Representation theory of finite groups II: Linear algebra Justin Campbell July 8, 2015 1 Definitions and basic results In linear algebra we work over a field of scalars F . Feel free to assume F = Q, R, C, or any other example you're comfortable with. In high school we learn that a vector is something that has direction and magnitude. From an algebraic point of view, the essential characteristics of vectors are that they can be added together and multiplied by scalars. Definition 1.1. A vector space over F is an abelian group V , written additively, equipped with an additional operation F × V ! V called scalar multiplication and written (a; v) 7! av. Scalar multiplication is required to satisfy, for all a; b 2 F and v; w 2 V , • 1 · v = v, • (ab)v = a(bv), • (a + b)v = av + bv, and • a(v + w) = av + aw. We call elements of a vector space vectors. Example 1.2. Let n ≥ 1. We denote by F n the vector space consisting of n-tuples of elements of F , which we often think of as n × 1 matrices. Addition and scalar multiplication are defined componentwise. This standard vector space is the one encountered in matrix algebra courses. Example 1.3. The set of continuous functions R ! R forms a (huge) vector space over R. Example 1.4. If L is a vector space with the property that there exists nonzero v 2 L such that every vector in L has the form cv for some c 2 F , we call L one-dimensional or a line. For instance, F itself is a line (the case n = 1 of Example 1.2). As with any class of mathematical objects, morphisms between vector spaces are as important as vector spaces themselves (and more interesting). Definition 1.5. Let V and W be vector spaces over F . An F -linear map ' : V ! W is a homomorphism of the underlying abelian groups which is also compatible with scalar multiplication, i.e. for all a 2 F and v 2 V we have '(av) = a'(v): A bijective linear map is called an isomorphism. Example 1.6. For any vector space V and any a 2 F , the map V ! V given by v 7! av is linear. Exercise 1.7. Show that if L is a line then any linear map ' : L ! L has the form '(v) = av for some a 2 F . 1 Exercise 1.8. Show that the composition of two linear maps is linear, and that the inverse of a bijective linear map is linear. The set of all linear maps V ! W is itself a vector space, which we denote by Hom(V; W ). When V = W we write End(V ) := Hom(V; V ) (short for endomorphism), which has the composition operation along with addition. If V 6= 0 then End(V ) is not a group under composition, since then the zero map V ! V is not invertible. Write GL(V ) := Aut(V ) for the group (under composition) of automorphisms of V , i.e. bijective linear maps V ! V , called the general linear group of V . Exercise 1.9. Check that Hom(V ) is a vector space and GL(V ) is a group. Definition 1.10. The dual space of a vector space V is the vector space V ∗ := Hom(V; F ). Given a linear map ' : V ! W , there is a corresponding dual or transpose linear map '∗ : W ∗ ! V ∗ which sends λ : W ! F to λ ◦ ' : V ! F . There is a canonical linear map ev : V ! (V ∗)∗ defined by ev(v)(λ) := λ(v); where v 2 V and λ 2 V ∗. Proposition 1.11. The map ev : V ! (V ∗)∗ is injective. Proof. We give the proof for finite-dimensional V . Suppose v 2 ker ev, meaning λ(v) = 0 for all λ 2 V ∗. ∗ If v 6= 0 then there is a basis v1; ··· ; vn of V with v1 = v, hence a dual basis λ1; ··· ; λn of V . But then λ1(v) = 1 6= 0, a contradiction. 2 Subspaces and matrices In this section we continue our discussion of the basic concepts. Definition 2.1. A subspace W of a vector space V is a subset which is closed under addition and scalar multiplication and itself forms a vector space. To see that a subset W ⊂ V is a subspace, it is enough to check that W is an additive subgroup which is closed under scalar multiplication. In symbols, this means that for any v; w 2 W and a 2 F we have • 0 2 W , • v + w 2 W , • −v 2 W , • av 2 W . Any vector space V has the zero subspace and the subspace V which contains all vectors. Example 2.2. The image of any linear map ' : V ! W is a subspace im ' ⊂ W . The kernel ker ' ⊂ V , meaning the vectors which ' sends to 0, is a subspace of V . Sometimes im ' is called the range of ' and ker ' is called the nullspace. Note that a linear map ' : V ! W is injective if and only if ker ' = 0. Example 2.3. Given an arbitrary subset S ⊂ V of a vector space, the span of S is the subspace of V consisting of vectors of the form a1v1 + ··· + anvn for some v1; ··· ; vn 2 S and a1; ··· ; an 2 F . Such an expression is called a linear combination of vectors in S. If the span of S is V we say that S spans V . If V is spanned by a finite set we call V finite-dimensional. 2 If S is any set, not necessarily given as a subset of a vector space, we can construct the universal vector space spanned by S as follows (the meaning of the word universal will be explained). This vector space is ⊕S denoted by F and as a set consists of collections (ai)i2S of elements of F indexed by elements of S, with the property that ai = 0 for all but finitely many i 2 S. Addition and scalar multiplication are performed ⊕S th componentwise. Thus for each i 2 S we have a vector ei 2 F , defined to be 1 in the i place and 0 ⊕S everywhere else, and vectors in F are by definition linear combinations of the ei. In particular there is a ⊕S canonical injection S ! F given by i 7! ei. Note that if S = f1; ··· ; ng then F ⊕S = F n. Proposition 2.4. For any vector space V , restriction along the canonical injection S ! F ⊕S defines a bijection from Hom(F ⊕S;V ) to the set of maps S ! V . Proof. If f : S ! V is any map, then we can define ' : F ⊕S ! V by the formula X '((ai)i2S) := aif(i); i2S which makes sense because all but finitely many ai = 0. This is the inverse construction. Exercise 2.5. Check that the span of a subset S ⊂ V is the image of the linear map F ⊕S ! V corresponding to the inclusion S ! V . For any m; n ≥ 1 let Matm×n(F ) denote the vector space of m × n matrices with coefficients in F . As is the convention in matrix algebra, we can view vectors in F n as n × 1 matrices and multiply them on the left by m × n matrices to obtain m × 1 matrices, i.e. vectors in F m. Exercise 2.6. Check that matrix multiplication, as described above, defines a linear map n m Matm×n(F ) ! Hom(F ;F ): The next result is fundamental. It is really a special case of Proposition 2.4. Proposition 2.7. The linear map from Exercise 2.6 is an isomorphism. Moreover, under this isomorphism multiplication of matrices goes to composition of linear maps. n m m Proof. Let ' : F ! F be a linear map. The corresponding m × n matrix has as columns the '(ei) 2 F , 1 ≤ i ≤ n. This is the inverse construction. n m Choose A 2 Matm×n(F ) and B 2 Matn×p(F ) with coefficients aij and bjk, and write ' : F ! F and : F m ! F p for the corresponding linear maps. Then for any 1 ≤ i ≤ n we have X BAei = ( aijbjk)1≤k≤p = ('(ei)): j Since ◦' and the linear map corresponding to BA agree on the basis e1; ··· ; en, they are equal by Exercise 2.8 below. Exercise 2.8. Let V and W be vector spaces and suppose S spans V . Show that if '; : V ! W are two linear maps satisfying '(v) = (v) for all v 2 S, then ' = . 3 Bases and dimension If we are given a subset S ⊂ V which spans V , we might ask whether S contains redundancies, i.e. whether there is a proper subset of S which still spans V . This can be detected using the notion of linear dependence. 3 Definition 3.1. A subset S ⊂ V of a vector space is called linearly dependent provided that there exists n ≥ 1, vectors v1; ··· ; vn 2 S, and nonzero a1; ··· ; an 2 F with the property that a1v1 + ··· + anvn = 0: If S is not linearly dependent we call it linearly independent. If S is linearly independent and spans V we say that S is a basis for V . Example 3.2. A basis of a line L is just a nonzero vector v 2 L. Observe that if S ⊂ V is linearly independent then 0 2= S. Exercise 3.3. In the notation of Proposition 2.4, check that (i) the morphism F ⊕S ! V is injective if and only if S is linearly independent, (ii) the morphism F ⊕S ! V is surjective if and only if S spans V , and (iii) the morphism F ⊕S ! V is an isomorphism if and only if S is a basis of V .