MIT 18.701 — Lecture 17

Classifying Hermitian Forms, Sylvester's Law of Inertia

§ Classifying Hermitian Forms

Today's lecture will look a lot like Lecture 8 on the Jordan Normal Form. In Lecture 8, we proved:

Theorem. (Jordan Normal Form) Let $T: V \to V$ be a linear transformation with $n$ eigenvalues $\{\lambda_1, \dots, \lambda_n\}$ . Then with an appropriate choice of basis, the matrix corresponding to $T$ can be made to look like:

M_T = \left[\begin{smallmatrix} J_{n_1}(\lambda_1) & 0 & \cdots & 0 \\ 0 & J_{n_2}(\lambda_2) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & J_{n_k}(\lambda_k) \end{smallmatrix}\right].

In other words, we showed that all linear transformations—with an appropriate choice of basis—can be made to look like a matrix of Jordan Blocks. Today is dedicated to proving the following analogous theorem:

Theorem. (Hermitian Form Classification) Let $\langle \bullet, \bullet \rangle : V \times V \to \mathbb{C}$ be a Hermitian form over a $\mathbb{C}$ -vector space $V$ , with signature $(n_1, n_0, n_{-1})$ . Then with an appropriate choice of basis, $\langle \bullet, \bullet \rangle$ can be made to look like:

\langle x, y \rangle = x^{\dagger}My ~~ \text{ where } ~~ M = \begin{bmatrix} I_{n_1} & 0 & 0 \\ 0 & -I_{n_{-1}} & 0 \\ 0 & 0 & 0_{n_0} \end{bmatrix} ~ \text{ (where $0_{n_0}$ is the $n_0 \times n_0$ zero matrix.)}

In other words, we'll show that all Hermitian forms—with an appropriate choice of basis—can be made to look like a diagonal matrix with only $1$ s, $-1$ s, and $0$ s. (We'll define “signature” later.)

Remark. Here's another way to see the analogy. Lecture 8 answered the following question:

Take the group action of $GL_n(F)$ on $F^{n \times n}$ , where $A \in GL_n(F)$ acts on $M \in F^{n\times n}$
via $M \mapsto A^{-1}MA$ . What are the orbits of this group action?

Today's lecture will answer the following alternative question:

Take the group action of $GL_n(\mathbb{C})$ on $\mathrm{Herm}_n(\mathbb{C})$ , where $A \in GL_n(\mathbb{C})$ acts on $M \in \mathrm{Herm}_n(\mathbb{C})$
via $M \mapsto A^{\dagger}MA$ . What are the orbits of this group action?

Notably, because $\mathbb{R} \subseteq \mathbb{C}$ , our classification of Hermitian forms over $\mathbb{C}$ also works equally well toward classifying symmetric forms over $\mathbb{R}$ : all symmetric forms over $\mathbb{R}^n$ under some basis look diagonal with entries $-1$ , $1$ , and $0$ .

§ Orthogonality and Nondegenerate Subspaces

Our proof relies on a couple definitions.

Definition (Orthogonal Subspace). Let $V$ be a finite-dimensional $\mathbb{C}$ -vector space with Hermitian form $\langle \bullet, \bullet \rangle$ . Given a subspace $W \subseteq V$ , its orthogonal subspace is $W^{\perp} := \{ v \in V \mid \langle w, v \rangle = 0 ~~ \forall \ w \in W\}$ .

Example. Depending on our choice of Hermitian form, orthogonality can be very counter-intuitive!

Take $V = \mathbb{C}^2$ with Hermitian form $\langle x, y \rangle = x^{\dagger} \left[\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right] y$ . Then $\left(\begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\right)$ is orthogonal to itself!
Take $V = \mathbb{C}^2$ with Hermitian form $\langle x, y \rangle = x^{\dagger} \left[\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}\right] y$ . Then $V^{\perp}$ is nontrivial: it's $\left\{ \left(\begin{smallmatrix} 0 \\ \lambda \end{smallmatrix}\right) \mid \lambda \in \mathbb{C}\right\}$ .

Definition (Nondegenerate). A Hermitian form $\langle x, y \rangle$ over a $\mathbb{C}$ -vector space $V$ is nondegenerate if $V^{\perp} = \{0\}$ . More generally, a subspace $W \subseteq V$ is nondegenerate if $W \cap W^{\perp} = \{0\}$ .

Example. As we saw in the previous example, $V = \mathbb{C}^2$ with $\langle x, y \rangle = x^{\dagger} \left[\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}\right]y$ is degenerate.

Example. Also as in the previous example, consider $V = \mathbb{C}^2$ with Hermitian form $\langle x, y \rangle = x^{\dagger} \left[\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right] y$ .

This Hermitian form is nondegenerate.
The subspace $\left\{ \left(\begin{smallmatrix} \lambda \\ 0 \end{smallmatrix}\right) \mid \lambda \in \mathbb{C}\right\}$ is also nondegenerate.
The subspace $W = \left \{ \left(\begin{smallmatrix} \lambda \\ \lambda \end{smallmatrix}\right) \mid \lambda \in \mathbb{C} \right\}$ is degenerate! In particular, $W^{\perp} = W$ .

Theorem. (Nondegenerate $\iff$ Invertible) Let $V$ be a $\mathbb{C}$ -vector space with a chosen basis and a Hermitian form $\langle x, y \rangle = x^{\dagger}Ay$ . Then $\langle \bullet, \bullet \rangle$ is nondegenerate if and only if $A$ is invertible.

Proof: Suppose $\langle \bullet, \bullet \rangle$ is degenerate, so there is a nonzero $y \in V$ such that $\langle x, y \rangle = 0$ for all $x \in V$ . Equivalently, for all $x \in V$ , we have $x^{\dagger}Ay = 0$ . This is possible if and only if $Ay = 0$ , that is, $A$ is not invertible. $\blacksquare$

The previous theorem will be useful for our proof of Hermitian Classification. This next one is just a fun fact.

Theorem. (Nondegenerate $\implies$ Injectivity) Let $V$ be a $\mathbb{C}$ -vector space with nondegenerate Hermitian form. Suppose $v, v' \in V$ satisfy $\langle v, w \rangle = \langle v', w \rangle$ for all $w \in V$ . Then $v = v'$ .

Proof: Observe $\langle v, w \rangle = \langle v', w \rangle$ implies $\langle v - v', w \rangle = 0$ for all $w \in V$ , which means $v - v' \in V^{\perp}$ . But the Hermitian form was nondegenerate, so $V^{\perp} = \{0\}$ , meaning $v - v' = 0$ . $\blacksquare$

§ Proving Hermitian Classification

Definitions aside, here's the driving theorem of our proof of Hermitian form classification.

Theorem. (Orthogonal Decomposition) Let $W$ be a nondegenerate subspace of a nondegenerate $\mathbb{C}$ -vector space $V$ . Then the decomposition $V = W \oplus W^{\perp}$ holds. Furthermore, $W^{\perp}$ is also nondegenerate.

Assuming this theorem to be true, we propose the following proof of Hermitian form classification:

Proof: We may assume $V$ is nondegenerate, as the degenerate subspace $V^{\perp}$ may be “taken out”.

Argue by induction on $\dim V$ , where the base case of $\dim V = 1$ is trivial. Now pick a nondegenerate subspace $W \subseteq V$ and set the basis of $V$ to the union of bases of $W$ and $W^{\perp}$ .

Because the “Orthogonal Decomposition” theorem holds, the $M$ in $\langle x, y \rangle = x^{\dagger}My$ must now look like:
$M = \begin{bmatrix} M_1 & 0 \\ 0 & M_2 \end{bmatrix} ~~ \text{ where } ~ ~\langle x, y \rangle = x^{\dagger}M_1y ~ \text{ in } W, ~ \text{ and } \langle x, y \rangle = x^{\dagger}M_2y ~ \text{ in } W^{\perp}.$
The zero-entries are guaranteed because $\langle e_1, e_2 \rangle = 0$ for any basis vectors $e_1 \in W$ and $e_2 \in W^{\perp}$ .

Apply the inductive hypothesis on $W$ and $W^{\perp}$ so $M_1$ and $M_2$ are diagonal with only $-1$ s, $0$ s, and $1$ s. Rearrange the diagonal entries to finish. $\blacksquare$

Pretty simple, right? It remains to justify the statements written in teal. Starting with the first…

Theorem. (Justification #1: Degeneracy) If the Hermitian form classification holds for all nondegenerate $V$ , then it also holds for all degenerate $V$ .

Proof: Set the basis of $V$ by first choosing a basis for $V^{\perp}$ and then extending it to all of $V$ . This way, $V = V^{\perp} \oplus V^*$ for some nondegenerate vector space $V^*$ . Then the $M$ in $\langle x, y \rangle = x^{\dagger}My$ looks like:

M = \begin{bmatrix} \ \boxed{0} & \boxed{0} \\ \ \boxed{0} & M^* \end{bmatrix}\!, ~ \text{ where } ~ \boxed{0} ~ \text{ indicates a submatrix of zeroes.}

Then since $V^*$ is nondegenerate, $M^*$ may be written in the desired form, at which point $M$ is too. $\blacksquare$

Now to justify the base case…

Theorem. (Justification #2: The Base Case) Let $\langle \bullet, \bullet \rangle$ be a Hermitian form over a one-dimensional $\mathbb{C}$ -vector space. Under an appropriate choice of basis, $\langle \bullet, \bullet \rangle$ can be made to look like either $\left[\begin{smallmatrix} 1 \end{smallmatrix}\right]$ , $\left[\begin{smallmatrix} -1 \end{smallmatrix}\right]$ , or $\left[\begin{smallmatrix} 0 \end{smallmatrix}\right]$ .

Proof: In the one-dimensional case, any change of basis $M \mapsto A^{\dagger}MA$ looks like the following:

\text{Given } M = \left[\begin{smallmatrix} m \end{smallmatrix}\right] \text{ and } A = \left[\begin{smallmatrix} \lambda \end{smallmatrix}\right], ~ \text{we have } A^{\dagger}MA = \left[\begin{smallmatrix} |\lambda|^2 m \end{smallmatrix}\right].

And $|\lambda|^2$ can be any positive real number, so $|\lambda|^2m$ can be made to equal either $1$ , $-1$ , or $0$ as needed. $\blacksquare$

For Justification #3, we'll use the fact that $V$ is nondegenerate… (Do you see where?)

Theorem. (Justification #3: Picking a Nondegenerate Subspace) Let $V$ be a nondegenerate $\mathbb{C}$ -vector space with Hermitian form $\langle \bullet, \bullet \rangle$ . Then $V$ has a $1$ -dimensional nondegenerate subspace $W$ .

Proof: We begin with the following lemma.

Lemma. There exists some $v \in V$ with $\langle v, v \rangle \neq 0$ .

Proof: Pick some $x, y \in V$ such that $\langle x, y \rangle \neq 0$ . We can also pick some $\lambda \in \mathbb{C}$ so that $\langle x, \lambda y \rangle = \lambda \langle x, y \rangle$ is real and nonzero. Because of this real-ness, $\langle x, \lambda y \rangle = \langle \lambda y, x \rangle$ , so the Polarization Identity holds:
$\langle x, \lambda y \rangle = \dfrac{\langle x + \lambda y, x + \lambda y \rangle - \langle x, x \rangle - \langle \lambda y, \lambda y \rangle}{2}.$
Since $\langle x, \lambda y \rangle \neq 0$ , at least one of $\{x + \lambda y, \ x, \ \lambda y \}$ must be a valid choice for $v \in V$ with $\langle v, v \rangle \neq 0$ . $\square$

Take $v$ as in the above lemma. Then for any nonzero $a, b \in \mathbb{C}v$ , we have $\langle a, b \rangle \neq 0$ , so $\mathbb{C}v$ is nondegenerate. $\blacksquare$

And finally, the most crucial step: justifying the “Orthogonal Decomposition” theorem…

Theorem. (Justification #4: Orthogonal Decomposition) Let $W$ be a nondegenerate subspace of a nondegenerate $\mathbb{C}$ -vector space $V$ . Then the decomposition $V = W \oplus W^{\perp}$ holds, and $W^{\perp}$ is nondegenerate.

Proof: Suppose $n = \dim V$ and $k = \dim W$ . We claim the following:

Claim. We always have $\dim W^{\perp} \geq n - k$ (regardless of whether $W$ or $V$ are degenerate or not!).

Proof: The only constraints on elements $w^{\perp} \in W^{\perp}$ are that $\langle w^{\perp}, w \rangle = 0$ for all $w \in W$ . It's good enough to ensure this constraint holds for exactly $k$ choices of $w$ , where these $k$ choices together span $W$ .

So $W^{\perp}$ is the set of vectors $w^{\perp} \in V$ satisfying $k$ linear constraints of the form $\langle w^{\perp}, w \rangle = 0$ , which must have dimension at least $n - k$ . $\square$

However, $W \cap W^{\perp} = \{0\}$ because $W$ is nondegenerate. Thus, $V = W \oplus W^{\perp}$ follows for dimension reasons.

It remains to show $W^{\perp}$ is nondegenerate; this is where the nondegeneracy of $V$ becomes necessary. Suppose, for the sake of contradiction, that some nonzero $v \in W^{\perp}$ were also in $(W^{\perp})^{\perp}$ . Then:

Since $v \in (W^{\perp})^{\perp}$ , it follows that $v$ is orthogonal to all of $W^{\perp}$ .
Since $v \in W^{\perp}$ , it follows that $v$ is orthogonal to all of $W$ .

But $V = W \oplus W^{\perp}$ , so the above implies $v$ is orthogonal to all of $V$ , contradicting the nondegeneracy of $V$ . $\blacksquare$

And that's the proof!

§ Signatures and Sylvester's Law of Inertia

Okay, one more thing… does every Hermitian form have a unique representation in this classification?

Definition (Signature). Take a Hermitian matrix $M$ in the form described by this classification. Its signature is the ordered triple $(n_1, n_0, n_{-1})$ with the number of $+1$ s, $0$ s, and $-1$ s, respectively, along the diagonal of $M$ .

Theorem. (Sylvester's Law of Inertia) Every Hermitian form has a unique signature, irrespective of basis.

Proof: Suppose a Hermitian form has two signatures $(n_1, n_0, n_{-1})$ and $(n_1', n_0', n_{-1}')$ . Then:

The signature $(n_1, n_0, n_{-1})$ yields a decomposition $V = W_1 \oplus W_0 \oplus W_{-1}$ .
The signature $(n_1', n_0', n_{-1}')$ yields a decomposition $V = W_1' \oplus W_0' \oplus W_{-1}'$ .

Suppose, without loss of generality, that $n_1' > n_1$ .

Then $W_1' \cap (W_0 \oplus W_{-1}) \neq \{0\}$ for size reasons. But this means there is some nonzero $v \in V$ such that:

We have $v \in W_1'$ , so $v$ is a nonzero vector with positive norm.
We have $v \in W_0 \oplus W_{-1}$ , so $v$ is a nonzero vector with nonpositive norm.

And obviously, that's a contradiction. $\blacksquare$

§ Foreshadowing: Signatures are Critical Points…

To conclude, we provide some handwavy foreshadowing on some applications of today's lecture.

Recall from last lecture (Section 16.3) that the Hessian is a matrix that encodes the “quadratic”-ness of a function $f: \mathbb{R}^n \to \mathbb{R}$ . It turns out the signature of the Hessian describes the behavior of the critical point!