Hermitian Inner Products, the Complex Spectral Theorem, and Classifying Bilinear Forms — Studies in Algebra and Group Theory

17 Hermitian Inner Products, the Complex Spectral Theorem, and Classifying Bilinear Forms

10 October 2025 · Denis Auroux

We’ve studied real inner product spaces and seen that a self-adjoint or orthogonal operator can be expressed in a particularly simple form (diagonal, resp. block diagonal) in a well-chosen orthonormal basis.

Hermitian inner products

Now on to the analogue of all this for complex vector spaces: Hermitian inner products.

As previously noted, a bilinear form on a complex vector space can’t be positive definite, since . Solution: abandon -linearity in one of the two variables, and only require “conjugate linear”.

Definition 17.1. A Hermitian form on a complex vector space is such that is sesquilinear:

, ;
, however (conjugate: );

and is conjugate-symmetric: .

Note. Caution: this is as in Artin. Axler has for the second input.

Conjugate symmetry .

Definition 17.2. A Hermitian inner product is a positive definite (conjugate-symmetric) Hermitian form, i.e. for all , and .

Remark. , , is now a complex antilinear map ! (.)

Still, various things carry over from the real case:

positive definite nondegenerate (i.e. ).
Given a subspace , its orthogonal is also a subspace, and . (-antilinearity doesn’t affect being a -subspace; positive definite implies .)

Definition 17.3. An orthonormal basis of with a Hermitian inner product is a basis such that

Theorem 17.4. admits an orthonormal basis.

Same proof as in the real case (by induction on : first pick with , then take an orthonormal basis of ) (or Gram–Schmidt…).

Corollary 17.5. Every finite-dimensional Hermitian inner product space is isomorphic to with the standard Hermitian inner product,

In matrix form: , where is the conjugate transpose.

Example 17.6 (Not-quite-example (Fourier series)). Let , the infinitely differentiable functions ( 1-periodic functions ). Define Then the functions are orthogonal, .

is not a basis of ; their span is the space of trigonometric polynomials. We can think of Fourier series as orthogonal projection onto . (This will make more sense with some analysis… or even better, Hilbert spaces.)

Adjoint, self-adjoint, unitary, normal

Definition 17.7. Let be a complex vector space, a Hermitian inner product, .

The adjoint of is such that for all ( for all ).
is self-adjoint if , i.e. for all .
is unitary if for all , i.e. .
is normal if .

Unitary operators form a subgroup (). Note (multiplication by any complex number of norm 1).
Note: in an orthonormal basis, . This is because for all . (Equivalently: the -entry of is the -component of , i.e.

So: self-adjoint complex operators are described by Hermitian matrices, .

The complex spectral theorem

Theorem 17.8 (The complex spectral theorem). Let be a finite-dimensional complex vector space, a Hermitian inner product, and self-adjoint () or unitary () or normal (). Then there exists an orthonormal basis consisting of eigenvectors of , i.e. is diagonalizable in an orthonormal basis, with eigenvalues if self-adjoint / (unit circle) if unitary.

Proof. Let be an eigenvector (it exists since is algebraically closed), , .

Claim 1: is also an eigenvector of , . Indeed (writing ),

Claim 2: is invariant under and . Indeed, and ( for all ✓).

So, proof by induction: take . ∎

Back to nondegenerate symmetric bilinear forms

Back to (not necessarily definite) nondegenerate symmetric bilinear forms:

Suppose is a finite-dimensional vector space over and is a nondegenerate symmetric bilinear form. Can we classify such ?

Remark. is something called a quadratic form. We can recover from if :

Classification approach: find some vector such that , and then look at (, so when ). Then study …

Proposition 17.9. Over , any nondegenerate symmetric bilinear form admits a basis such that

Note. Caution: Hermitian forms are what most “normal” people care about, however.

Proof.

Since one of , , is nonzero, nonzero implies the existence of such that .
Let . Then consider . We have since , and .
The restriction of to is nondegenerate because the matrix of in the basis , is invertible (rank ) iff is invertible (rank ).
Complete the proof by induction on dimension (assuming the result holds in dimension , take + a basis of such that ).

∎

Proposition 17.10. Over , any nondegenerate symmetric bilinear form admits a basis such that i.e. we can assume

We say has signature . (The case = positive definite.) ( = the maximal dimension of a subspace such that is positive definite, = the maximal dimension of a subspace such that is negative definite.)

Proof. Same as in the complex case, except we can’t always scale to ; instead we can only force . ∎

Over , things get much harder: number theory enters!

Example 17.11. , . There is no such that : clearing denominators, we get with with no common factor (in particular not all even). However and , so necessarily all are even: contradiction.

Whereas does have such that ().

Notes not reached in this lecture, covered at the start of Lecture 18

The skew-symmetric case

What about the skew-symmetric case? (Suppose .)

We can still find a “standard basis” for a finite-dimensional vector space with a nondegenerate skew-symmetric bilinear form (a.k.a. a symplectic form), but the process is slightly different since for all .

Instead: pick any nonzero ; since is nondegenerate, is nonzero, so there exists such that , and we can make it by scaling . Now we find (if has then ), so Then study the restriction of to the latter subspace (induction on dimension).

Proposition 17.12. Let be finite-dimensional over , , and a nondegenerate skew-symmetric bilinear form . Then is even, and has a basis such that i.e. the matrix of is block diagonal:

The group of linear transformations preserving is called the symplectic group .

Next time: tensor product & multilinear algebra. This gives us a way to think of bilinear (or multilinear) maps as linear maps from a new vector space .