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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.

Expand the increment of the quadratic form, using $A^T=A$:The last term obeys $|h^{T}Ah|\le \Vert A\Vert \Vert h{\Vert }^2=o(\Vert h\Vert )$. Thus $q$ is differentiable andThis is the quadratic form gradient. The derivative is a linear functional of the increment $h$, whereas the gradient is its representing vector under the Euclidean inner product.

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 $C_0\setminus \{0\}$.

For a real symmetric matrix $A$ and nonzero real vector $x$, the quotient rule and the quadratic form gradient give the displayed gradient. It vanishes exactly when $Ax=R_A(x)x$, so the stationary points are eigenvectors. Equivalently, the Lagrange multiplier condition for optimizing $x^{T}Ax$ on the unit sphere is $Ax=\lambda x$. The smallest and largest eigenvalues give the global extrema; other eigenvectors can be stationary without being extrema, as a direction mixing larger and smaller eigenvalues changes the quotient in opposite ways.