puts any symmetric form intowith $r$ further zero coordinates. Sylvester's law of inertia says that $(p,q,r)$ is independent of the diagonalizing basis. To prove this, let $P$ be the span of the positive coordinate vectors and let $N\oplus Z$ be the span of the negative and zero vectors. If $P^{\prime }$ is positive definite for another diagonalization and $\dim P^{\prime }>p$, thenso the two spaces intersect nontrivially. A vector in the intersection would have both positive and nonpositive square, a contradiction. Thus $p^{\prime }\le p$; symmetry gives $p=p^{\prime }$. Applying the same argument to $-\varphi$ gives $q=q^{\prime }$, and then $r=r^{\prime }$.

For the nondegenerate form on $V$, write its inertia as $(p,q,0)$, so $p+q=2n$. The restriction vanishes identically on $E$: polarization gives $phi(u,v)=0$ for $u,v\in E$. Projection of $E$ to the positive coordinate space is injective, since a vector with zero positive projection cannot be isotropic unless it is zero. Hence $k\le p$; projection to the negative space likewise gives $k\le q$. Therefore $k\le \min (p,q)\le n$.

The matrix of $l^2$ is $l{l}^T$. If $l\ne 0$, it has rank one and inertia $(1,0,n-1)$, hence signature one; if $l=0$, its rank and signature are zero. The coefficientwise product $(l^2,s^2)$ has matrix $t{t}^T$, where $t_i=l_i{s}_i$, so it has the same rank-one conclusion when $t\ne 0$ and is zero otherwise.

Finally, diagonalization writes every positive semidefinite form as a sum of squares, say $f={\sum }_a{l}_a^2$ and $g={\sum }_b{s}_b^2$. Bilinearity of coefficientwise multiplication givesa sum of positive semidefinite rank-at-most-one forms. Thus $(f,g)$ is positive semidefinite.

A bilinear form $B$ is symmetric when $B(v,w)=B(w,v)$ for every pair of vectors $v,w$. Equivalently, its matrix of a bilinear form in any basis is a symmetric matrix, since $A_{ij}=B(e_i,e_j)=B(e_j,e_i)=A_{ji}$. Symmetry makes the relation $B(v,w)=0$ reciprocal, giving the bilinear orthogonal complement a geometric interpretation.

Over the real numbers, set $q(v)=B(v,v)$. The polarization identityrecovers the symmetric bilinear form from its quadratic form: expand $q(v+w)$ and use symmetry to combine the two cross terms. For $q(x,y)=x^2-y^2$, take $v=(2,1)$ and $w=(1,3)$. Then $q(v+w)=-7$, $q(v)=3$, and $q(w)=-8$, so $B(v,w)=(-7-3+8)/2=-1=2\cdot 1-1\cdot 3$. The same formula works whenever the characteristic is not two.^[ref]

Three real diagonal matrices show how much variety symmetry allows. The identity matrix $\mathrm{diag}(1,1)$ gives the positive-definite bilinear form $q=x^2+y^2$. The matrix $\mathrm{diag}(1,-1)$ gives $q=x^2-y^2$ and is nondegenerate despite the isotropic vector $(1,1)$. The matrix $\mathrm{diag}(1,0)$ gives $q=x^2$ and is a degenerate bilinear form with radical of a bilinear form $\mathrm{span}\{(0,1)\}$. Their level sets make the distinction visible: a circle, two branches of a hyperbola, and two parallel lines for $q=1$.

The inertia of a bilinear form records the numbers of positive, negative, and zero directions after matrix congruence diagonalization. By Sylvester's law of inertia, these counts survive change of basis; the three examples have inertia of a bilinear form $(2,0,0)$, $(1,1,0)$, and $(1,0,1)$. Historically, Sylvester stated this invariance in an 1852 paper on reducing real quadratic forms to sums of positive and negative squares. Ghys and Ranicki reproduce his statement and distinguish it from his subsequent 1853 proof.^[ref]