Inner Products — Studies in Algebra and Group Theory

So far we have studied vector spaces equipped only with their linear structure. Adding geometric notions—length, angle, orthogonality—requires additional data: an inner product. This structure transforms abstract linear algebra into something closer to Euclidean geometry, while revealing that many geometric intuitions extend far beyond .

We begin with the more general notion of bilinear forms, then specialize to the symmetric positive-definite case (inner products over ) and its complex analogue (Hermitian inner products over ).

9.1.1 Bilinear Forms

Before defining inner products, we study the broader class of bilinear forms. This generality clarifies which properties of inner products come from bilinearity alone and which require symmetry or positivity.

Definition 9.1.1 (Bilinear form). Let be a vector space over a field . A bilinear form on is a map that is linear in each variable:

for all , .
for all .
for all .

Note. A bilinear form is not a linear map . If we view as a vector space (the external direct sum ), a linear map would satisfy , i.e., . But bilinearity gives . The tensor product is the correct domain for linearizing bilinear maps.

Definition 9.1.2 (Symmetric and skew-symmetric forms). A bilinear form is:

Symmetric if for all .
Skew-symmetric (or alternating) if for all .

Example 9.1.1 (The standard dot product). On , the map is a symmetric bilinear form. This is the “dot product” familiar from calculus.

Example 9.1.2 (A skew-symmetric form). On , define . This equals and is skew-symmetric. Geometrically, it computes signed area.

Matrix Representation

Given a basis of , a bilinear form is determined by the values . By bilinearity: Writing and as column vectors, this is: where is the matrix of with respect to the chosen basis.

Proposition 9.1.1 (Symmetry and the matrix).

is symmetric if and only if .
is skew-symmetric if and only if .

Proof. We have and . So for all iff for all , iff . The skew-symmetric case is similar. ∎

Bilinear Forms and the Dual Space

There is a natural connection between bilinear forms and linear maps to the dual space.

Proposition 9.1.2 (The map ). Given a bilinear form , define by i.e., is the linear functional . Then is a linear map.

Proof. For and : So . ∎

Conversely, any linear map determines a bilinear form .

Theorem 9.1.1 (Bilinear forms and ). The map defines an isomorphism of vector spaces where denotes the space of bilinear forms on .

Definition 9.1.3 (Rank and nondegeneracy). The rank of a bilinear form is . Equivalently, it is the rank of the matrix in any basis.

The form is nondegenerate if is an isomorphism, i.e., if . Equivalently, is nondegenerate iff is invertible.

Note. For finite-dimensional, , so is an isomorphism iff it is injective iff it is surjective. Nondegeneracy means: if for all , then .

Proposition 9.1.3 (Dimension of the space of bilinear forms). If , then .

Proof. We have . Since , we get . Alternatively, a bilinear form is determined by an matrix. ∎

9.1.2 Inner Products over the reals

To do geometry, we need more than bilinearity. The key additional properties are symmetry (so that ) and positive-definiteness (so that “length” is always nonnegative).

Definition 9.1.4 (Inner product (real case)). Let be a vector space over . An inner product on is a bilinear form satisfying:

Symmetry: for all .
Positive-definiteness: for all , with equality iff .

A vector space equipped with an inner product is called an inner product space.

Note. Positive-definiteness only makes sense over an ordered field where “” is meaningful. In practice, this means . We cannot directly define inner products over or finite fields using this definition—we will see the complex workaround shortly.

Example 9.1.3 (The standard inner product on ). The dot product is an inner product. Symmetry is clear, and with equality iff all .

Example 9.1.4 (A weighted inner product). On , for positive constants , define This is an inner product (check!). Different weights emphasize different coordinates.

Proposition 9.1.4 (Inner products are nondegenerate). Every inner product is nondegenerate.

Proof. Suppose for all . Taking , we get . By positive-definiteness, . ∎

Note. The converse is false: a nondegenerate symmetric bilinear form need not be positive-definite. For example, the Minkowski metric on is nondegenerate and symmetric, but can be negative.

9.1.3 Hermitian Inner Products over the complexes

Why can’t we directly use the same definition over ? If is a bilinear form on a complex vector space with , consider . So if , then . Bilinearity and positive-definiteness are incompatible over .

The solution is to weaken bilinearity: we require linearity in one variable and conjugate-linearity in the other.

Definition 9.1.5 (Sesquilinear form). A map on a complex vector space is sesquilinear if:

and .
for .
for .

That is, is linear in the first argument and conjugate-linear in the second.

Note. Some authors (including Axler) use the opposite convention: conjugate-linear in the first argument, linear in the second. The choice is a matter of convention; we follow Artin here.

Definition 9.1.6 (Hermitian form). A sesquilinear form is Hermitian (or conjugate-symmetric) if for all .

Note. Conjugate-symmetry implies , so for all . This makes positive-definiteness meaningful.

Definition 9.1.7 (Hermitian inner product). A Hermitian inner product on a complex vector space is a sesquilinear form that is:

Conjugate-symmetric: .
Positive-definite: for all , with equality iff .

Example 9.1.5 (The standard Hermitian inner product on ). Define where is the conjugate transpose. Then:

.
, with equality iff .

Proposition 9.1.5 (The map ). For a Hermitian form , the map is a conjugate-linear map : If is positive-definite, then is injective (and hence bijective when is finite-dimensional).

Note. Various properties carry over from the real case. A positive-definite Hermitian form is nondegenerate. Given a subspace , its orthogonal complement is a subspace, and when is positive-definite.

9.1.4 Norms and the Cauchy-Schwarz Inequality

An inner product induces a notion of length via the norm.

Definition 9.1.8 (Norm). Let be an inner product space (real or Hermitian). The norm of a vector is This is well-defined since .

Definition 9.1.9 (Orthogonality). Vectors are orthogonal, written , if .

Proposition 9.1.6 (Basic properties of the norm).

, with equality iff .
for .
(Pythagorean theorem) If , then .

Proof. (i) and (ii) follow from the definitions. For (iii): since (using symmetry/conjugate-symmetry). ∎

The Pythagorean theorem suggests that orthogonal vectors behave like legs of a right triangle. More generally:

This motivates defining the angle between nonzero vectors. But first we need to know that so that the ratio lies in .

Theorem 9.1.2 (Cauchy–Schwarz inequality). For all in an inner product space, with equality iff and are linearly dependent.

Proof. The inequality is trivial if . Assume ; by scaling, we may assume .

Decompose along : let . Since is nondegenerate and , we have . Write Then , so by the Pythagorean theorem: Taking square roots gives , which is the inequality for .

Equality holds iff , i.e., . ∎

Definition 9.1.10 (Angle between vectors). For nonzero vectors in a real inner product space, the angle between them is defined by The Cauchy–Schwarz inequality ensures this ratio lies in .

Corollary 9.1.1 (Triangle inequality). For all in an inner product space,

Proof. ∎

9.1.5 Orthogonality and Orthogonal Complements

Definition 9.1.11 (Orthogonal complement). Let be an inner product space and a subset. The orthogonal complement of is

Proposition 9.1.7 (Basic properties of orthogonal complements).

is a subspace of (even if is not).
.
.
and .

Proof. (i) If and , then for any : (ii)–(iv) are immediate from the definition. ∎

Note. For a general bilinear form (not necessarily symmetric), we must distinguish “left-orthogonal” ( for all ) from “right-orthogonal” ( for all ). For symmetric forms and inner products, these coincide.

The key result is that an inner product space decomposes as for any subspace .

Lemma 9.1.1 (Dimension of orthogonal complement). Let be a nondegenerate bilinear form on a finite-dimensional space , and let be a subspace. Then If is an inner product (positive-definite), then .

Proof. Consider the composition where and is restriction. The kernel of is exactly .

By rank-nullity: .

Since is an isomorphism (nondegeneracy) and is surjective, is surjective. Thus , giving .

For the second claim: if , then . By positive-definiteness, . ∎

Theorem 9.1.3 (Orthogonal decomposition). Let be a finite-dimensional inner product space and a subspace. Then

Proof. By the lemma, and . The dimension formula for sums gives So , and the sum is direct since . ∎

Example 9.1.6 (Standard orthogonal complement in ). Let with the standard inner product . If is a subspace, then is the “usual” orthogonal complement: vectors perpendicular to every vector in .

For instance, if in , then .

Example 9.1.7 (Skew-symmetric forms can have ). Let with the skew-symmetric form . Let be any 1-dimensional subspace spanned by a nonzero vector .

For any , we have .

So . In particular, , and . This cannot happen for inner products.

9.1 Inner Products