Having established the structure of vector spaces, we now study the maps between them. Linear maps were defined in Chapter 4; in this chapter we organize them into vector spaces of their own, represent them by matrices once bases are chosen, and see how the matrix changes when the bases change. We end with the determinant, the most important function of a square matrix, and with the determinant and trace of a linear operator. Throughout, is a field.
Recall (Definition 4.5.12) that a linear map between vector spaces over is a function satisfying and for all and ; it then preserves all linear combinations. Just as group homomorphisms preserve the group operation, linear maps preserve both vector space operations.
Example 5.1.1.
- The zero map , the identity , and the scaling for a fixed are linear.
- Differentiation , , is linear, since it acts coefficientwise. (This is the formal derivative; no limits are involved.)
- For , evaluation , , is linear.
- If is a basis of , the coordinate isomorphism , , is linear.
- If , the projection onto along , sending (, ) to , is linear, because the decompositions of and of are and .
Proposition 5.1.2 (Linear maps are determined by a basis). Let be a basis of , and let be arbitrary vectors. There is exactly one linear map with for all , namely
Proof. Uniqueness: a linear map with satisfies .
Existence: every has unique coordinates (Proposition 4.6.9), so the formula defines a function , and . If and , then and are the coordinate expressions of and . Hence and . ∎
Note. This is the universal property of a basis: specifying a linear map out of is the same as choosing where to send the basis vectors, and the choice is completely free. The same proof works for an arbitrary basis , finite or not: every function extends uniquely to a linear map .
Definition 5.1.3 (Space of linear maps). The set of all linear maps from to is denoted When , we write and call its elements endomorphisms of , or linear operators on .
Proposition 5.1.4 ( is a vector space). The set is a vector space over , with the operations
- addition: for all ;
- scalar multiplication: for all .
Proof. We must check that and are linear when and are. Let and . Then and similarly The vector space axioms for follow from those of , because the operations are defined pointwise. The zero vector is the zero map , and the additive inverse of is . ∎
Proposition 5.1.5 (Composition). Let , and . Then is linear, and Consequently is a ring, with addition as above, multiplication given by composition, and identity element .
Proof. Composites of linear maps are linear by Proposition 4.5.13. For , linearity of gives , which is the first identity; the second holds by the definition of ; and the third follows from . For : it is an abelian group under addition, composition of maps is associative with identity , and the first two identities are the distributive laws. ∎
The ring is not commutative when (Example 5.2.11). Its units are the invertible linear operators on , which form the group . We will see in Corollary 5.2.8 that when both spaces are finite-dimensional.
5.1.1 Dependence on the Scalar Field
The notions of basis, dimension and linearity depend on the choice of the field of scalars. The same set can be a vector space over different fields, with different dimensions.
Example 5.1.6. as a vector space over different fields:
- Over : , with basis . Every complex number is .
- Over : , with basis . Every complex number is the -linear combination , and the expression is unique.
More generally, if is a subfield of (for example or ), then any -vector space is also a -vector space by restriction of scalars: we only use scalar multiplication by elements of . In particular itself is a -vector space. A -linearly independent subset of is also -linearly independent, since a -linear relation is in particular a -linear relation; so .
Proposition 5.1.7 (Linearity over subfields). Let and be -vector spaces and a subfield. Then:
- and are -vector spaces by restriction of scalars;
- every -linear map is also -linear.
Thus .
Proof. The axioms for scalar multiplication by elements of are special cases of those for , and for all holds in particular for . ∎
The inclusion can be strict:
Example 5.1.8. Complex conjugation is -linear but not -linear. Consider It is -linear: for and we have and , since . It is not -linear: for and ,
So the natural maps between spaces depend on which scalars we allow. Complex conjugation is an -linear automorphism of that is not -linear, because it does not commute with multiplication by .