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Let $\varphi$ be a twice continuously differentiable local diffeomorphism, and suppose $\nabla f(\varphi (u_{\ast }))=0$. The chain rule gives the displayed transformation of the Hessian matrix: terms containing second partial derivatives of $\varphi$ are multiplied by first partial derivatives of $f$ and vanish at the critical point. By Sylvester's law of inertia, the numbers of positive, negative and zero eigenvalues are unchanged. Independently of nondegeneracy, the local diffeomorphism preserves the critical point's status as a local minimum, local maximum or saddle point of a scalar function, because it maps neighbourhoods to neighbourhoods.

The inertia of a real symmetric bilinear form is the triple giving the numbers of positive, negative, and zero squares in a diagonalization. Sylvester's law of inertia makes it independent of the chosen basis.

Its symmetric matrix is $11^T-I$. The line spanned by $1$ has eigenvalue $n-1$ and its orthogonal complement has eigenvalue $-1$. Thus for $n>1$ it is nondegenerate with inertia $(1,n-1,0)$; for $n=1$ it vanishes. This gives a useful explicit example for Sylvester's law of inertia.

First work over a field of characteristic different from two. If a nonzero symmetric bilinear form had $\varphi (v,v)=0$ for every vector, polarization would givea contradiction. Hence choose $e$ with $\varphi (e,e)\ne 0$. Every $v$ decomposes uniquely aswith the second summand orthogonal to $e$. Induct on the orthogonal complement to obtain a diagonal basis. If the restricted form becomes zero, any basis of that remaining space completes the diagonalization. In this basis the rank of a quadratic form is exactly the number of nonzero diagonal coefficients; the remaining basis vectors span its radical of a bilinear form.

Over the reals, rescale the nonzero basis vectors so the diagonal entries are $p$ copies of $+1$, $q$ copies of $-1$ and $z=n-p-q$ zeros. Define the signature of a quadratic form by $\sigma =p-q$, so $r=p+q$. To prove it is basis independent, note that $p$ is the largest dimension of a linear subspace on which the form is positive-definite. The positive coordinate linear subspace attains dimension $p$; on any other positive linear subspace, projection to those positive coordinates is injective, since its kernel has nonpositive quadratic value. Thus its dimension is at most $p$. The same argument for the negative form characterizes $q$. This proves Sylvester's law of inertia and that the signature is well-defined.

Let $e_i$ and $f_i$ be positive and negative normalized basis vectors, and let $Z$ be the radical of a bilinear form. The span of $Z$ and $e_i+f_i$ for $1\le i\le \min (p,q)$ is a totally isotropic subspace: all mutual bilinear pairings vanish. Its dimension is $z+\min (p,q)$. Conversely, project any totally isotropic subspace $U$ to the nondegenerate positive-plus-negative quotient. The projected linear subspace is still null, and its projections to both the positive and negative coordinate spaces are injective: a vector with one component zero would have strictly signed quadratic value unless it vanished. Its dimension is therefore at most $\min (p,q)$, while the kernel of the quotient map on $U$ has dimension at most $z$. HenceThis proves both construction and sharpness of the maximum dimension of a totally isotropic subspace.

For the five-variable example, the symmetric coefficient matrix isIts cyclic structure makes the vectors $(1,{\zeta }^k,{\zeta }^{2k},{\zeta }^{3k},{\zeta }^{4k})$, $\zeta =e^{2\pi i/5}$, eigenvectors with eigenvalues ${\zeta }^k+{\zeta }^{-k}=2\cos (2\pi k/5)$. Thus the eigenvalues areTheir product is $2$, and three are positive and two negative. ThereforeThe real and imaginary parts of each conjugate pair of eigenvectors give the corresponding real two-dimensional eigenspaces, consistent with the real inertia calculation.

The Morse lemma states that if $p$ is a nondegenerate critical point of a smooth real function $f$ on an $m$-dimensional smooth manifold, there are smooth coordinates $y_1,\dots ,y_m$, centered at $p$, such thatHere $\lambda$ is the Morse index, the number of negative directions of the Hessian matrix at $p$. The equality is exact on a neighborhood, not merely a second-order expansion.

To prove it, start with any coordinates putting $p=0$. Since $df(0)=0$, integral Taylor theorem givesThe matrix $A(x)=(a_{ij}(x))$ is smooth and symmetric, with $A(0)=\frac{1}{2}\mathrm{Hess}f(0)$. By Sylvester's law of inertia, a constant invertible linear coordinate change makes $A(0)=\mathrm{diag}(-I_{\lambda },I_{m-\lambda })$.

We perform completion of squares with coefficients depending smoothly on the original $x$. The first pivot $a_{11}(x)$ stays nonzero and keeps its sign on a sufficiently small neighborhood. Algebraically,The remaining coefficient matrix is the Schur complement. At zero it is the remaining signed diagonal matrix, so its first pivot is also nonzero nearby. Continue recursively, shrinking the neighborhood finitely many times. Every pivot $d_i(x)$ is smooth and nonzero, with sign ${ϵ}_i=-1$ for $i\le \lambda$ and $+1$ thereafter. Absorb $|d_i(x)|$ into the square by its smooth positive square root. This produces a smooth triangular matrix $B(x)$ with nonzero diagonal such thatAll coefficients here are functions of the original $x$; no circular coordinate substitution is involved. Define $\Phi (x)=B(x)x$. Its derivative at zero is $D\Phi (0)=B(0)$, which is invertible. The inverse function theorem makes $y=\Phi (x)$ a smooth coordinate system and gives the asserted exact quadratic form. The number of negative squares agrees with the Hessian matrix signature and hence with the Morse index. This is the smooth completing-square proof of the Morse lemma.

Associate to $Q$ its symmetric bilinear form by the polarization identity,We first construct a diagonal basis by induction on the dimension. A zero quadratic form is already diagonal. Otherwise choose $v$ with $Q(v)\ne 0$. Every $x$ has the decompositionwhose second term lies in $W=\{w:B(w,v)=0\}$. Also $\mathbb{R}v\cap W=\{0\}$, so $V=\mathbb{R}v\oplus W$ and $\dim W=n-1$. Apply the induction hypothesis to the restricted quadratic form on $W$. The cross terms with $v$ vanish, giving a diagonal basis on $V$. Rescale each vector with diagonal coefficient $a\ne 0$ by $1/\sqrt{|a|}$, and reorder positive, negative and zero coefficients. This yieldsNo nondegeneracy assumption was used.

For uniqueness, let $V_{+}$ be the positive coordinate subspace and $V_{\le 0}$ the span of all negative and zero coordinate vectors. Their dimensions are $p$ and $n-p$. Any subspace $W$ on which $Q$ has positive definiteness intersects $V_{\le 0}$ only in zero. The dimension formula for a sum of subspaces therefore gives $\dim W\le p$. Conversely $V_{+}$ itself has dimension $p$ and positive restriction. Thus $p$ is the largest possible dimension of a positive subspace, an intrinsic description independent of the basis. Applying the same argument to $-Q$ shows that $q$ is the largest dimension of a negative subspace. This proves Sylvester's law of inertia, including uniqueness of the zero count $n-p-q$.

The extremal subspaces themselves need not be unique. For $Q(x,y)=x^2-y^2$, the coordinate axes give one positive and one negative line. For any $t\ne 0$, putThen $Q(u)=1$, $Q(v)=-1$, $B(u,v)=0$, and the determinant of the change of basis is one. The distinct lines $\mathbb{R}u$ and $\mathbb{R}v$ give another positive/negative decomposition, with the same intrinsic counts $p=q=1$.

Sylvester's law of inertia says that any real quadratic form has coordinates in which it is a sum of $p$ positive squares minus $m$ negative squares, with $z$ unused coordinates, and that the numbers $(p,m,z)$ are invariant under every invertible linear change of coordinates. Its rank is $p+m$; its signature of a quadratic form is commonly $p-m$, with $(p,m)$ also called its signature pair.

The symmetric matrix here is $J-I$, where $J=11^T$ and $1=(1,\dots ,1{)}^T$. On the line spanned by $1$, $J-I$ has eigenvalue $n-1$. If ${\sum }_i{x}_i=0$, then $Jx=0$, so that the orthogonal complement has eigenvalue $-1$ with multiplicity $n-1$. Thus the off-diagonal all-ones quadratic form hasFor $n=1$ the form vanishes, so rank and scalar signature are both zero, with inertia $(0,0,1)$. Including this case is necessary because the distinguished eigenvalue then becomes zero.

Sylvester's law of inertia states that a real quadratic form can be brought by an invertible real change of coordinates towith $r$ remaining zero coordinates, and that the integers $(p,q,r)$ are uniquely determined. To prove existence, represent it by a symmetric matrix. The allowed orthogonal diagonalization yields real eigenvalues, and scaling each nonzero eigen-coordinate by the square root of its absolute eigenvalue changes its coefficient to $+1$ or $-1$.

For uniqueness, $p$ is the largest dimension of a subspace on which the form is positive definite. The positive coordinate subspace attains $p$. Any subspace of dimension greater than $p$ intersects the $(q+r)$-dimensional span of the negative and zero coordinates nontrivially, by the dimension formula. A nonzero vector in that intersection has nonpositive quadratic value, contradicting positive definiteness. Similarly, $q$ is the maximal dimension of a negative-definite subspace. These descriptions are independent of coordinates, and $r=\dim V-p-q$ is then fixed. This proves existence and uniqueness, including degenerate forms.

For the final application, nonsingularity gives $p+q=2m$. In diagonal coordinates write $v=(v_{+},v_{-})$, with $Q(v)=|v_{+}{|}^2-|v_{-}{|}^2$. If $v\in U$ projects to zero in the positive coordinates, $Q(v)=0$ forces $v_{-}=0$, so projection $U\to {\mathbb{R}}^p$ is injective. The same argument applies to the negative coordinates. Hence $m\le p$ and $m\le q$, and their sum is $2m$. ThereforeThe inertia signature is $(m,m)$, or signature difference $p-q=0$ if that convention is used. This is the isotropic dimension bound from real inertia.

Two real symmetric bilinear forms are congruent if an invertible linear map takes one to the other; in matrices, $B=S^{\mathsf{T}}AS$ for some invertible $S$. The spectral theorem for real symmetric matrices gives an orthogonal eigenbasis. Rescaling each basis vector with nonzero eigenvalue then reduces any such form toSylvester's law of inertia asserts that $(p,q,r)$ is independent of the congruence reduction. To prove it, $p$ is the greatest dimension of a subspace on which the form is positive definite: the positive coordinate subspace attains it, whereas any subspace of dimension greater than $p$ meets the nonpositive coordinate subspace of dimension $n-p$ and therefore cannot be positive definite. Applying the same argument to the negative of the form determines $q$. Then $r=n-p-q$ is determined too.

The rank of a quadratic form is $p+q$, and its signature of a quadratic form is $p-q$ under the signed-difference convention. They determine $p,q,r$ and hence the congruence class. Counting all pairs $p,q\ge 0$ with $p+q\le n$ givesFor $n\ge 1$, exactly two classes have every level set bounded: the positive-definite and negative-definite classes. Their nonempty nonzero levels are ellipsoids, and their zero level is the origin. A degenerate form has an entire null line in its zero level. An indefinite nondegenerate form also has a nonzero isotropic vector, for example the sum of one positive and one negative normalized basis vector, whose scalar multiples make the zero level unbounded.

For the displayed three-dimensional form,Its leading principal minors are $3,14,32$, all positive. Sylvester's criterion makes it positive definite, so the surface is a bounded ellipsoid. Directly, its quadratic expression iswhich also proves positivity and boundedness of the unit level.

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 real vector space of symmetric matrices has dimension $n(n+1)/2$. On its invertible part the derivative of the determinant isAt a determinant-one symmetric matrix, this derivative is nonzero: the symmetric direction $C=B$ gives value $n$. The regular level set theorem therefore yieldsThis applies to every signature component, not only positive-definite matrices.

The special linear congruence action on symmetric matrices preserves symmetry because $(AB{A}^T{)}^T=AB{A}^T$, and preserves determinant because $\det (AB{A}^T)=(\det A{)}^2\det B=1$. The identity acts trivially andThese verify a smooth left Lie group action of the special linear group. By Sylvester's law of inertia, it also preserves the numbers of positive and negative eigenvalues.

For $n=2$, parametrize the symmetric matrices byThus $M$ is exactly the two-sheeted hyperboloidThe two sheets consist respectively of positive-definite and negative-definite matrices. Each is preserved by the special linear congruence action on symmetric matrices. They are individually transitive: on the positive sheet $B=A{A}^T$ with $A=B^{1/2}$ and $\det A=1$; on the negative sheet use $B=-A{A}^T$. The stabilizer of either $I$ or $-I$ is $SO(2)$.

Differentiating the left Lie group action along $\exp (sX)$ gives the fundamental vector fieldFor the three matrices in the question, their coordinate components areA sign convention matters here. With the usual Lie bracket of vector fields $[U,V]f=U(Vf)-V(Uf)$, the fundamental fields of this left action obey $[u_X,u_Y]=-u_{[X,Y]}$. Indeed, for linear coordinate fields $u_X(z)=R_{X}z$, the bracket has coefficient $(R_Y{R}_X-R_X{R}_Y)z$. To obtain a Lie algebra representation rather than an anti-representation, use the infinitesimal left-action sign conventionAn explicit answer is thereforeThese are tangent to the two-sheeted hyperboloid: applying each to $Q=t^2-x^2-y^2$ gives zero. For example,They can be written on either sheet using $t=\pm \sqrt{1+x^2+y^2}$:Here the omitted ${\partial }_t$ component is determined by tangency, and derivatives of $t(x,y)$ must be included when computing brackets in these coordinates.

Direct matrix multiplication givesDirect differentiation of the displayed vector fields gives exactlyFor instance, in ambient $(t,x,y)$ coordinates, $[v_1,v_3]=(-2x,-2t,0)=v_2$. These are the defining relations of the sl2 Lie algebra. The three fields are linearly independent over constant real coefficients: if $a{v}_1+b{v}_2+c{v}_3$ vanishes on a sheet, its ${\partial }_x$ coefficient $(-a+c)y-2bt$ forces $b=0,c=a$, and its ${\partial }_y$ coefficient then equals $-2at$, forcing $a=c=0$. Thus the representation is faithful. Their pointwise span need only have dimension two, consistent with the dimension of each sheet.

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.

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

By the real spectral theorem, the real symmetric matrix $A$ has an orthonormal basis of eigenvectors. In that basis the associated quadratic form isRescaling the coordinates belonging to nonzero eigenvalues changes every positive coefficient to $+1$ and every negative coefficient to $-1$. Hence the diagonal normal form has one positive square for each positive eigenvalue and one negative square for each negative eigenvalue. By Sylvester's law of inertia, these counts do not depend on the diagonalizing basis, so

The numerical eigenvalues are not invariant under a general change of basis, because the matrix changes by congruence rather than similarity. For example, on a one-dimensional space let $q(e)=1$. Its matrix in the basis $(e)$ is $(1)$, whereas in the basis $(2e)$ it is $(4)$. The eigenvalue changes from $1$ to $4$, although its sign, and therefore the signature, is unchanged.

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.

The spectral theorem for real symmetric matrices supplies an orthonormal basis of real eigenvectors. Put these in the columns of $Q$, so $Q^{T}Q=I$ and $Q^{T}AQ=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. In the question's coordinate convention choose $O=Q^T$ and $y=Ox$. Since $x=O^{T}y$,Reorder the eigenvalues with $m$ positive, $\ell -m$ negative and $n-\ell$ zero. For nonzero eigenvalues set $z_i=\sqrt{|{\lambda }_i|}{y}_i$, and retain the remaining coordinates as $z_i=y_i$. This is an invertible real transformation $z=DOx$, where $D$ is the diagonal rescaling; the combined matrix $L=DO$ gives $z=Lx$. It giveswith the zero coordinates absent. The general allowed range is $0\le m\le \ell \le n$. The printed restriction $m\ge 1$ is not valid for every real symmetric matrix: a negative-definite or zero form has no positive square. The invariance of these counts is Sylvester's law of inertia.

For the specified form, a particularly simple change isSubtracting the first two squared coordinates verifies the identity directly, and $\det L=1$ proves invertibility. Thus $m=1$, $\ell =2$ and there is one zero direction. The final rescaling need not remain orthogonal.

Complete the square in the quadratic form:The real invertible linear transformationsets $x_1+2x_2=y_1$, $x_2=y_2$, $\sqrt{6}{x}_3=y_3$, giving $Q_1=y_1^2+y_2^2+y_3^2$. The same transformation givesIts signature is $(2,1)$. By Sylvester's law of inertia no real invertible change of variables can turn this second form into a sum of three positive squares.

Since $A$ is invertible, the quadratic form gradient vanishes exactly at the origin. At every nonzero point of $C_{\alpha }$, the scalar derivative is surjective, so the regular level set theorem gives a smooth submanifold of dimension $n-1$.

If $\alpha \ne 0$, the origin is absent. Therefore the largest relatively open smooth subset is all of $C_{\alpha }$.

If $\alpha =0$, the nontriviality assumption forces the form to be an indefinite quadratic form: an invertible definite form would have zero set consisting only of the origin. Every nonzero point is regular. To show that the origin really fails to be a manifold point, use Sylvester's law of inertia to reduce to $|u{|}^2-|w{|}^2=0$. With positive and negative coordinate vectors $e_i,f_j$, the null vectors $e_i+f_j$ and $e_i-f_j$ span ${\mathbb{R}}^n$. Each line in one of these directions lies in the cone.

If a relatively open manifold neighborhood of the origin existed, its tangent space would contain all these line directions, so would have dimension $n$. Such an embedded manifold would be open in ${\mathbb{R}}^n$, impossible for the zero set of a nonzero quadratic form. ConsequentlyThis gives the smooth locus of a nondegenerate real quadratic level set, including maximality at the singular vertex.

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.