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A quadratic form on a finite-dimensional real vector space is a function $q(v)=B(v,v)$ for a symmetric bilinear form $B$. Conversely $B(u,v)=\frac{1}{2}(q(u+v)-q(u)-q(v))$ recovers that bilinear form. Two quadratic forms are congruent if a linear isomorphism $T$ satisfies $q_2(v)=q_1(Tv)$; in coordinates this is matrix congruence, $A_2=P^T{A}_{1}P$, with $P$ invertible.

Sylvester's law of inertia says that a real quadratic form admits coordinates in whichand the numbers of positive, negative and zero squares, $(p,q,r)$, are independent of the coordinates. Here $p+q+r=\dim V$; $p$ and $q$ are the maximal dimensions of subspaces on which the quadratic form is positive definite and negative definite, while $r$ is the dimension of its radical of a bilinear form. These numbers are the inertia of a bilinear form. The rank of a quadratic form is $p+q$, and the signature of a quadratic form, in the signed-difference convention, is $p-q$.

Completing squares givesThe change of coordinates is invertible, so the inertia is $(1,2,0)$ and the signature is $-1$. If signature is recorded as an ordered pair instead, that pair is $(1,2)$.

The complexification of a Lie algebra $L_0$ is $L=L_0{\otimes }_{\mathbb{R}}\mathbb{C}$, with the Lie bracket extended complex-bilinearly. Equivalently write $L=L_0\oplus i{L}_0$, whereComplex conjugation is an antilinear map and a Lie algebra automorphism whose fixed subalgebra is $L_0$.

If the solvable radical of $L_0$ is nonzero, its complexification is a nonzero solvable ideal of $L$. Conversely the solvable radical $R$ of $L$ is preserved by complex conjugation, because it is the unique largest solvable ideal. Consequentlyfor $z\in R$, both $(z+\overline{z})/2$ and $(z-\overline{z})/(2i)$ belong to $R_0$. If $R\ne 0$, $R_0\ne 0$ is a solvable ideal of $L_0$. This proves $L_0$ is semisimple if and only if $L_0{\otimes }_{\mathbb{R}}\mathbb{C}$ is semisimple. Consistently, the complex Killing form is just the complex-bilinear extension of the real one; its determinant in a real basis is unchanged by extending scalars.

Take the sl2R Lie algebra ${\mathfrak{sl}}_2(\mathbb{R})$ and the special unitary Lie algebra $\mathfrak{su}(2)$. Both complexify to ${\mathfrak{sl}}_2(\mathbb{C})$. This is immediate for the former; for the latter, the real basis $ih,e-f,i(e+f)$ consists of traceless matrices that are skew-Hermitian matrices and is also a complex basis of ${\mathfrak{sl}}_2(\mathbb{C})$.

They are not isomorphic as real Lie algebras. Their Killing forms are $B(X,Y)=4\mathrm{tr}(XY)$, but on $\mathfrak{su}(2)$ this is negative definite. On ${\mathfrak{sl}}_2(\mathbb{R})$ the basis $h,e+f,e-f$ has a diagonal Gram matrix with entries $8,8,-8$, so the signature of a quadratic form is $(2,1)$. A Lie algebra isomorphism preserves the Killing form and therefore its signature.

A real split semisimple Lie algebra has a Cartan subalgebra whose Adjoint representation of a Lie algebra is simultaneously diagonalizable over $\mathbb{R}$, so its root-space decomposition is defined over $\mathbb{R}$. The example ${\mathfrak{sl}}_2(\mathbb{R})$ is split: the diagonal Cartan subalgebra $\mathbb{R}h$ has eigenvalues $0,2,-2$ and real root spaces $\mathbb{R}e,\mathbb{R}f$. The example $\mathfrak{su}(2)$ is not split. Invariance makes every $\mathrm{ad}x$ skew-adjoint for the positive definite inner product $-B$, so its eigenvalues are purely imaginary. If it were diagonalizable over $\mathbb{R}$, all these eigenvalues would be zero and $\mathrm{ad}x=0$. The center is zero, so only $x=0$ has this property; no nonzero split Cartan subalgebra exists. Thus ${\mathfrak{sl}}_2(\mathbb{R})$ is split and $\mathfrak{su}(2)$ is not.

The symmetric bilinear form associated with a quadratic form is obtained by polarization:To diagonalize it, argue by induction on $\dim V$. If $\varphi (v,v)=0$ for every $v$, polarization gives $\varphi =0$, so every basis works. Otherwise choose $e$ with $\varphi (e,e)\ne 0$. Thenand diagonalize the restriction to $e^{\perp }$ inductively. Thus some basis gives a diagonal matrix with positive, negative, and zero entries. By Sylvester's law of inertia, their counts $n_{+},n_{-},n_0$ are basis-independent. In the convention relevant here, the signature of a quadratic form is

Suppose $R$ lies in the radical of a bilinear form, so $\varphi (r,v)=0$ for all $r\in R,v\in V$. Then $q(r)=0$ andHence $q^{\prime }(v+R)=q(v)$ is a well-defined quadratic form on the quotient vector space $V/R$. In an adapted diagonal basis, quotienting by $R$ merely removes zero diagonal directions, so $n_{+}$ and $n_{-}$, and therefore the signature, are unchanged.

Now let $W=\mathrm{span}(e,f)$. Its Gram matrix iswhich has determinant $-1$. Thus $W$ is nondegenerate and has one positive and one negative square: it is a hyperbolic plane (quadratic form) and has signature zero. Nondegeneracy gives the orthogonal direct sumSignature is additive under orthogonal direct sums, so

The symmetric matrix of the quadratic form in the standard basis isbecause the off-diagonal entries contribute twice to $x^{T}Ax$.

Apply Gram-Schmidt orthogonalization for a symmetric bilinear form toThese vectors are pairwise orthogonal for $B(u,v)=u^{T}Av$, andThus the form is diagonal in the basis $(u_1,u_2,u_3)$, with matrixIt follows that its rank is $3$ and its signature is

For a real symmetric bilinear form, let $n_{+}$ and $n_{-}$ be the dimensions of its positive and negative diagonal subspaces. This pair gives the first two components of the inertia of a bilinear form and is independent of basis by the Sylvester's law of inertia. For a nondegenerate form their sum is the full dimension; otherwise the nullity is an additional invariant. Another convention calls the difference $n_{+}-n_{-}$ the signature of a quadratic form. The pair convention describes the positive and negative harmonic spaces of the real de Rham intersection form in dimension four.