Quadratic Forms and Sylvester's Law of Inertia

Definition Let $A \in \mathbb{C}^{n \times n}$ be Hermitian, so $A^{*} = A$. Its quadratic form is the function $x \mapsto x^{*}Ax$ on $\mathbb{C}^n$. It takes only real values, because the scalar $x^{*}Ax$ equals its own conjugate $(x^{*}Ax)^{*} = x^{*}A^{*}x = x^{*}Ax$. Under an invertible change of coordinates $x = Sy$ with $S \in \mathbb{C}^{n \times n}$,

so in the new coordinates the same form has the matrix $S^{*}AS$. Two matrices $A$ and $B$ are conjugate congruent when $B = S^{*}AS$ for some invertible $S$. This is an equivalence relation, it sends Hermitian matrices to Hermitian matrices, and it preserves rank, because multiplying by invertible matrices does not change rank. Over $\mathbb{R}$, with $A$ real symmetric and $S$ real invertible, $S^{*} = S^{\mathsf{T}}$ and the relation $B = S^{\mathsf{T}}AS$ is ordinary congruence. Congruence is not similarity. Similarity $S^{-1}AS$ describes one operator in a new basis and keeps the eigenvalues, while congruence describes one form in new coordinates and in general changes them. The two formulas agree for every $A$ exactly when $S^{*} = S^{-1}$, that is, when $S$ is unitary or real orthogonal.

Definition Let $\mathbb{F}$ be $\mathbb{R}$ or $\mathbb{C}$, and let $A \in \mathbb{F}^{n \times n}$ be real symmetric when $\mathbb{F} = \mathbb{R}$ or Hermitian when $\mathbb{F} = \mathbb{C}$. Then there are an invertible $S \in \mathbb{F}^{n \times n}$ and whole numbers $p, q, z \ge 0$ with $p + q + z = n$ such that

To build $S$, use the spectral theorem to write $A = Q\Lambda Q^{*}$, with $Q$ unitary (real orthogonal when $\mathbb{F} = \mathbb{R}$) and $\Lambda$ real diagonal, ordering the orthonormal eigenvectors so that the positive eigenvalues come first, then the negative ones, then the zeros. Let $D$ be the real diagonal matrix with $d_{ii} = 1/\sqrt{\lvert \lambda_i \rvert}$ when $\lambda_i \ne 0$ and $d_{ii} = 1$ when $\lambda_i = 0$. Then $S = QD$ is invertible and $S^{*}AS = D\Lambda D$, whose diagonal entries are $\lambda_i / \lvert \lambda_i \rvert = \pm 1$ when $\lambda_i \ne 0$ and $0$ when $\lambda_i = 0$. In this construction $p$ and $q$ are the numbers of positive and negative eigenvalues counted with multiplicity, and $z = \dim \ker A$. The matrix $S$ is far from unique. Completing squares, for instance, gives other choices.

Definition Let $A \in \mathbb{F}^{n \times n}$ be real symmetric or complex Hermitian, and let $S^{*}AS = \operatorname{diag}(I_p, -I_q, 0_z)$ for an invertible $S$. The inertia of $A$, and of its quadratic form, is the triple $(p, q, z)$ of the numbers of positive, negative and zero diagonal entries. Sylvester’s law of inertia, below, shows that this triple does not depend on which $S$ is used. Consequently $p$ and $q$ are the numbers of positive and negative eigenvalues of $A$ counted with multiplicity, and

Definition Let $A \in \mathbb{F}^{n \times n}$ be real symmetric or complex Hermitian. A subspace $L$ of $\mathbb{F}^n$ is positive for $A$ if $x^{*}Ax > 0$ for every nonzero $x \in L$, and negative if $x^{*}Ax < 0$ for every nonzero $x \in L$. A positive subspace of the largest possible dimension is a maximal positive subspace. If $A$ has inertia $(p, q, z)$, this largest dimension is exactly $p$. To see it, take $S$ with $S^{*}AS = \operatorname{diag}(I_p, -I_q, 0_z)$ and columns $s_1, \dots, s_n$. For $x = Sy$,

So the span of $s_1, \dots, s_p$ is positive and has dimension $p$. The span $N$ of $s_{p+1}, \dots, s_n$ has dimension $n - p$, and $x^{*}Ax \le 0$ on it. If a positive subspace $L$ had $\dim L > p$, the dimension formula $\dim(L + N) + \dim(L \cap N) = \dim L + \dim N$ together with $\dim(L + N) \le n$ would give

A nonzero vector of $L \cap N$ would have a value both positive and at most $0$, which is impossible. The same argument applied to $-A$ shows that $q$ is the largest dimension of a negative subspace. A positive subspace that is not contained in any larger positive subspace also turns out to have dimension $p$, so the word maximal can be read either way. When $0 < p < n$, maximal positive subspaces are not unique, only their dimension is. When $p = n$ the only one is $\mathbb{F}^n$, and when $p = 0$ it is $\{0\}$.

Definition Let $\mathbb{F}$ be $\mathbb{R}$ or $\mathbb{C}$, and let $A, B \in \mathbb{F}^{n \times n}$ both be real symmetric or both be complex Hermitian. Then $B = S^{*}AS$ for some invertible $S \in \mathbb{F}^{n \times n}$ if and only if $A$ and $B$ have the same inertia. In particular, whenever an invertible $S$ makes $S^{*}AS$ diagonal, with any real diagonal entries, the numbers of positive, negative and zero entries are $p$, $q$ and $z$. If $B = S^{*}AS$, then $y^{*}By = (Sy)^{*}A(Sy)$, so $S$ carries each subspace that is positive for $B$ onto a subspace of the same dimension that is positive for $A$, and $S^{-1}$ does the reverse. The largest such dimension $p$ is therefore the same for both matrices, the same holds for $q$ with negative subspaces, and $z = n - p - q$ agrees as well. Conversely, if both have inertia $(p, q, z)$, choose invertible $S_1$ and $S_2$ with $S_1^{*}AS_1 = N = S_2^{*}BS_2$, where $N = \operatorname{diag}(I_p, -I_q, 0_z)$. Then

So the inertia classifies these forms completely under congruence.

Definition Let $A \in \mathbb{F}^{n \times n}$ be real symmetric or complex Hermitian with inertia $(p, q, z)$. Its form is positive definite if $x^{*}Ax > 0$ for every $x \ne 0$, which happens exactly when $p = n$, and positive semidefinite if $x^{*}Ax \ge 0$ for every $x$, which happens exactly when $q = 0$. Negative definite and negative semidefinite forms have the inequalities reversed and correspond to $q = n$ and $p = 0$. The form is indefinite if it takes both a positive and a negative value, which happens exactly when $p \ge 1$ and $q \ge 1$. A nonzero vector with $x^{*}Ax = 0$ need not lie in $\ker A$. Every indefinite form has such vectors, for instance $s_1 + s_{p+1}$ in the notation above, even when $A$ is invertible.