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 brings abstract linear algebra closer to Euclidean geometry, while revealing that many geometric intuitions extend far beyond .
We begin with the general notion of a bilinear form, then specialize to the symmetric positive-definite case (inner products over ) and its complex analogue (Hermitian inner products over ). We then study the operators adapted to an inner product—adjoints and self-adjoint, normal, orthogonal, and unitary operators—and prove the spectral theorems. Next we return to bilinear forms without any positivity: symmetric forms, which we classify over and , and skew-symmetric (symplectic) forms. The chapter ends with tensor products and multilinear algebra, which turn bilinear and multilinear maps into linear ones.
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 correct domain for linearizing bilinear maps is the tensor product , constructed in Section 9.8.
Definition 9.1.2 (Symmetric, skew-symmetric, and alternating forms). A bilinear form on is:
- Symmetric if for all .
- Skew-symmetric if for all .
- Alternating if for all .
Note. Every alternating form is skew-symmetric: expanding gives . Conversely, if , a skew-symmetric form is alternating, since gives . In characteristic the two notions differ: there , so skew-symmetric means symmetric, and the dot product on is skew-symmetric but not alternating ().
Example 9.1.3 (The standard dot product). On , the map is a symmetric bilinear form. This is the “dot product” familiar from calculus.
Example 9.1.4 (A skew-symmetric form). On , define . This equals and is alternating, hence skew-symmetric. Over , it computes the signed area of the parallelogram spanned by the two vectors.
Matrix Representation
Let be finite-dimensional with a basis . 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. Conversely, every matrix defines a bilinear form by this formula, so the values can be chosen freely.
Proposition 9.1.5 (Symmetry and the matrix).
- is symmetric if and only if .
- is skew-symmetric if and only if .
Proof. We have and , since a matrix equals its transpose. So for all iff for all , iff (take , ). The skew-symmetric case is the same with a sign. ∎
The matrix of a bilinear form depends on the basis, and it changes differently from the matrix of an operator.
Proposition 9.1.6 (Change of basis for bilinear forms). Let and be bases of , and let be the change-of-basis matrix, so that . If and are the matrices of a bilinear form in the bases and , then
Note. Matrices related by with invertible are called congruent. Compare with for operators (Corollary 5.3.4): the two notions agree when , which will be the case for change of basis between orthonormal bases (Section 9.5). Since is invertible, congruent matrices have the same rank, and .
There is a natural connection between bilinear forms and linear maps to the dual space.
Proposition 9.1.7 (The map ). Given a bilinear form , define by i.e., is the linear functional . Then is a linear map.
Proof. For and , and every : So . ∎
Theorem 9.1.8 (Bilinear forms and ). Let denote the vector space of bilinear forms on . The map is an isomorphism of vector spaces Its inverse sends to the bilinear form .
Proof. The map is linear: Given , the map is linear in because is linear, and linear in because ; so . By construction and , so the two maps are inverse to each other. ∎
Note. In the language of Chapter 8, this isomorphism is natural. A linear map induces maps in the opposite direction, which make and contravariant functors on (Definition 8.2.5). The isomorphisms form a natural transformation between them (Definition 8.2.9): for and ,
Definition 9.1.9 (Rank and nondegeneracy). Let be finite-dimensional. The rank of a bilinear form is . The form is nondegenerate if is an isomorphism.
Note. Let be a basis of with dual basis , and let be the matrix of . Then , so the matrix of in these bases is . Hence , and is nondegenerate iff is invertible. Since , is an isomorphism iff it is injective, so nondegeneracy means: if for all , then . As is invertible iff is, this is equivalent to the same condition with the roles of the two variables exchanged.
Proposition 9.1.10 (Dimension of the space of bilinear forms). If , then .
Proof. We have by Theorem 9.1.8, and since . Alternatively, is an isomorphism from to the space of matrices, since the entries of can be chosen freely. ∎
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 9.1.11 (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 real vector space equipped with an inner product is called a (real) 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 9.1.12 (The standard inner product on ). The dot product is an inner product. Symmetry is clear, and with equality iff all .
Example 9.1.13 (A weighted inner product). On , for constants , define This is an inner product: it is clearly bilinear and symmetric, and , with equality iff all because every . Different weights emphasize different coordinates.
Proposition 9.1.14 (Inner products are nondegenerate). If is an inner product on , the map , , is injective. In particular, if is finite-dimensional, every inner product on 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 form on is nondegenerate and symmetric, but can be negative. Such forms are classified in Section 9.6.
9.1.3 Hermitian Inner Products over the Complex Numbers
Why can’t we directly use the same definition over ? If is a bilinear form on a complex vector space, then . So if , then : bilinearity and positive-definiteness are incompatible over .
The solution is to weaken bilinearity: we require linearity in one variable and only conjugate-linearity in the other.
Definition 9.1.15 (Sesquilinear form). A map on a complex vector space is sesquilinear if, for all and :
- and .
- .
- .
That is, is conjugate-linear in the first argument and linear in the second.
Note. This is the convention of Artin (and of most physics texts). Some authors, including Axler, use the opposite convention: linear in the first argument and conjugate-linear in the second. The two conventions are exchanged by swapping the arguments, , so every formula in one convention has a counterpart in the other; but formulas from the two conventions must not be mixed.
Definition 9.1.16 (Hermitian form). A sesquilinear form on a complex vector space is Hermitian (or conjugate-symmetric) if for all .
Note. Conjugate-symmetry implies , so for all . This makes positive-definiteness meaningful.
Definition 9.1.17 (Hermitian inner product). A Hermitian inner product on a complex vector space is a Hermitian form that is positive-definite: for all , with equality iff . A complex vector space with a Hermitian inner product is a (complex) inner product space.
Example 9.1.18 (The standard Hermitian inner product on ). Define where is the conjugate transpose (a row vector). Then is sesquilinear, and:
Note. If is a Hermitian form on a finite-dimensional space with basis , let and . Then and conjugate-symmetry says , i.e., . A matrix with is called Hermitian; conversely, every Hermitian matrix defines a Hermitian form on .
Convention. From now on, an inner product space is either a real vector space with an inner product or a complex vector space with a Hermitian inner product. We write in both cases and let be the scalar field. Over , complex conjugation is the identity, so every statement below specializes to the real case by ignoring the bars. In both cases,
The map is no longer linear, and we need a version of rank–nullity for such maps.
Lemma 9.1.19 (Conjugate-linear maps). Let and be complex vector spaces. A map is conjugate-linear if and for all , . If is conjugate-linear, then and are subspaces, and if is finite-dimensional, A composite of two conjugate-linear maps is linear, and a composite of a linear map and a conjugate-linear map (in either order) is conjugate-linear.
Proof. Let be the set with the same addition and with the new scalar multiplication . Since complex conjugation is a field automorphism of (, , ), the vector space axioms for follow from those of . A subset of is closed under all operations iff it is closed under all , i.e., under all scalar multiplications; so and have the same subspaces. A list is a basis of a subspace in iff it is one in , because and is a bijection; so dimensions agree. Finally, , so is linear, and the claims about and follow from rank–nullity (Theorem 6.1.12). The statements about composites are immediate: two conjugations cancel, . ∎
Proposition 9.1.20 (The map ). Let be a Hermitian form on a complex vector space . For , the map is linear, so lies in . The map is conjugate-linear: We call nondegenerate if is injective, i.e., if for all implies . A Hermitian inner product is nondegenerate, and if is finite-dimensional and is nondegenerate, then is bijective.
Proof. The functional is linear because is linear in its second argument, and for all , . If is positive-definite and for all , then , so . If is finite-dimensional and is injective, Lemma 9.1.19 gives , so is also surjective. ∎
Combining the real and complex cases, we obtain the fundamental fact that every linear functional on a finite-dimensional inner product space is “inner product with a vector.”
Theorem 9.1.21 (Riesz representation). Let be a finite-dimensional inner product space. For every there is a unique such that Equivalently, the map , , is a bijection. It is linear if and conjugate-linear if .
Proof. Over , for the bilinear form , which is nondegenerate by Proposition 9.1.14, so is an isomorphism. Over , is a conjugate-linear bijection by Proposition 9.1.20. ∎
9.1.4 Norms and the Cauchy–Schwarz Inequality
An inner product induces a notion of length via the norm.
Definition 9.1.22 (Norm). Let be an inner product space. The norm of a vector is This is well-defined since is a nonnegative real number.
Definition 9.1.23 (Orthogonality). Vectors are orthogonal, written , if (equivalently, ).
Proposition 9.1.24 (Basic properties of the norm).
- , with equality iff .
- for .
- (Pythagorean theorem) If , then .
Proof. (i) is positive-definiteness. (ii) . For (iii): since . ∎
The Pythagorean theorem suggests that orthogonal vectors behave like the legs of a right triangle. More generally, since , In a real inner product space, comparing with the law of cosines suggests defining the angle between nonzero vectors by . For this we first need to know that , so that the ratio lies in .
Theorem 9.1.25 (Cauchy–Schwarz inequality). For all in an inner product space, with equality iff and are linearly dependent.
Proof. If , both sides are and are dependent. Assume . Replacing by multiplies both sides by and does not change linear dependence, so we may assume .
Decompose along : Then , so and hence . By the Pythagorean theorem, Taking square roots gives .
Equality holds iff , i.e., iff , i.e. (as ) iff and are linearly dependent. ∎
Definition 9.1.26 (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 9.1.27 (Triangle inequality). For all in an inner product space,
9.1.5 Orthogonal Complements
Orthogonality makes sense for any bilinear or Hermitian form, not only for inner products.
Definition 9.1.28 (Orthogonal complement). Let be a bilinear form on a vector space , or a Hermitian form on a complex vector space , and let be a subset. The orthogonal complement of is For an inner product, .
Proposition 9.1.29 (Basic properties of orthogonal complements).
- is a subspace of (even if is not).
- .
- If is symmetric, skew-symmetric, or Hermitian, then .
- , and . For an inner product, .
Proof. (i) If and is a scalar, then for any , where for a bilinear form and for a Hermitian form.
(ii) is immediate. (iii) If and , then equals , , or , which is in each case; so . (iv) The first two statements are the definitions, and the last is Proposition 9.1.14 (or Proposition 9.1.20). ∎
Note. For a general bilinear form (neither symmetric nor skew-symmetric), we must distinguish “left-orthogonal” ( for all ) from “right-orthogonal” ( for all ). For symmetric, skew-symmetric, and Hermitian forms these coincide.
The key result is that an inner product space decomposes as for any subspace . We first count dimensions.
Lemma 9.1.30 (Dimension of the orthogonal complement). Let be a nondegenerate bilinear form, or a nondegenerate Hermitian form, on a finite-dimensional space , and let be a subspace. Then If is an inner product, then moreover .
Proof. Consider the composition where and is restriction, . The restriction is the dual map of the inclusion , which is injective, so is surjective (Corollary 6.2.16). The kernel of is exactly .
Since is bijective (by nondegeneracy) and is surjective, is surjective. It is linear for a bilinear form and conjugate-linear for a Hermitian form, so in either case rank–nullity (Theorem 6.1.12, or Lemma 9.1.19) gives
For the second claim: if , then , so by positive-definiteness. ∎
Theorem 9.1.31 (Orthogonal decomposition). Let be a finite-dimensional inner product space and a subspace. Then
Proof. An inner product is nondegenerate (Proposition 9.1.14 or Proposition 9.1.20), so by the lemma, and . Hence So , and the sum is direct since . ∎
Corollary 9.1.32 (Double orthogonal complement). Let be a nondegenerate symmetric, skew-symmetric, or Hermitian form on a finite-dimensional space (for instance an inner product). Then for every subspace .
Proof. By Proposition 9.1.29(iii), , and by Lemma 9.1.30 applied twice, . ∎
Example 9.1.33 (Standard orthogonal complement in ). Let with the standard inner product. If is a subspace, then is the “usual” orthogonal complement: the vectors perpendicular to every vector in .
For instance, if in , then .
Example 9.1.34 (Skew-symmetric forms can have ). Let with the skew-symmetric form of Example 9.1.4, which is nondegenerate (its matrix is invertible). Let be the line spanned by a nonzero vector .
For any , we have .
So . In particular, , and . This cannot happen for inner products.