Representation Theory of Finite Groups II: Linear Algebra

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 ∈ F and v, w ∈ 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 ∈ L such that every vector in L has the form cv for some c ∈ 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 ϕ : V → W is a homomorphism of the underlying abelian groups which is also compatible with scalar multiplication, i.e. for all a ∈ F and v ∈ V we have ϕ(av) = aϕ(v). A bijective linear map is called an isomorphism. Example 1.6. For any vector space V and any a ∈ 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 ϕ : L → L has the form ϕ(v) = av for some a ∈ 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 ϕ : V → W , there is a corresponding dual or transpose linear map ϕ∗ : W ∗ → V ∗ which sends λ : W → F to λ ◦ ϕ : V → F . There is a canonical linear map ev : V → (V ∗)∗ deﬁned by

ev(v)(λ) := λ(v),

where v ∈ V and λ ∈ V ∗. Proposition 1.11. The map ev : V → (V ∗)∗ is injective. Proof. We give the proof for ﬁnite-dimensional V . Suppose v ∈ ker ev, meaning λ(v) = 0 for all λ ∈ 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 ∈ W and a ∈ F we have

• 0 ∈ W , • v + w ∈ W , • −v ∈ W , • av ∈ 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 ϕ : V → W is a subspace im ϕ ⊂ W . The kernel ker ϕ ⊂ V , meaning the vectors which ϕ sends to 0, is a subspace of V . Sometimes im ϕ is called the range of ϕ and ker ϕ is called the nullspace. Note that a linear map ϕ : V → W is injective if and only if ker ϕ = 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 ∈ S and a1, ··· , an ∈ 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)i∈S of elements of F indexed by elements of S, with the property that ai = 0 for all but finitely many i ∈ S. Addition and scalar multiplication are performed ⊕S th componentwise. Thus for each i ∈ S we have a vector ei ∈ 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 = {1, ··· , n} 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 ϕ : F ⊕S → V by the formula X ϕ((ai)i∈S) := aif(i), i∈S

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 ϕ : F → F be a linear map. The corresponding m × n matrix has as columns the ϕ(ei) ∈ F , 1 ≤ i ≤ n. This is the inverse construction. n m Choose A ∈ Matm×n(F ) and B ∈ Matn×p(F ) with coeﬃcients aij and bjk, and write ϕ : F → F and ψ : F m → F p for the corresponding linear maps. Then for any 1 ≤ i ≤ n we have X BAei = ( aijbjk)1≤k≤p = ψ(ϕ(ei)). j

Since ψ ◦ϕ 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 ϕ, ψ : V → W are two linear maps satisfying ϕ(v) = ψ(v) for all v ∈ S, then ϕ = ψ.

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 ∈ S, and nonzero a1, ··· , an ∈ 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 ∈ L. Observe that if S ⊂ V is linearly independent then 0 ∈/ 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 .

The next result is foundational to linear algebra. Proposition 3.4. Any vector space admits a basis. We will not prove this result for infinite-dimensional vector spaces, since we will mostly work with finite- dimensional vector spaces in this class. Also, in general this fact requires (is even equivalent to) the axiom of choice, and the argument involves some unenlightening set theory. Proof. Assume V is finite-dimensional. If V = 0 then the empty set is a basis, so assume V 6= 0. Since V has a finite spanning set, by the well-ordering principle there is a minimal n such that there exist v1, ··· , vn which span V . We claim that v1, ··· , vn is a basis for V , and it suffices to prove linear independence. Suppose that a1v1 + ··· + anvn = 0

for some a1, ··· , an ∈ F . If there exists 1 ≤ k ≤ n such that ak 6= 0, then

−1 vk = −ak (a1v1 + ··· + ak−1vk−1 + ak+1vk+1 + ··· anvn).

It follows that v1, ··· , vk−1, vk+1, ··· vn spans V , but this contradicts the minimality of n.

Corollary 3.4.1. Any ﬁnite-dimensional vector space is isomorphic to F n for some n. Proof. Just combine Proposition 3.4 and Exercise 3.3.

∗ If v1, ··· , vn is a basis of V , then there is a natural dual basis λ1, ··· , λn of V , defined by λi(vj) = δij for all 1 ≤ i, j ≤ n. Here δij is the Dirac delta, defined by δii = 1 and δij = 0 for i 6= j. In particular dim V ∗ = dim V . Corollary 3.4.2. For any two finite-dimensional vector spaces V and W , a choice of bases for both deter- mines a linear isomorphism Hom(V,W )−→ ˜ Matm×n(F ). Composition of linear maps corresponds to matrix multiplication. Proof. Apply Proposition 2.7.

Recall that if A is a matrix with coeﬃcients aij, then the transpose of A is the matrix obtained by switching the rows and columns, i.e. its coeﬃcients are aji.

4 Proposition 3.5. Let ϕ : V → W be a linear map and ﬁx bases of V and W , so ϕ corresponds to a matrix. Then the matrix of ϕ∗ : W ∗ → V ∗ with respect to the dual bases of W ∗ and V ∗ is the transpose of the matrix of ϕ.

Proof. Let v1, ··· , vm and w1, ··· , wn be bases of V and W respectively, and λ1, ··· , λm and µ1, ··· , µn the ∗ ∗ P dual bases of V and W . Let the matrix of ϕ have coeﬃcients aij, i.e. ϕ(vi) = j aijwj. Then we have

∗ X X ϕ (µj)(vi) = µj(ϕ(vi)) = µj(aikwk) = aij = ( aijλi)(vi) k i

∗ P for all i, j, so ϕ (µj) = i aijλi as desired.

So all ﬁnite-dimensional vector spaces and linear maps between them become familiar objects from matrix algebra after choosing bases.

Definition 3.6. A finite-dimensional vector space V has dimension n ≥ 0 provided that V =∼ F n. We write dim V = n. Since all vector spaces have bases, they have dimensions as well. The next result says that dimension is is well-defined.

Theorem 3.7. There is an isomorphism F m =∼ F n if and only if m = n. We deduce the theorem from the following technical lemma.

Lemma 3.8. Let V be a ﬁnite-dimensional vector space. If v1, ··· , vm ∈ V are linearly independent and w1, ··· , wn span V , then m ≤ n. Proof. See Theorem 2.6 in Sheldon Axler’s Linear Algebra Done Right.

Proof of Theorem 3.7. The non-obvious assertion amounts to the following: if v1, ··· , vm and w1, ··· , wn are bases of V , then m = n. Since v1, ··· , vm are linearly independent and w1, ··· , wn span, we have m ≤ n by the lemma. Symmetrically n ≤ m, so m = n.

4 Direct sums, quotient spaces, and the rank-nullity theorem

In this section we discuss some of the basic operations on vector spaces. Deﬁnition 4.1. The direct sum V ⊕ W of two vector spaces V and W is the direct sum of the underlying abelian groups with scalar multiplication deﬁned by

a · (v, w) = (a · v, a · w).

Notice that there are natural surjections πV : V ⊕ W → V and πW : V ⊕ W → W given by πV (v, w) = v and πW (v, w) = w, as well as natural injections ιV : V → V ⊕W and ιW : W → V ⊕W given by ιV (v) = (v, 0) and ιW (w) = (0, w). Proposition 4.2. Let U be a third vector space. Then the map

Hom(U, V ⊕ W ) −→ Hom(U, V ) ⊕ Hom(U, W ) given by ϕ 7→ (πV ◦ ϕ, πW ◦ ϕ) and the map

Hom(V ⊕ W, U) −→ Hom(V,U) ⊕ Hom(W, U) given by ϕ 7→ (ϕ ◦ ιV , ϕ ◦ ιW ) are linear isomorphisms.

5 Proof. It is not hard to check that both maps are linear. The inverse of the ﬁrst map is

(ϕ, ψ) 7→ ιV ◦ ϕ + ιW ◦ ψ,

the second (ϕ, ψ) 7→ ϕ ◦ πV + ψ ◦ πW .

Proposition 4.3. If V and W are ﬁnite-dimensional, then so is V ⊕W and dim(V ⊕W ) = dim V +dim W .

Proof. If v1, ··· , vm and w1, ··· , wn are bases for V and W respectively, then

(v1, 0), ··· , (vm, 0), (0, w1), ··· , (0, wn) is a basis for V ⊕ W .

Given a collection of vector spaces {Vi}i∈I where I is any indexing set, finite or infinite, we can endow the direct product Y Vi := {(vi)i∈I | vi ∈ Vi} i∈I with componentwise addition and scalar multiplication to make it into a vector space. The direct sum L Q i∈I Vi is the subspace of i∈I Vi consisting of I-tuples with vi = 0 for all but finitely many I. This is the most general kind of vector space direct sum. Q Exercise 4.4. Generalize the first universal property from Proposition 4.2 to Vi, and the second to L i∈I i∈I Vi. Note that the latter also generalizes Proposition 2.4. Now we discuss vector space quotients. Suppose that W is a subspace of V . Exercise 4.5. Show that the formula a·(v+W ) = a·v+W , where a ∈ F and v ∈ W , is a well-defined scalar multiplication operation on the quotient abelian group V/W which makes it into a vector space. Observe that the natural surjection π : V → V/W given by π(v) = v + W is linear. Definition 4.6. Call V/W with the scalar multiplication defined above the quotient vector space. The universal property of the vector space quotient is analogous to that of the group quotient. Exercise 4.7. Show that the projection π : V → V/W has the following universal property. If U is another vector space and ϕ : V → U is a linear map with the property that W ⊂ ker ϕ, then there is a unique linear map ϕ0 : V/W → U satisfying ϕ0 ◦ π = ϕ. Proposition 4.8 (Isomorphism theorem for vector spaces). Let ϕ : V → W be a linear map. Denote by π : V → V/ ker ϕ the projection and ι : im ϕ → W the inclusion. Then there is a unique isomorphism ϕ0 : V/ ker ϕ→˜ im ϕ satisfying ι ◦ ϕ0 ◦ π = ϕ. Proof. The existence and uniqueness of the linear map ϕ0 follows from Exercise 4.7, and the isomorphism theorem for groups implies that ϕ0 is an isomorphism.

For finite-dimensional vector spaces, the quotient operation subtracts dimensions. Proposition 4.9. If V is finite-dimensional then so is W , and dim(V/W ) = dim V − dim W. Proof. The first assertion is Proposition 2.7 in Linear algebra done right. For the second, choose a basis v1, ··· , vm of W and extend it to a basis v1, ··· , vn of V . We claim that

vm+1 + W, ··· , vn + W is a basis of V/W .

6 To see the linear independence, suppose that

am+1(vm+1 + W ) + ··· + an(vn + W ) = 0. It follows that am+1vm+1 + ··· + anvn ∈ W,

so there exist a1, ··· , am ∈ F such that

am+1vm+1 + ··· + anvn = a1v1 + ··· + amvm.

The linear independence of the v1, ··· , vn implies that a1 = ··· = an = 0. It remains to prove that these vectors span V/W . Any vector v ∈ V can be written v = a1v1 +···+anvn, and then we have

v + W = (am+1vm+1 + W ) + ··· + (anvn + W ) = am+1(vm+1 + W ) + ··· + an(vn + W ).

Corollary 4.9.1 (Rank-nullity theorem). If V is a ﬁnite-dimensional vector space and ϕ : V → W is a linear map, then im ϕ is ﬁnite-dimensional and

dim V = dim ker ϕ + dim im ϕ.

Proof. Combine Propositions 4.9 and 4.8.

Sometimes dim ker ϕ is called the nullity of ϕ and dim im ϕ the rank. Corollary 4.9.2. Suppose that V and W are ﬁnite-dimensional and that dim V = dim W . Then if a linear map ϕ : V → W is either injective or surjective, it is bijective. Proof. If ϕ is injective then dim im ϕ = dim V = dim W , which implies im ϕ = W , i.e. ϕ is surjective. If ϕ is surjective then dim ker ϕ = dim V − dim im ϕ = dim V − dim W = 0, so ϕ is injective.

This result is extremely useful. Compare to the fact that if two ﬁnite sets have the same cardinality, a injective or surjective map between them is automatically bijective. Corollary 4.9.3. The map ev : V → (V ∗)∗ is an isomorphism. Proof. Proposition 1.11 says that ev is injective. Now apply Corollary 4.9.2.

5 Tensor products

If the direct sum construction “adds” vector spaces, then the tensor product “multiplies” them. The study of tensor products is the beginning of multilinear algebra. Deﬁnition 5.1. If V , W , and U are vector spaces over F , then an F -bilinear map B : V × W → U is a map which is F -linear in each variable separately. That is, for any w ∈ W the map V → U given by v 7→ B(v, w) is F -linear, and for any v ∈ V the map W → U given by w 7→ B(v, w) is F -linear.

Example 5.2. Let Pn be the vector space of polynomials in one variable of degree at most n with coeﬃcients in F , i.e. n Pn := {a0 + a1x + ··· + anx | a0, ··· , an ∈ F }.

Then multiplication is a bilinear map Pm × Pn → Pm+n.

7 Example 5.3. For any vector space V there is a natural evaluation map V ∗ × V → F deﬁned by (λ, v) 7→ λ(v), which is bilinear. The best way to deﬁne the tensor product is to use its universal property: it is the universal recipient of a bilinear map. Then it is still necessary to prove that the tensor product exists.

Deﬁnition 5.4. The tensor product of two vector spaces V and W is a vector space V ⊗F W (hereafter abbreviated as V ⊗ W ) which receives a bilinear map B : V × W → V ⊗ W with the following property: for any vector space U and any bilinear map B0 : V × W → U, there is a unique linear map ϕ : V ⊗ W → U satisfying ϕ ◦ B = B0. Proposition 5.5. The tensor product of any two vector spaces exists.

Proof. We give a proof for ﬁnite-dimensional V and W ; the general case is similar. Choose bases v1, ··· , vm and w1, ··· , wn for V and W respectively, and put M V ⊗ W := F · (vi, wj), i,j

i.e. V ⊗ W is the vector space with a basis consisting of ordered pairs (vi, wj) with 1 ≤ i ≤ m, 1 ≤ j ≤ n. P P Deﬁne B : V × W → V ⊗ W as follows: given v ∈ V and w ∈ W write v = i aivi, w = j bjwj, and put X B(v, w) := aibj(vi, wj). i,j

It is not diﬃcult to verify that B has the desired universal property.

A vector in V ⊗ W is called a tensor. Given v ∈ V and w ∈ W , the tensor v ⊗ w is by deﬁnition the image of (v, w) under the universal bilinear map V × W → V ⊗ W (called B in the proof of Proposition 5.5). A tensor of the form v ⊗ w is called a pure tensor, and a general tensor is a sum of pure tensors. Exercise 5.6. For any vector spaces V , W , and U, construct a linear isomorphism

Hom(V ⊗ W, U)−→ ˜ Hom(V, Hom(W, U)).

The tensor product operation is functorial, meaning it extends naturally to linear maps.

Exercise 5.7. If ϕ1 : V1 → W1 and ϕ2 : V2 → W2 are linear maps, show that the formula

(ϕ1 ⊗ ϕ2)(v1 ⊗ v2) := ϕ1(v1) ⊗ ϕ2(v2)

deﬁnes a linear map ϕ1 ⊗ ϕ2 : V1 ⊗ V2 → W1 ⊗ W2. For ψ1 : W1 → U1 and ψ2 : W2 → U2, prove that

(ψ1 ⊗ ψ2) ◦ (ϕ1 ⊗ ϕ2) = (ψ1 ◦ ϕ1) ⊗ (ψ2 ◦ ϕ2).

The remaining basic properties of the tensor product are summarized in the following two results. Proposition 5.8. For any vector spaces V , W , and U, there are canonical isomorphisms

(i) F ⊗ V →˜ V ⊗ F →˜ V , (ii) V ⊗ W →˜ W ⊗ V , and (iii) (V ⊗ W ) ⊗ U →˜ V ⊗ (W ⊗ U).

8 Proof. For (i), observe that the scalar multiplication map F ×V → V is bilinear and therefore factors through a linear map F ⊗ V → V . It is not hard to see that v 7→ 1 ⊗ v is inverse to this map. That F ⊗ V →˜ V ⊗ F will follow from (ii). As for (ii), the map V × W → W ⊗ V given by (v, w) 7→ w ⊗ v is bilinear, hence extends to a linear map V ⊗ W → W ⊗ V . Symmetrically, there is a linear map W ⊗ V → V ⊗ W given by w ⊗ v 7→ v ⊗ w, which is inverse to the ﬁrst. To prove (iii), observe that (V ⊗ W ) ⊗ U and V ⊗ (W ⊗ U) have the same universal property: they are both the universal recipient of a trilinear map from V × W × U. It follows that they are canonically isomorphic.

Proposition 5.9. Suppose V and W are ﬁnite-dimensional. If v1, ··· , vm and w1, ··· , wn are any two bases of V and W respectively, then the vi ⊗ wj with 1 ≤ i ≤ m, 1 ≤ j ≤ n form a basis of V ⊗ W . In particular dim(V ⊗ W ) = dim V · dim W . Proof. Both assertions follow directly from our construction of V ⊗ W in the proof of Proposition 5.5.

Associativity of the tensor product means that we can unambiguously deﬁne the tensor product

V1 ⊗ · · · ⊗ Vn ⊗n of several vector spaces V1, ··· ,Vn. In particular we have the tensor power V for any n ≥ 1, i.e. the tensor product of V with itself n times. It is convenient to define V ⊗0 := F . It follows from Proposition 5.9 that if V is finite-dimensional, then V ⊗n has dimension (dim V )n. Example 5.10. Let V and W be two vector spaces. Then there is a natural bilinear map V ∗ × W −→ Hom(V,W ), which sends a pair (λ, w) to the linear map ϕ : V → W defined by ϕ(v) = λ(v)w. This defines a linear map V ∗ ⊗ W → Hom(V,W ). We claim is that if V and W are finite-dimensional then this is an isomorphism (actually this is is still true if W is infinite-dimensional, and when V is infinite-dimensional the map is injective but not surjective). Choose bases v1, ··· , vm and w1, ··· , wn of V and W respectively, and let λ1, ··· , λm be the dual basis of ∗ P V . To see the injectivity, suppose that aijλi(v)wj = 0 for all v ∈ V . Taking v = vi for some 1 ≤ i ≤ m, P i,j this implies that j aijwj = 0, so aij = 0 because the w1, ··· , wn are linearly independent. Now it follows that the map is an isomorphism by Corollary 4.9.2. Exercise 5.11. For finite-dimensional V , W , and U, show that under the isomorphism from Example 5.10, composition Hom(W, U) ⊗ Hom(V,W ) −→ Hom(V,U) agrees with the map W ∗ ⊗ U ⊗ V ∗ ⊗ W −→ V ∗ ⊗ U given by µ ⊗ u ⊗ λ ⊗ w 7→ µ(w) · λ ⊗ u. Definition 5.12. For a finite-dimensional vector space V , the trace map is defined as the composition tr : End(V )−→ ˜ V ∗ ⊗ V −→ F where the second map is induced by the evaluation map from Example 5.3.

Exercise 5.13. Fix ϕ ∈ End(V ) and choose a basis v1, ··· , vn of V . If the aij are the coeﬃcients of the matrix of ϕ with respect to v1, ··· , vn, then show that n X tr ϕ = aii. i=1

In particular, tr idV = dim V .

9 Exercise 5.14. Use Exercise 5.11 or Exercise 5.13 to prove that tr(ψ ◦ ϕ) = tr(ϕ ◦ ψ) for any linear endomorphisms ϕ, ψ ∈ End(V ).

6 Symmetric and exterior powers

Symmetric and antisymmetric multilinear maps give rise to corresponding variations of the tensor power operation. Deﬁnition 6.1. A bilinear map B : V × V → W is called symmetric (respectively antisymmetric) provided that B(v1, v2) = B(v2, v1) (respectively B(v1, v2) = −B(v2, v1)) for all v1, v2 ∈ V . Proposition 6.2. There is a vector space Sym2 V with a symmetric bilinear map V × V → Sym2 V which has the property that any symmetric bilinear map V × V → W factors uniquely through it. Similarly, there is a vector space V2 V and a universal alternating bilinear map V × V → V2 V .

Proof. Let R1 and R2 be the subspaces of V ⊗V spanned by vectors of the form v⊗w−w⊗v and v⊗w+w⊗v respectively, where v, w ∈ V . Put

2 2 ^ Sym V := (V ⊗ V )/R1 and V := (V ⊗ V )/R2.

The bilinear map V × V → V ⊗ V is universal among all bilinear maps V × V → W . It is not hard to check that a bilinear form is symmetric (respectively antisymmetric) if and only if the corresponding linear map V ⊗ V → W vanishes on R1 (respectively R2).

More generally, a multilinear map M : V × · · · × V → W is symmetric (respectively antisymmetric) if interchanging two variables does not change the map (respectively introduces a minus sign). Equivalently, for any v1, ··· , vk and any σ ∈ Sk, we have

M(vσ(1), ··· , vσ(k)) = M(v1, ··· , vk) (respectively M(vσ(1), ··· , vσ(k)) = sgn(σ) · M(v1, ··· , vk)).

There are corresponding symmetric and antisymmetric powers Symk V and Vk V which receive universal symmetric and antisymmetric multilinear maps from V × · · · × V . Exercise 6.3. Exercise 5.7 immediately implies that the tensor power operation is functorial. Use this to show that the symmetric and exterior power operations are functorial. This means that a linear endomorphism ϕ : V → V gives rise to linear endomorphisms Symk ϕ and Vk ϕ of Symk V and Vk V respectively, and that if ψ : V → V is another endomorphism then Symk(ψ◦ϕ) = Symk ψ◦Symk ϕ and Vk(ψ◦ϕ) = Vk ψ◦Vk ϕ. k Vk For any v1, ··· , vk ∈ V , the image of v1 ⊗ · · · ⊗ vk in Sym V (respectively V ) is denoted by v1 ··· vk (respectively v1 ∧ · · · ∧ vk). Note that if vi = vj for any i 6= j, then v1 ∧ · · · vk = 0. Proposition 6.4. If V has ﬁnite dimension n, then

n + k − 1 ^k n dim Symk V = and dim V = . k k

Proof. Let v1, ··· , vn be a basis of V . One checks that the vi1 ∧ · · · ∧ vik 1 ≤ i1 < ··· < ik ≤ n form a basis Vk n of V , and there are k choices of k distinct indices. Similarly, the vi1 ··· vik where 1 ≤ i1 ≤ · · · ≤ ik ≤ n form a basis of Symk V , and this basis has the desired cardinality.

Exercise 6.5. Check the details of Proposition 6.4. In particular we have dim Vk V = dim Vn−k V and dim Vn V = 1. A nonzero element of the line Vn V has the form v1 ∧ · · · ∧ vn, where v1, ··· , vn is a basis of V .

10 Deﬁnition 6.6. If V is a vector space of dimension n, we deﬁne the associated determinant line as det V := Vn V . For a linear endomorphism ϕ : V → V , its determinant is ^n det ϕ := ϕ ∈ End(det V ) = F.

Exercise 6.3 implies that det(ψ ◦ ϕ) = det ψ ◦ det ϕ for any ϕ, ψ ∈ End(V ).

Exercise 6.7. Fix ϕ ∈ End(V ) and choose a basis v1, ··· , vn of V . Then if aij denote the matrix coeﬃcients of ϕ with respect to the chosen basis, show that X det ϕ = sgn(σ)aiσ(i).

σ∈Sn In particular, det is a degree n homogeneous polynomial in the matrix coeﬃcients.

Given a matrix A ∈ Matn×n(F ) with coeﬃcients aij, the (i, j) minor Mij of A is the determinant of the (n − 1) × (n − 1) matrix obtained by deleting the ith row and jth column of A. The adjugate adj(A) of A is i+j the n × n matrix whose (i, j) entry is (−1) Mji. Exercise 6.8. Show that an invertible linear map has nonzero determinant. Prove that

A adj(A) = det(A) · I, where A is any square matrix and I denotes the identity matrix. Use this to prove that a linear map with nonzero determinant is invertible.

7 Eigenvalues and eigenvectors

The deeper structure theory of operators starts with the study of eigenvalues and eigenvectors. Deﬁnition 7.1. Let V be a vector space and ϕ ∈ End(V ). An eigenvector of ϕ is a vector v ∈ V such that ϕ(v) = λv for some λ ∈ F , and λ is the associated eigenvalue.

For a given λ ∈ F , observe that v ∈ V is an eigenvector with eigenvalue λ if and only if v ∈ ker(λ·idV −ϕ). In particular, the eigenvectors of ϕ with eigenvalue λ form a subspace

Vλ := ker(λ · idV −ϕ).

For example, we have V0 = ker ϕ. L Proposition 7.2. The morphism λ∈F Vλ → V induced by the inclusions Vλ → V is injective. In particular, a set of eigenvectors with pairwise distinct eigenvalues is linearly independent.

Proof. Say we are given pairwise distinct λ1, ··· , λn ∈ F , and for each 1 ≤ i ≤ n an eigenvector vi ∈ Vλ , P i satisfying i vi = 0. Assume, for the sake of contradiction, that vi 6= 0 for all 1 ≤ i ≤ n. Reorder if necessary so that for some 1 ≤ m < n the set v1, ··· , vm is linearly independent, but v1, ··· , vm, vk is linearly dependent for any m < k ≤ n. In particular there exist c1, ··· , cm ∈ F such that

c1v1 + ··· + cmvm = vm+1,

and by applying ϕ we obtain

c1λ1v1 + ··· + cmλmvm = λm+1vm+1 = λm+1(c1v1 + ··· + cmvm).

Since λm+1 6= λi for any 1 ≤ i ≤ m, this implies that c1 = ··· = cm = 0 and hence vm+1 = 0, a contradiction.

Recall that a square matrix A with coeﬃcients aij is called diagonal if aij = 0 for all i 6= j. If there exists an invertible matrix B such that BAB−1 is diagonal, then A is called diagonalizable.

11 Proposition 7.3. The following are equivalent:

(i) the matrix of ϕ with respect to any basis of V is diagonalizable,

(ii) there is a basis of V with respect to which the matrix of ϕ is diagonal, (iii) there is a basis of V consisting of eigenvectors of ϕ, L (iv) the canonical morphism λ∈F Vλ → V is an isomorphism.

Proof. To see that (i) implies (ii), let v1, ··· , vn be a basis of V and A the matrix corresponding to ϕ. Then (i) says that there exists an invertible matrix B such that BAB−1 is diagonal, say with entries λ1, ··· , λn. Let ψ ∈ GL(V ) be the operator corresponding to B (with respect to the chosen basis). Then −1 −1 ψ (v1), ··· , ψ (vn) is a basis of V because ψ is invertible, and we have

−1 −1 −1 −1 −1 ϕ(ψ (vi)) = ψ ((ψ ◦ ϕ ◦ ψ )(vi)) = ψ (λivi) = λiψ (vi).

It is not hard to see that (ii) implies (i). That (ii) implies (iii) is clear. L Now (iii) says that the eigenvectors span V , i.e. λ∈F Vλ → V is surjective. Proposition 7.2 says that this map is injective, whence (iv). Finally, we claim (iv) implies (ii). For this, choose a basis of each Vλ, so (iv) implies that the union of these bases as λ varies is a basis of V .

One reason diagonal matrices are useful is because they can easily be raised to a large power, since one only has to raise all the diagonal entries to the same power. Since

(BAB−1)k = BAkB−1,

diagonalizable matrices can be treated similarly. We are immediately faced with a computational problem, namely the determination of the eigenvalues and eigenvectors of a given operator ϕ ∈ End(V ). From now on we will assume V has finite dimension n. Definition 7.4. The characteristic polynomial of ϕ is defined by the formula

χϕ(x) := det(x · idV −ϕ).

Exercise 6.7 implies that χϕ(x) is a monic polynomial in x of degree n.

k Exercise 7.5. Show that the coeﬃcient of x in χϕ(x) is given by

^n−k (−1)n−k tr( ϕ).

In particular the constant term is (−1)n det ϕ, and the coeﬃcient of xn−1 is − tr ϕ. The next result explains the relationship with eigenvalues.

Proposition 7.6. The roots of χϕ(x) are the eigenvalues of ϕ.

Proof. Observe that λ ∈ F is an eigenvalue of ϕ if and only if the operator λ · idV −ϕ is singular, meaning not invertible. This in turn is equivalent to

χϕ(λ) = det(λ · idV −ϕ) = 0.

Corollary 7.6.1. If F is algebraically closed (e.g. F = C) then any linear operator on a nonzero vector space has an eigenvalue.

12 Recall that a square matrix A with coeﬃcients aij is called upper triangular if i > j implies aij = 0.

Exercise 7.7. Check that if A is upper triangular, we have det A = a11 ··· ann. Corollary 7.7.1. If F is algebraically closed, then any square matrix is similar to an upper triangular matrix.

n n Proof. Let A be an n×n matrix and ϕ the corresponding operator on F . Now ϕ has an eigenvector v1 ∈ F , n i.e. it stabilizes the line F · v1. In particular ϕ is an operator on F /(F · v1), and we can ﬁnd an eigenvector n n there; let v2 ∈ F be a representative. Since ϕ stabilizes F · v1 + F · v2, it acts on F /(F · v1 + F · v2) and hence has an eigenvector there, etc. Continuing in this way, we obtain a basis v1, ··· , vn of V with the property that vk+1 projects to an eigenvector of ϕ in

n F /(F · v1 + ··· + F · vk).

A straightforward computation shows that ϕ is upper triangular with respect to this basis.

From a more geometrical perspective, this result says that if F is algebraically closed, then any linear operator ϕ on a ﬁnite-dimensional vector space V stabilizes a complete ﬂag, i.e. a sequence of subspaces

0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V such that dim(Vk/Vk−1) = 1 for all 1 ≤ k ≤ n. In the proof of the proposition

Vk = F · v1 + ··· + F · vk.

Exercise 7.8. Show that if F = R, then any operator ϕ on V has an invariant subspace of dimension one or two. Deduce that if V has odd dimension, then ϕ has an eigenvalue.