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At a nondegenerate critical point placed at the origin, integral Taylor theorem gives $f(x)-f(0)=x^{T}A(x)x$, where $A(x)$ is a smooth symmetric matrix and $A(0)=\frac{1}{2}\mathrm{Hess}f(0)$. A constant linear change makes $A(0)$ diagonal with entries $\pm 1$. Successive completion of squares, with the Schur complement at each step, produces a smooth invertible triangular matrix $B(x)$ such that $x^{T}A(x)x={\sum }_i{ϵ}_i(B(x)x{)}_i^2$. The pivots keep their nonzero signs near zero. The map $x\mapsto B(x)x$ has invertible derivative at zero, so the inverse function theorem makes it a coordinate system. The signs are the Hessian matrix signature by Sylvester's law of inertia.

The quadratic form gradient gives $D{q}_x(h)=2x^{T}Ah$. Since $A$ is invertible, this linear map is nonzero exactly when $x\ne 0$. The regular level set theorem therefore makes every nonzero point of $C_{\alpha }$ a smooth point of dimension $n-1$. If $\alpha \ne 0$, the entire level set is smooth.

If $\alpha =0$ and the level set contains a nonzero point, the form is an indefinite quadratic form. Its only singular point is the origin. To prove this rather than merely observe a vanishing derivative, use Sylvester's law of inertia to write $q=|u{|}^2-|v{|}^2$ with both sign spaces nonzero. The null vectors $e_i\pm f_j$ span the whole ambient space. Their straight-line curves through the origin lie in the cone. If a relatively open neighborhood of the origin in the cone were an embedded submanifold, its tangent space would contain all these directions and hence have dimension $n$. It would then be open in the ambient space, impossible for the zero set of this nonzero quadratic form. Thus the smooth locus is $C_0\setminus \{0\}$.

The special linear group acts on determinant-one real symmetric matrices by matrix congruence. Symmetry and determinant are preserved, and signatures are preserved by Sylvester's law of inertia. The determinant-one set is a regular level set of dimension $n(n+1)/2-1$, with tangent directions $C=C^T$ satisfying $\mathrm{tr}(B^{-1}C)=0$. Its infinitesimal left-action fields are $XB+B{X}^T$; converting them to a Lie algebra representation requires the infinitesimal left-action sign convention.

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]