Maximal Positive Subspaces and Inertia

Positive Directions and Subspaces

A real quadratic form can be positive on some vectors, negative on others, and zero on still others. Its positive index counts the positive entries in a diagonal representation of the form. That count has a geometric interpretation: it is the largest dimension a subspace can have while the form remains strictly positive on every nonzero vector in that subspace. Establishing this interpretation explains why the positive index is an invariant of the form, rather than a feature of a particular diagonalizing basis.

Let \(D=\operatorname{diag}(d_1,\dots,d_n)\) be a real diagonal matrix, and let \(q:\mathbb{R}^n\to\mathbb{R}\) be the quadratic form \(q(\mathbf{x})=\mathbf{x}^{\top}D\mathbf{x}\). Reordering coordinates if necessary, write its entries so the positive ones come first, then the negative ones, then the zero entries. Let \(p\) be the number of positive entries, \(m\) the number of negative entries, and \(r\) the number of zero entries, so \(p+m+r=n\). In these coordinates, write \(\mathbf{x}=(\mathbf{x}_+,\mathbf{x}_-,\mathbf{x}_0)\). Then

The zero coordinates do not appear in the value of \(q\). The positive index is \(p\); the negative index is \(m\), and the number of zero entries is \(r\). We will focus on the positive index.

The condition on a subspace concerns every nonzero vector in it, not merely a selected list of vectors. This matters because combinations of vectors can change the value of the form. In particular, showing that each vector in a basis has positive value does not by itself show that the subspace they span is positive definite.

The Largest Possible Dimension

Proof. First we construct a positive-definite subspace of dimension \(p\). Let \(U_+\) be the span of the standard basis vectors in the \(p\) coordinates corresponding to positive entries. If \(\mathbf{x}\in U_+\) is nonzero, then at least one of its first \(p\) coordinates is nonzero, and

because each \(d_i>0\) and each \(x_i^2\ge 0\). Thus \(U_+\) is positive definite, and its dimension is \(p\).

Now let \(U\) be any positive-definite subspace. Project each vector onto its first \(p\) coordinates, defining the linear map \(\pi_+:U\to\mathbb{R}^p\) by \(\pi_+(\mathbf{x})=\mathbf{x}_+\). We show this map is injective. If \(\mathbf{x}\in U\) satisfies \(\pi_+(\mathbf{x})=\mathbf{0}\), then its positive coordinates all vanish, so

since every \(d_i\) in this sum is negative and every square is nonnegative. If \(\mathbf{x}\ne\mathbf{0}\), positive definiteness of \(U\) instead requires \(q(\mathbf{x})>0\). Therefore \(\mathbf{x}=\mathbf{0}\), and \(\pi_+\) is injective. An injective linear map sends a basis of \(U\) to a linearly independent list in \(\mathbb{R}^p\), which has at most \(p\) vectors. Hence \(\dim U\le p\). Together, the constructed subspace and this upper bound show that the largest possible dimension is exactly \(p\). If \(p=0\), the same argument maps \(U\) injectively into the zero-dimensional space \(\mathbb{R}^0\), so \(U=\{\mathbf{0}\}\); the conclusion still holds. \(\square\)

The construction proves that the bound can be attained, while the projection argument proves that no positive-definite subspace can be larger. The upper bound is the key step: negative and zero coordinates cannot provide additional independent directions while preserving strict positivity on every nonzero vector.

Worked Examples

Why the Positive Index Is Intrinsic

A diagonal matrix represents a form in a chosen coordinate system. Changing basis for a bilinear form changes its matrix by congruence: if \(\mathbf{P}\) is the change-of-coordinates matrix, the new matrix is \(\mathbf{P}^{\top}D\mathbf{P}\). The vector coordinates change, but the underlying subspaces and the values of the form correspond under the invertible coordinate map. The maximal-dimension characterization therefore gives a way to compare positive counts from different diagonal representations.

Proof. Let \(D\) and \(D'\) be the two diagonal matrices, with coordinates related by an invertible matrix \(\mathbf{P}\), so \(D'=\mathbf{P}^{\top}D\mathbf{P}\). For each coordinate vector \(\mathbf{y}\) in the \(D'\) coordinates, its corresponding vector in the \(D\) coordinates is \(\mathbf{P}\mathbf{y}\), and

The map \(\mathbf{y}\mapsto\mathbf{P}\mathbf{y}\) is a linear isomorphism: it sends subspaces to subspaces of the same dimension, and the displayed equality shows that a subspace is positive definite in one coordinate system exactly when its image is positive definite in the other. Consequently, the largest positive-definite subspace dimension is the same for both diagonal matrices. By the maximal positive dimension theorem, that dimension is the number of positive entries in each matrix. The two counts are therefore equal. \(\square\)

This is the positive-index part of the invariance behind Sylvester’s law of inertia. The same maximal-dimension idea can be applied to \(-q\) to interpret the negative index. The zero count requires a separate argument; equality of positive counts alone does not establish it.

Common Pitfall: Checking Only Basis Vectors

Consider \(q(x,y)=x^2-y^2\), with diagonal matrix \(\operatorname{diag}(1,-1)\). Let \(\mathbf{u}=(1,0)^{\top}\) and \(\mathbf{v}=(2,1)^{\top}\). These vectors are independent: if \(a\mathbf{u}+b\mathbf{v}=\mathbf{0}\), then \((a+2b,b)=(0,0)\), which gives \(b=0\) and then \(a=0\). Both basis vectors have positive value:

A mistaken test would conclude that their span is positive definite. But the combination \(\mathbf{v}-2\mathbf{u}=(0,1)^{\top}\) is nonzero and

So the span is not positive definite. This example shows why positivity must be checked for every nonzero linear combination, or established by a valid argument that covers them all. Here the positive index is \(1\), and the theorem also rules out any positive-definite subspace of dimension \(2\).

Also distinguish a subspace of maximal dimension from one that is merely maximal under inclusion. The theorem concerns the greatest dimension among all positive-definite subspaces. It does not assert that the positive coordinate subspace is the only subspace attaining that dimension; the second worked example exhibited another one.