BASIC LINEAR ALGEBRA
AN EXERCISE APPROACH
Gabriel Nagy
Kansas State University
c Gabriel Nagy CHAPTER 1
Vector spaces and linear maps
In this chapter we introduce the basic algebraic notions of vector spaces and linear maps.
1. Vector spaces Suppose k is a field. (Although the theory works for arbitrary fields, we will eventually focus only on the choices k = R, C.) Definition. A k-vector space is an abelian group (V, +), equipped with an external operation1 k × V 3 (λ, v) 7−→ λv ∈ V, called scalar multiplication, with the following properties: • λ · (v + w) = (λ · v) + (λ · w), for all λ ∈ k, v, w ∈ V . • (λ + µ) · v = (λv) + (µv), for all λ, µ ∈ k, v ∈ V . • (λ · µ)v = λ · (µ · v), for all λ, µ ∈ k, v ∈ V . • 1 · v = v, for all v ∈ V . The elements of a vector space are sometimes called vectors. Examples. The field k itself is a k-vector space, with its own multiplication as scalar multiplication. A trivial group (with one element) is always a k-vector space (with the only possible scalar multiplication).
1 Suppose V is a k-vector space. Prove that 0 · v = 0, for all v ∈ V. (The zero in the left-hand side is the field k. The zero in the right-hand side is the neutral element in the abelian group V .) Use the above fact to conclude that for any v ∈ V , the vector −v (the inverse of v in the abelian group V ) can also be described by −v = (−1) · v.
2 Fix a field k, a non-empty set I, and a family (Vi)i∈I of k-vector spaces. Q Consider the product i∈I Vk, equipped with the operations: • (vi)i∈I + (wi)i∈I = (vi + wi)i∈I ; • λ · (vi)i∈I = (λvi)i∈I .
1 When convenient, the symbol · may be omitted.
1 2 1. VECTOR SPACES AND LINEAR MAPS
Q Prove that i∈I Vk is a k-vector space. This structure is called the k-vector space direct product of the family (Vi)i∈I .
Definition. Suppose V is a k-vector space. A subset X ⊂ V is called a k-linear subspace, if • Whenever x, y ∈ X, we have x + y ∈ X. • Whenever x, y ∈ Xand λ ∈ k, we have λx ∈ X.
3 If X is a k-linear subspace of the k-vector space V , then X itself is a k-vector space, when equipped with the operations “inherited” from V . Prove than any linear subspace of V contains the zero vector 0 ∈ V
4 Let (Vi)i∈I be a family of k-vector spaces (indexed by a non-empty set I). For Q an element v = (vi)i∈I ∈ i∈I Vi let us define the set
bvc = {i ∈ I : vi 6= 0}. Prove that the set Y {v ∈ Vi : bvc is finite } i∈I Q is a linear subspace of Vi. This space is called the k-vector space direct sum i∈I L of the family (Vi)i∈I , and is denoted by i∈I Vi.
Definition. Suppose we have a family (Vi)i∈I of vector spaces. For a fixed L index j ∈ I, define the map εj : Vj → i∈I Vi as follows. For a vector v ∈ Vj we construct εj(v) = (wi)i∈I , where v if i = j w = i 0 if i 6= j. L We call the maps εj : Vj → i∈I Vi, j ∈ I, the standard inclusions. The maps Y πj : Vi 3 (vi)i∈I 7−→ vj ∈ Vj i∈I are called the coordinate maps.
5 Let (Vi)i∈I be a family of vector spaces. Prove that the standard inclusions εi, i ∈ I. are injective. In fact prove that πi ◦ εi = IdV . Prove that any element L i v ∈ i∈I Vi is given as X v = (εi ◦ πi)(v). i∈bvc P In other words, if v = (vi)i∈I , then v = i∈bvc εi(vi).
6 Suppose (Xj)j∈J is a family of k-linear subspaces of V . Prove that the inter- T section j∈J Xj is again a k-linear subspace of V . 1. VECTOR SPACES 3
Definition. Let V be a k-vector space, and let M ⊂ V be an arbitrary subset of V . Consider the family F = {X : X k-linear subspace of V , and X ⊃ M}. The set \ Spank(M) = X, X∈F which is a linear subspace of V by the preceding exercise, is called the k-linear span of M in V .
Convention. Spank(∅) = {0}. Example. The linear span of a singleton is described as
Spank({v}) = kv(= {λv : λ ∈ k}).
7 Prove that if M and N are subsets of a k-vector space V , with M ⊂ N, then we also have the inclusion Spank(M) ⊂ Spank(N). Give an example where M ( N, but their spans coincide.
8 Let V be a k-vector space, and M be a subset of V . For an element v ∈ V , prove that the following are equivalent:
(i) v ∈ Spank(M); (ii) there exists an integer n ≥ 1, elements x1, . . . , xn ∈ M, and scalars 2 λ1, . . . , λn ∈ k such that v = λ1x1 + ··· + λnxn.
Hint: First prove that the set of elements satisfying property (ii) is a linear subspace. Second, prove that the linear span of M contains all elements satisfying (ii).
Notation. Suppose V is a vector space, and A1,...,An are subsets of V . We define
A1 + ··· + A2 = {a1 + ··· + an : ak ∈ Ak, k = 1, . . . , n}.
9 Let V be a k-vector space. Suppose A1,...,An are k-homogeneous, in the sense that for every k = 1, . . . , n we have the equality:
Ak = {λx : λ ∈ k, x ∈ Ak}. Prove the equality
Span(A1 ∪ · · · ∪ An) = Span(A1) + ··· + Span(An).
10 Let V be a k-vector space, and (Xj)j∈J be a family of linear subspaces of V . For an element v ∈ V , prove that the following are equivalent: S (i) v ∈ Spank j∈J Xj); S (ii) there exists an integer n ≥ 1 and x1, . . . , xn ∈ j∈J Xj, such that v = x1 + ··· + xn.
2 From now on we will use the usual convention which gives the scalar multiplication prece- dence over addition. 4 1. VECTOR SPACES AND LINEAR MAPS
Comment. If X1,...,Xn are linear subspaces of the vector space V , then using the notation preceding Exercise ??, and the above result, we get
Span(X1 ∪ · · · ∪ Xn) = X1 + ··· + Xn.
11 In general, a union of linear subspaces is not a linear subspace. Give an 2 example of two linear subspaces X1,X2 ⊂ R , such that X1 ∪ X2 is not a linear subspace.
12 Prove that the union of a directed family of linear subspaces is a linear subspace. That is, if (Xj)j∈J is a family of k-linear subspaces of the k-vector space V , with the property
• For any j, k ∈ J there exists some ` ∈ J such that Xj ⊂ X` ⊃ Xk, S then j∈J Xj is again a k-linear subspace of V . Hint: Use the preceding exercise.
Definition. Suppose V is a k-vector space. A set M ⊂ V is said to be k- linearly independent, if (compare with Exercise ??) for every strict subset P ( M, one has the strict inclusion Spank(P ) ( Spank(M).
13 Let M be a subset in the k-vector space V . Prove that the following are equivalent: (i) M is linearly independent. (ii) If n ≥ 1 is an integer, if x1, . . . , xn ∈ M are different elements, and if λ1, . . . , λn ∈ k satisfy
λ1x1 + ··· + λnxn = 0,
then λ1 = ··· = λn = 0.
Hint: To prove (i) ⇒ (ii) show that if one has a relation as in (ii), then one of the x’s can be eliminated, without changing the linear Span.
14 Prove that a linearly independent set M cannot contain the zero element. Prove that a subset of a linearly independent set is again linearly independent.
15 Prove that the union of a directed family of linearly independent sets is again a linearly independent set. That is, if (Xj)j∈J is a family of k-linearly independent subsets of the k-vector space V , with the property
• For any j, k ∈ J there exists some ` ∈ J such that Xj ⊂ X` ⊃ Xk, S then j∈J Xj is again a k-linearly independent subset of V .
16 Suppose V is a vector space, and x1, x2,... is a sequence (finite or infinite) of different non-zero vectors in V . Prove that the following are equivalent:
(i) The set M = {x1, x2,... } is linearly independent. (ii) The sequence of susbspaces Wk = Span({x1, . . . , xk}) is strictly increasing, in the sense that we have strict inclusions
W1 ( W2 ( .... 1. VECTOR SPACES 5
Hint: The implication (i) ⇒ (ii) is clear from the definition.
Conversely, if M were not linearly independent, there exist scalars λ1, . . . , λn ∈ k such that λ1x1 + ··· + λnxn = 0, and at least one of the λ’s non-zero. If we take k = max{j : 1 ≤ j ≤ n and λj 6= 0}, then we get xk ∈ Span{x1, . . . , xk−1, which proves that Wk = Wk−1.
Definition. Let V be a k-vector space. A subset B ⊂ V is called a k-linear basis for V , if:
• Spank(B) = V ; • B is k-linearly independent.
17 Prove the following: Theorem 1.1.1. Let V be a k-vector space, and let P ⊂ M ⊂ V be subsets, with P linearly independent, and Spank(M) = V . Then there exists a linear basis B such that P ⊂ B ⊂ M.
Sketch of proof: Consider the set B = {B ⊂ V : B linearly independent, and P ⊂ B ⊂ M}, equipped with the inclusion as the order relation. Use Zorn’s Lemma to prove that B has a maximal element, and then prove that such a maximal element must span the whole space. (One key step is the checking of the hypothesis of Zorn’s Lemma. Use Exercise 13??.)
18 Suppose B is a linear basis for the vector space V , and P and M are subsets of V , such that one has strict inclusions P ( B ( M. Prove that P and M are no longer bases (although P is linearly independent and Span(M) = V ).
19 Let W be a linear subspace of the vector space V , and let A be a linear basis for W . Prove that there exists a linear basis B for V , with B ⊃ A. Hint: Use Theorem 1.1.1.
Definition. A vector space is said to be finite dimensional, if it has a finite basis.
20 Prove the following Lemma 1.1.1. (Exchange Lemma) Suppose A and B are linear bases for the vector space V . Prove that for every a ∈ A, there exists some b ∈ B, such that the set (A r {a}) ∪ {b} is again a linear basis.
Hint: Write a = β1b1 + ··· + βnbn, for some b1, . . . , bn ∈ B, and some β1, . . . , βn ∈ k r {0}. At least one of the b’s does not belong to Span (A r {a}). Choose this as the exchange for a. We have the inclusion (A r {a}) ∪ {b} ⊂ A ∪ {b}, with the first set linearly independent, but the second one not. Using the fact that Span (A ∪ {b}) = V , Theorem 1.1.1 will force (A r {a}) ∪ {b} to be the basis.
21 Prove that any two linear bases, in a finite dimensional vector space, have the same number of elements.
Hint: Fix a finite basis A = {a1, . . . , an}, and let B be an arbitrary basis. Use the Exchange Lemma 1.1.1 to construct inductively a sequence of elements b1, . . . , bn ∈ B, such that
• For every k = 1, . . . , n, the set {b1, . . . , bk−1, ak, . . . , an} is a linear basis. 6 1. VECTOR SPACES AND LINEAR MAPS
Note that all the b’s must be different. We end up with a linear basis B0 = {b1, . . . , bn}, with |B0| = n. Use Exercise ?? to conclude that B0 = B.
22* Prove the following generalization of the above result. Theorem 1.1.2. Any two linear bases in a vector space have the same car- dinality. (In cardinal arithmetic two sets have the same cardinality if there is a bijection between them.)
Sketch of proof: Suppose A and B are two bases for V . If either A or B is finite, we apply the previous exercise. So we can assume that both A and B are infinite. For every a ∈ A we write it (uniquely) as a = β1b1 + ··· + βnbn, for some b1, . . . , bn ∈ B, and some β1, . . . , βn ∈ k r {0}, and we set PB (a) = {b1, . . . , bn}, so that PB (a) is the smallest subset of P ⊂ B, with Span(P ) 3 a. If we denote by F in(B) the set of all finite subsets of B, we have now a map Π : A 3 a 7−→ PB (a) ∈ F in(B). This map is not injective. However, for every P ∈ F in(B), we have Π−1(P ) ⊂ Span(P ), which means that Π−1(P ) is a linearly independent set in the finite dimensional vector space Span(P ). By Theorem 1.1.1, combined with the previous exercise, this forces each of the sets Π−1(P ), P ∈ F in(B), to be finite. S −1 Then A is a disjoint union of preimages A = P ∈F in(B) Π (P ), each of the sets in this union being finite. Since F in(B) is infinite, we get [ Card(A) = Card Π−1(P ) ≤ Card F in(B) = Card(B). P ∈F in(B) By symmetry, we also have Card(B) ≤ Card(A), and we are done.
Definition. Given a k-vector space V , the above Theorem states that the cardinality of a linear basis for V is independent of the choice of the basis. This “number” is denoted by dimkV , and is called the dimension of V . In the finite dimensional case, the dimension is a non-negative integer. If V is the trivial (zero) space, we define its dimension to be zero.
23 Let n ≥ 1 be an integer. Define, for each j ∈ {1, . . . , n}, the element ej = n n (δij)i=1 ∈ k . (Here δij stands for the Kronecker symbol, defined to be 1, if i = j, and 0 if i 6= j.) Prove that the set {e1,..., en} is a linear basis for the vector space kn, therefore we have dim kn = n.
24 Generalize the above result, by proving that M dim k = Card(I). i∈I
25 Suppose W is a linear subspace of the vector space V . Prove that dim W ≤ dim V . (This means that if A is a linear basis for W , and B is a linear basis for V , then Card(A) ≤ Card(B), which in turn means that there exists an injective map f : A → B.) Hint: Find another basis B0 for V , with B0 ⊃ A. Then use Theorem 1.1.2.
26 Suppose V is a vector space. Prove that the following are equivalent: (i) V is finite dimensional. (ii) Whenever W is a linear subspace of V , with dim W = dim V , it follows that W = V . 1. VECTOR SPACES 7
(iii) Every infinite increasing sequence W1 ⊂ W2 ⊂ W3 ⊂ · · · ⊂ V is an infinite of linear subspaces is stationary, in the sense that there exists some k ≥ 1, such that Wn = Wk, for all n ≥ k.
Definition. Suppose W = (Wj)j∈J is a family of non-zero linear subspaces of a vector space V . We say that W is linearly independent, if for every k ∈ J one has [ Wk ∩ Span Wj = {0}. j∈Jr{k}
27 Let V be a vector space, and X be a subset of V r {0}. The following are equivalent: (i) The set X is linearly independent (as a subset of V ). (ii) The family (kx)x∈X is a linearly independent family of linear subspaces of V .
28 Let W = (Wi)i∈I be a linearly independent family of linear subspaces of V . Suppose X = (Xi)i∈I is another family of linear subspaces, such that Xi ⊂ Wi, for all i ∈ I. Define J = j ∈ I : Xj 6= {0} . Prove that XJ = (Xj)j∈J is also a linearly independent family.
29 Suppose V is a vector space, and W1,W2,... is a sequence (finite or infinite) of non-zero linear subspaces of V . Prove that the following are equivalent:
(i) The sequence (W1,W2,... ) is linearly independent family. (ii) For every k we have
Wk ∩ (W1 + ··· + Wk) = {0}.
Hint: Argue along the same lines of the hint given to exercise ??.
30 Let V be a vector space, and let W1 and W2 be two non-zero linear subspaces. Prove that (W1,W2) is a linearly independent pair of linear subspaces, if and only if W1 ∩ W2 = {0}.
31 Let W be a linear subspace of the vector space V . Prove that there exists a linear subspace X of V , such that W ∩ X = {0}, and W + X = V . Hint: Start with some linear basis A for W . Find a linear basis B for V , with B ⊃ A. Take X = Span(B r A).
32 Suppose W = (Wj)j∈J is a family of linear subspaces of a vector space V . Prove that the following are equivalent (i) W is linearly independent. (ii) If n ≥ 1 is an integer, if j1, . . . , jn ∈ J are different indices, and if w1 ∈
Wj1 , . . . , wn ∈ Wjn are elements such that w1 + ··· + wn = 0, it follows that w1 = ··· = wn = 0. (iii) There exists a choice, for each j ∈ J, of a linear basis Bj for Wj, such that (α) the sets (Bj)j∈J are mutually disjoint, i.e. for any j, k ∈ I with j 6= k, we have Bj ∩ Bk = ∅; S (β) the set j∈J Bj is linearly independent. 8 1. VECTOR SPACES AND LINEAR MAPS
(iii’) For any choice, for each j ∈ J, of a linear basis Bj for Wj, we have the properties (α) and (β) above.
33 Let (Wj)j∈J be a linearly independent family of linear subspaces of V . If we S choose, for each j ∈ J, a linear basis Bj for Wj, then j∈J Bj is linear basis for S the vector space Span j∈J Wj .
34 Let W1,...,Wn be a finite linearly independent family of linear subspaces of V . Prove that
dim (W1 + ··· + Wn) = dim W1 + ··· + dim Wn.
35 Let V be a finite dimensional vector space. Suppose W1,...,Wn are linear subspaces of V , with
• W1 + ··· + Wn = V ; • dim W1 + ··· + dim Wn. n Prove that (Wk)k=1 is a linearly independent family. Hint: For each k ∈ {1, . . . , n}, let Bk be a linear basis for Wk. Put B = B1 ∪ · · · ∪ Bn. On the one hand, we have
(1) |B| ≤ |B1| + ··· + |Bn| = dim W1 + ··· + dim Wn = dim V. On the other hand, we clearly have Span(B) = V , so we must have |B| ≥ dim V . This means that we must have equality in (1), so in particular the sets B1,...,Bn are mutually disjoint, and their union is a linear basis for V . Then the result follows from exercise ?? L 36 Let (Vi)i∈I be a family of vector spaces, and let εj : Vj → i∈I Vi, j ∈ I, be the standard inclusions. L (i) εj(Vj) j∈I is a linearly independent family of linear subspaces of i∈I Vi. (ii) Suppose for each j ∈ I, we choose a linear basis Bj for Vj. Then εj(Bj) is a linear basis for Vj.
Using the fact that εj are injective, conclude that dim εj(Vj) = dim Vj, for all j ∈ I As an application of the above, prove (use exercise ??) that the dimension of a finite direct sum is given as n n M X dim Vi = dim Vi. i=1 i=1
Definition. Suppose X is a k-linear subspace of the k-vector space V . In particular X is a subgroup of the abelian group (V, +) We can then define the quotient group V/X, which is again an abelian group. Formally, the quotient is defined as the set of equivalence classes modulo X: v ≡ w (mod X), if and only if v − w ∈ X.
The addition is defined as [v]X + [w]X = [v + w]X . (Here [v]X stands for the equivalence class of v.) We can also define the scalar multiplication by λ · [v]X = [λv]X . With these operations V/X becomes a vector space, called the quotient vector space. For a subset A ⊂ V , we denote by [A]X the set {[a]X : a ∈ A} ⊂ V/X. 1. VECTOR SPACES 9
37 Verify the statements made in the above definition.
Definitions. Suppose X is a k-linear subspace of the k-vector space V . • For a set M ⊂ V we define its k-linear span relative to X to be the linear subspace
Spank(M; X) = Spank(M ∪ X) = X + Spank(M). • A set P ⊂ V is said to be k-linearly independent relative to X, if the map P 3 p 7−→ [p]X ∈ V/X is injective, and the set [P ]X is linearly independent in the quotient vector space V/X. • A set B ⊂ V is called a k-linear X-basis for V , - Span(B; X) = V ; - B is linearly independent relative to X.
38 Let X be a linear subspace of the vector space V . A. If P ⊂ V is linearly independent relative to X, then P is linearly inde- pendent. B. For a set P ⊂ V , the following are equivalent: (i) P is linearly independent relative to X. (ii) If n ≥ 1 is an integer, if p1, . . . , pn ∈ P are different elements, and if λ1, . . . , λn ∈ k satisfy
λ1p1 + ··· + λnpn ∈ X,
then λ1 = ··· = λn = 0. (iii) There exists a linear basis B for X, such that P ∪ B is linearly independent. (iii’) If B is any linear basis for X, then P ∪ B is linearly independent. (iv) P is linearly independent, and X ∩ Span(P ) = {0}.
39 Let X and Y be linear subspaces of the vector space V , such that X ⊂ Y . A. Prove that for every set M ⊂ V , we have the inclusion Span(M; X) ⊂ Span(M; Y ). B. Prove that if a set P is linearly independent relative to Y , then P is also linearly independent relative to X.
40 Let X be a linear subspace of the vector space V . For a set B ⊂ V prove that the following are equivalent: (i) B is a linear X-basis for V . (ii) The map B 3 b 7−→ [b]X ∈ V/X is injective, and [B]X is a linear basis for V/X. In this situation, prove the equality dim V/X = Card(B).
41 Let X be a linear subspace of the vector space V , and let A be a linear basis for X. Suppose B ⊂ V has the property A ∩ B = ∅. Prove that the following are equivalent: (i) B is a linear X-basis for V . 10 1. VECTOR SPACES AND LINEAR MAPS
(ii) A ∪ B is a linear basis for V . Use this fact, in conjunction with Exercise ??, to prove the equality (2) dim X + dim V/X = dim V.
42 Let W1 and W2 be linear subspaces of the vector space V . Let B ⊂ W1 be an arbitrary subset. Prove that the following are equivalent:
(i) B is a linear (W1 ∩ W2)-basis for W1. (i) B is a linear W2-basis for W1 + W2. Use this fact to prove the following “Paralellogram Law”
(3) dim W1 + dim W2 = dim (W1 + W2) + dim (W1 ∩ W2).
Hint: Assume B satisfies condition (i). Prove first that Span(B; W2) = W1 + W2. It suffices to prove that W1 ⊂ Span(B; W2), which is pretty easy. Second, prove that B is linearly independent relative to W2. Argue that, if v = β1b1 + ··· + βnbn ∈ W2, then in fact we have v ∈ W1 ∩ W2, which using (i) forces β1 = ··· = βn = 0. Use Exercise ?? to conclude that B is a linear W2-basis for W1 + W2. If B satisfies condition (ii), then it is clear that B is also linearly independent relative to W1 ∩ W2. To prove the equality Span(B; W1 ∩ W2) = W1, it suffices to prove only inclusion “⊃.” Start with some element w1 ∈ W1. Using (ii) there exist b ∈ Span(B) and w2 ∈ W2, such that w1 = b + w2. This forces w2 = w1 − b to belong to W1 ∩ W2, so w1 ⊂ Span B ∪ (W1 ∩ W2) , thus proving the desired inclusion. To prove the equality (3), we use the above equivalence, combined with exercise ?? to get dim W1/(W1 ∩ W2) = Card(B) = dim (W1 + W2)/W2 .
By adding dim(W1 ∩ W2) + dim W2 to both sides, and using exercise ??, the result follows.
2. Linear maps Definition. Suppose V and W are k-vector spaces. A map T : V → W is said to be k-linear, if: • T is a group homomorphism, i.e. T (x + y) = T (x) + T (y), for all x, y ∈ V ; • T is compatible with the scalar multiplication, i.e. T (λx) = λT (x), for all λ ∈ k, x ∈ V .
43 Suppose V is a k-vector space. For a scalar µ ∈ k, we define the multiplication map Mµ : V 3 v 7−→ µv ∈ V.
Prove that Mµ is k-linear. If we take µ = 1, then Mµ = IdV , the identity map on V .
44 Let V and W be vector spaces, and let T : V → W be a linear map. For any subset M ⊂ V , prove the equality Span T (M) = T Span(M).
45 Prove that the composition of two linear maps is again linear. Prove that the inverse of a bijective linear map is again linear. A bijective linear map is called a linear isomorphism. 2. LINEAR MAPS 11
46 Let X be a linear subspace of the vector space V , and let T : V → W be a linear map. Prove that the restriction T X : X → W is again a linear map.
47 Let X be a k-linear subspace of a k-vector space V . Prove that the quotient map (defined on page 8) V 3 v 7−→ [v]X ∈ V/X is k-linear. Also prove that the inclusion ι : X,→ V is linear.
48 Let T : V1 → V2 be a linear map, let X1 be a linear subspace of V1, and let X2 be a linear subspace of V2. Prove the Lemma 1.2.1. (Factorization Lemma) The following are equivalent:
(i) There exists a linear map S : V1/X1 → V2/X2 such that the diagram quotient map V1 −−−−−−−−→ V1/X1 T y yS quotient map V2 −−−−−−−−→ V2/X2
is commutative. (This means that, if we denote by qk : Vk → Vk/Xk, k = 1, 2 the quotient maps, then we have the equality q2 ◦ T = S ◦ q1.) (ii) T (X1) ⊂ X2. Moreover, in this case the map S, described implicitly in (i), is unique.
49 Let (Vi)i∈I be a family of vector spaces. For each j ∈ I, denote by εj : L Q Vj → i∈I Vi the standard inclusion, and by πj : i∈I Vi → Vj the coordinate projection. Prove that εj and πj are linear maps.
50 Prove the following:
Theorem 1.2.1. (Universal Property of the direct sum) Let (Vi)i∈I be a family L of vector spaces. For each j ∈ I, denote by εj : Vj → i∈I Vi the standard inclusion. Let W be a vector space, and let (Tj : Vj → W )j∈I be a collection of L linear maps. Then there exists a unique linear map T : i∈I Vi → W , such that
T ◦ εj = Tj, for all j ∈ J. L Hint: We know that any element v = (vi)i∈I ∈ i∈I Vi is represented as X X v = (εj ◦ πj )(v) = εj (vj ). j∈bvc j∈bvc P P Define T (v) = j∈bvc(Tj ◦ πj )(v) = j∈bvc Tj (vj ).
Corollary 1.2.1. Let V be a k-vector space and let W = (Wi)j∈J be a family L of linear subspaces of V . Then there exists a unique linear map ΓW : i∈I Wi → V , such that, for every i ∈ I, we have
(ΓW ◦ εi)(w) = w, for all w ∈ Wi. L For any w = (wi)i∈I ∈ i∈I Wi, we have X ΓW (w) = wi. i∈bwc 12 1. VECTOR SPACES AND LINEAR MAPS
Comment. A particular case of the above result is when all the Wj’s have dimension one (or zero). In this case the family W is represented by an element Q b = (bj)j∈J ∈ j∈J V , by Wj = kbj. Following the above construction, we have V L a linear map denoted Γb : j∈J k → V , which is defined as follows. For every L V P λ = (λj)j∈J ∈ j∈J k, we have Γb (λ) = j∈bλc λjbj ∈ V .
51 Let V be a k-vector space and let W = (Wi)j∈J be a family of linear subspaces L of V . Let ΓW : j∈J Wj → V be the linear map defined in exercise ??. A. Prove that ΓW is injective, if and only if W is a linearly independent family. S B. Prove that ΓW is surjective, if and only if V = Span j∈J Wj . Conclude that ΓW is an isomorphism, if and only if both condition below are true: (i) W is a linearly independent family; S (ii) V = Span j∈J Wj .
Definition. A family W = (Wj)j∈J of linear subspaces of V , satisfying the conditions (i) and (ii) above, is called a direct sum decomposition of V . Comment. A particular case of the above result can be derived, along the same lines as in the comment following exercise ??. Specifically, if V is a vector Q space, and if we have a system b = (bj)j∈J ∈ j∈J V , then V L A. Γb : j∈J k → V is surjective, if and only if V = Span {bj : j ∈ J} . V L B. Γb : j∈J k → V is injective, if and only if all the bj’s are different and the set {bj : j ∈ J} is linearly independent. V L In particular, Γb : j∈J k → V is a linear isomorphism, if and only if all the bj’s are different and the set {bj : j ∈ J} is a linear basis for V .
52 Let (Vj)j∈J be a direct sum decomposition for a vector space V . Let W be a vector space, and assume that, for each j ∈ J, we are given a linear map Tj : Vj ∈ W . Prove that there exists a unique linear map T : V → W , such that
T = Tj, for all j ∈ J. Vj
53 Let V and W be vector spaces, and let T : V → W be a linear map. Since T is a group homomorphism, we know that Ker T ⊂ V and Ran T ⊂ W are subgroups. Prove that they are in fact linear subspaces.
Comment. We know from group theory that the triviality of the kernel is equivalent to the injectivity of the homomorphism. As a particular case, we get the following: • A linear map is injective, if and only if its kernel is the zero subspace.
54 Generalize the above fact as follows. Let V and W be vector spaces, let T : V → W be a linear map, and let Z be a linear subspace of V . Prove that the following are equivalent:
(i) The restriction T Z : Z → W is injective. (ii) Z ∩ Ker T = {0}. 2. LINEAR MAPS 13
55 Let X be a linear subspace of the vector space V , and let q : V → V/X denote the quotient map. A. Prove that Ker q = X. Use this to show that, given a linear subspace W of
V , the restriction q : W → V/X is injective, if and only if W ∩X = {0}. W B. Given a linear subspace W of V , the restriction q W : W → V/X is surjective, if and only if W + X = V .
56* Prove the following technical result. Lemma 1.2.2. (Descent Lemma) Let V be a vector space, and let T : V → V be a linear map. For every Define the subspaces W0 = {0} and
Wk = Ker(T ◦ · · · ◦ T ), | {z } k factors for all k ≥ 1.
(i) One has the inclusions W0 ⊂ W1 ⊂ W2 ⊂ ... . −1 (ii) For every k ≥ 1, one has Wk = T (Wk−1) (iii) For every k ≥ 1, there exists a unique linear map Sk : Wk+1/Wk → Wk/Wk−1 such that the diagram quotient map Wk+1 −−−−−−−−→ Wk+1/Wk T y ySk
Wk −−−−−−−−→ Wk/Wk−1 quotient map is commutative. (iv) All linear maps Sk : Wk+1/Wk → Wk/Wk−1 are injective. (v) Suppose there exists some `, such that W` = W`+1. Then Wk = W`, for all k > `.
Sketch of proof: Parts (i) and (ii) are obvious. From (ii) we get T (Wk) ⊂ Wk−1, and then part (iii) follows from the Factorization Lemma ??.
To prove part (iv), start with some x ∈ Ker Sk. This means that x = [w]Wk , for some w ∈ Wk+1, and T (w) ∈ Wk−1. This forces w ∈ Wk, so x = 0 in the quotient space Wk+1/Wk. This means that Ker Sk = {0}, and we are done. To prove part (v), observe first that the given condition forces W`+1/W` = {0}. Use then induction, based on (iv).
57 Let V and Z be vector spaces, let W be a linear subspace of V , and let T : W → Z be a linear map. Prove that there exists a linear map S : V → Z, such that S W = T . Hint: It suffices to prove this in the particular case when Z = W and T = Id. By exercise ??,we can choose a linear subspace X of V , with X ∩ W = {0} and X + W = V . If we take q : V → V/X the quotient map, then its restriction q W : W → V/X is a linear isomorphism. −1 Put S = (q W ) ◦ q.
58 Let V be a non-zero k-vector space, and let v ∈ V r {0}. Prove the existence of a k-linear map φ : V → k with φ(v) = 1. 14 1. VECTOR SPACES AND LINEAR MAPS
Hint: Take W = Span(v) = kv, and define the map ψ : k 3 λ 7−→ λv ∈ W . Prove that
ψ : k → W is a linear isomorphism. Take φ0 : W → k to be the inverse of ψ, and apply the previous exercise.
59 Let V and W be vector spaces, and let T : V → W be a linear map. For a subset M ⊂ V , prove the equality T −1 Span T (M) = Span(M; Ker T ).
60 Let V and W be vector spaces, and let T : V → W be a linear map. For a subset M ⊂ V , prove that the following are equivalent: (i) M is linearly independent relative to Ker T .
(ii) The restriction T M : M → W is injective and the subset T (M) ⊂ W is linearly independent.
61 Let V and W be vector spaces, and let T : V → W be a linear map. Prove that the following are equivalent: (i) T is injective.
(ii) There exists a linear basis B for V , such that the restriction T B : B → W is injective, and T (B) is linearly independent.
(ii’) If B is any linear basis for V , then the restriction T B : B → W is injective, and T (B) is linearly independent.
62 Let V and W be vector spaces, and let T : V → W be a linear map. Prove that the following are equivalent: (i) T is surjective. (ii) There exists a linear basis B for V , such that the restriction Span T (B) = W . independent. (ii’) If B is any linear basis for V , Span T (B) = W .
63 Let T : V → W be a linear isomorphism from the vector space V into the vector space W . Prove that a set B ⊂ V is linear basis for V , if and only if T (B) is a linear basis for W . Conclude that dim V = dim W .
64 Let V and W be vector spaces, and let T : V → W be a linear map. Prove that the following are equivalent: (i) T is a linear isomorphism.
(ii) There exists a linear basis B for V , such that the restriction T B : B → W is injective, and T (B) is a linear basis for W .
(ii’) If B is any linear basis for V , then the restriction T B : B → W is injective, and T (B) is a linear basis for W .
65 Let V and W be vector spaces, and let T : V → W be a linear map. Prove the following. 2. LINEAR MAPS 15
Theorem 1.2.2. (Isomorphism Theorem) There exists a unique bijective linear map Tb : V/Ker T → Ran T such that the following diagram is commutative quotient map V −−−−−−−−→ V/Ker T T y yTb W ←−−−−− Ran T inclusion (this means that T = ι ◦ Tˆ ◦ q, where ι is the inclusion, and q is the quotient map). Use exercise ?? to conclude that (4) dim(Ker T ) + dim(Ran T ) = dim V.
Hint: To get the existence, and uniqueness of Tˆ, use the Factorization Lemma ??, with the subspaces Ker T ⊂ V and {0} ⊂ W .
66 Let V and W be finite dimensional vector spaces, with dim V = dim W , and let T : V → W be a linear map. A. Prove the “Fredholm alternative” dim(Ker T ) = dim (W/Ran T ). B. Prove that the following are equivalent: (i) T is a linear isomorphism. (ii) T is surjective. (iii) T is injective.
Hint: Part A follows immediately from (4) and (2). To prove the equivalence in part B, use the fact that for a linear subspace Y ⊂ W , the condition dim(W/Y ) = 0 is equivalent to the fact that Y = W .
67* Give an infinite dimensional counterexample to the “Fredholm alternative.” L∞ Specifically, consider the infinite dimensional vetor space V = n=1 k and con- struct a non-invertible surjective linear map T : V → V .
Notation. Suppose V and W are k-vector spaces. We denote by Link(V,W ) the set of all k-linear maps V → W .
68 Let V and W be k-vector spaces. For T,S ∈ Lin(V,W ) we define the map T + S : V → W by (T + S)(v) = T (v) + S(v), for all v ∈ V. Prove that T + S is again a linear map. Prove that Lin(V,W ) is an abelian group, when equipped with the addition operation. The neutral element is the null map 0(v) = 0.
69 Let V and W be k-vector spaces. For T ∈ Lin(V,W ) and λ ∈ k, we define the map λT : V → W by (λT )(v) = λT (v), for all v ∈ V. Prove that λT is a linear map. Prove that Lin(V,W ), when equipped with this multiplication, and the addition defined above, is k-vector space. 16 1. VECTOR SPACES AND LINEAR MAPS
70 The above construction is a special case of a more general one. Start with a vector space W and a set X. Consider the set Map(X,W ) of all functions X → W . For f, g ∈ Map(X,W ) we define f + g ∈ Map(X,W ) by (f + g)(x) = f(x) + g(x), for all x ∈ X, and for f ∈ Map(X,W ) and λ ∈ k, we define λf ∈ Map(X,W ) by (λf)(x) = λf(x), for all x ∈ X. The Map(X,W ) becomes a k-vector space. Prove that if V is a vector space, then Lin(V,W ) is a linear subspace in Map(V,W ).
Q Comment. The set Map(X,W ) is precisely the product x∈X W . The above vector space structure on Map(X,W ) is precisely the direct product vector space structure.
71 Suppose V , W , X and Y are vector spaces, and T : X → V and S : W → Y are linear maps. Prove that the map Lin(V,W ) 3 R 7−→ S ◦ R ◦ T ∈ Lin(X,Y ) is linear.
72 Let V be a k-vector space. Prove that the map
Link(k,V ) 3 T 7−→ T (1) ∈ V is a k-linear isomorphism.
73 Let V and W be vector spaces. Suppose Z ( V is a proper linear subspace. Assume we have elements v ∈ V r Z and w ∈ W . Prove that there exists a linear map T ∈ Lin(V,W ), such that T Z = 0, and T (v) = w. Hint: By assumption, the element vb = [v]Z is non-zero in the quotient space V/Z. Use exercise ?? to find a linear map φ : V/Z → k, with φ(vb) = 1. Take σ : K → W to be unique linear map with σ(1) = w. Now we have a composition σ ◦ φ ∈ Lin(V/Z, W ), with (σ ◦ φ)(vb) = w. Finally, compose this map with the quotien map V → V/Z.
74 Suppose V is a vector space, and X is a finite set. Prove that dim Map(X,V ) = Card(X) · dim V.
L Hint: Use exercise ??, plus the identification Map(X,V ) = x∈X V .
75 Let V and W be vector spaces. Fix a subset X ⊂ V , and consider the map
ΣX : Lin(V,W ) 3 T 7−→ T X ∈ Map(X,W ).
Prove that ΣX is a linear map.
76* Use same notations as in the previous exercise. Assume both V and W are non-zero.
A. Prove that ΣX is injective, if and only if V = Span(X). B. Prove that ΣX is surjective, if and only if X is linearly independent. 2. LINEAR MAPS 17
Conclude that ΣX is an isomorphism, if and only if X is a linear basis for V . Hint: A. If Span(X) V , choose an element v ∈ V Span(X), and an element w ∈ W {0}. ( r r Use exercise ?? to produce a linear map T : V → W , such that T (v) = w and T X = 0. Such an element is non-zero, but belongs to Ker ΣX , so ΣX is not injective. Conversely, any T ∈ Ker ΣX will have the property that X ⊂ Ker T . In particular, if Span(X) = V , this will force Ker T = V , hence T = 0. B. Assume X is not linearly independent. This means that there exist different elements x1, . . . , xn ∈ X and scalars α1, . . . , αn ∈ k r {0}, such that α1x1 + ··· + αnxn = 0. Choose a an element w ∈ W r {0} and define the map f : X → W by w if x = x f(x) = 1 0 if x 6= x1
Prove that f does not belong to Ran ΣX , so ΣX is not surjective. Conversely, assume X is linearly Q independent, and let f : X → W , thought as an element b = f(x) x∈X ∈ x∈X W . Define the W L linear map Γb : x∈X k → W , as in exercise ??. Likewise, if we put Z = Span(X), we can use Q Z L Z the element a = (x)x∈X ∈ x∈X Z, to define a linear map Γa : x∈X k → Z. This time, Γa W Z −1 is a linear isomorphism. If we consider the composition T = Γb ◦ (Γa ) ∈ Lin(Z,W ), we have managed to produce a linear map with T (x) = f(x), for all x ∈ X. We now extend T to a linear map S : V → W , which will satisfy ΣX (S) = f.
77 Suppose V and W are vector spaces. A. Prove that the following are equivalent: (i) dim V ≤ dim W ; (ii) there exists an injective linear map T : V → W ; (iii) there exists a surjective linear map S : W → V . B. Prove that the following are equivalent: (i) dim V = dim W ; (ii) there exists a linear isomorphism T : V → W .
Hint: Fix A a linear basis for V , and B a linear basis for W , so that ΣA : Lin(V,W ) → Map(A, W ) and ΣB : Lin(W, V ) → Map(B,V ) are linear isomorphisms. To prove A (i) ⇒ (ii), we use the definition of cardinal inequality Card(A) ≤ Card(B) ⇔ ∃ f : A → B injective .
For f as above, if we take T ∈ Lin(V,W ) with the property that ΣA(T ) = f, it follows that T is injective. To prove the implication (ii) ⇒ (iii), start with some injective linear map T : V → W , and consider S : Ran T → V to be the inverse of the isomorphism T : V → Ran T . Use exercise ?? to extend S to a linear map R : W → V . Since S is surjective, so is R. The implication (iii) ⇒ (i) follows from the (2). B. The implication (i) ⇒ (ii) has already been discussed. To prove the converse, use the equivalence Card(A) ≤ Card(B) ⇔ ∃ f : A → B bijective .
For f as above, if we take T ∈ Lin(V,W ) with the property that ΣA(T ) = f, it follows that T is an isomorphism.
78 Assume V and W are vector spaces, with V finite dimensional. Prove that (5) dim Lin(V,W ) = dim V · dim W. As a particular case, we have
(6) dim Link(V, k) = dim V
Definition. For a k-vector space V , the vector space Link(V, k) is called the dual of V . 18 1. VECTOR SPACES AND LINEAR MAPS
79 Let V be a k-vector space. Suppose B is a linear basis for V . For every b ∈ B, define the map fb : B → k by 1 if x = b f = b 0 if x 6= b ∗ Use exercise ?? to find, for each b ∈ B, a unique element b ∈ Link(V, k), with ∗ ∗ ∗ ΣB(b ) = fb. The linear map b is uniquely characterized by b (b) = 1, and b∗(x) = 0, for all x ∈ B r {b}. ∗ ∗ (i) Prove that set B = {b : b ∈ B} is linearly independent in Link(V, k). (ii) Prove that the map B 3 b 7−→ b∗ ∈ B∗ is injective, so we have Card(B) = Card(B∗).
80 Let V be a k-vector space, and let B be a linear basis for V . Using the notations from the preceding exercise, prove that the following are equivalent ∗ (i) B is a linear basis for Link(V, k). (ii) V is finite dimensional. Q Hint: Use the isomorphism ΣB : Link(V, k) → b∈B k, we have ∗ Span ΣB (B ) = Span({fb : b ∈ B}). Q But when we identify Map(B, k) with b∈B k, we get M Span({fb : b ∈ B}) = k. b∈B ∗ Q L Therefore the condition Span(B ) = Link(V, k) is equivalent to the condition b∈B k = b∈B k, which is equivalent to the condition that B is finite.
Definition. Suppose V is a finite dimensional k-vector space, and B is a linear ∗ basis for V . Then (see the above result) the linear basis B for Link(V, k) is called the dual basis to B. In fact the notion of the dual basis includes also the bijection B 3 b 7−→ b∗ ∈ B∗ as part of the data.
3. The minimal polynomial Notations. From now on, the composition of linear maps will be written without the ◦ symbol. In particular, if V is a k-vector space, if n ≥ 2 is and n integer, and if T ∈ Link(V,V ), we write T instead of T ◦ T ◦ · · · ◦ T (n factors). Instead of Link(V,V ) we shall simply write Link(V ).
81 Suppose V is a k-vector space. The composition operation in Link(V,V ) is obviously associative. Prove that, for every T ∈ Link(V ), the maps
Link(V ) 3 S 7−→ ST ∈ Link(V )
Link(V ) 3 S 7−→ TS ∈ Link(V ) are linear. (In particular, the composition product is distributive with respect to the addition.)
Comment. The above result states that Link(V ) is a unital k-algebra. The term “unital” refers to the existence of a multiplicative identity element. In our case, this is the identity IdV , which from now on will be denoted by I (or IV when we need to specify the space). 3. THE MINIMAL POLYNOMIAL 19
82 If dim V ≥ 2, prove that the algebra Lin(V ) is non-commutative, in the sense that there exist elements T,S ∈ Lin(V ) with TS 6= ST . (Give an example when TS = 0, but ST 6= 0.
83 Let V be a vector space, and let S, T : V → V be linear maps. Suppose S and T commute, i.e. ST = TS. Prove m n m (i) For any integers m, n ≥ 1, the linear maps S and T commute: S Tn = T nSm. (ii) The Newton Binomial Formula holds: n n X n n−k k (S + T ) = k S T . k=0
Hints: For (i) it Suffices to analyze the case n = 1. Use induction on m For (ii) use (i) and induction. The following is a well-known result from algebra Theorem 1.3.1. Suppose L is a unital k-algebra. Then for any element X ∈ L, there exists a unique unital k-algebra homomorphism ΦX : k[t] → L, such that ΦX (t) = X. Here k[t] stands for the algebra of polynomials in a (formal) variable t, with coefficients in k. The unital algebra homomorphism condition means that
• ΦX is linear; • ΦX is multiplicative, in the sense that φx(PQ) = ΦX (P )ΦX (Q), for all p, q ∈ k[t]; • Φx(1) = I.
To be more precise, the homomorphism ΦX is constructed as follows. Start with n some polynomial P (t) = α0 + α1t + ··· + αnt . Then n ΦX (P ) = α0I + α1X + ··· + αnX . Definition. Suppose V is a k-vector space, and X : V → V is a linear map. If we apply the above Theorem for the algebra Link(V ), and the element X, the corresponding map ΦX : k[t] → Link(V ) is called the polynomial functional calculus of X. For a polynimial P ∈ k[t], the element Φx(P ) will simply be denoted by P (X). Remark 1.3.1. If P and Q are polynomials, then P (X)Q(X) = Q(X)P (X), simply because both are equal to (PQ)(X). This means that, although Link(V ) is not commutative, the sub-algebra
{P (X): P ∈ k[t]} ⊂ Link(V ) is commutative.
84 Let V be a vector space, and let S, T : V → V be linear maps. Assume S and T commute. Prove that, for any two polynomials P,Q ∈ k[t], the linear maps P (S) and Q(T ) commute.
85 Prove that the algebra k[t] is infinite dimensional, as a k-vector space. More explicitly, the set {1} ∪ {tn : n ≥ 1} is a linear basis for k[t]. 20 1. VECTOR SPACES AND LINEAR MAPS
86 Let V be a finite dimensional k-vector space. For every X ∈ Link(V ), prove that there exists a non-zero polynomial P ∈ k[t], such that P (X) = 0.
Hint: Consider the polynomial functional calculus ΦX : k[t] → Link(V ). On the one hand, we 2 notice that dim Link(V ) = (dim V ) < ∞ (by exercise ??). This will force dim(Ran ΦX ) < ∞. On the other hand, we know that
dim(Ker ΦX ) + dim(Ran ΦX ) = dim(k[t]).
But since k[t] is infinite dimensional, this forces Ker φX to be infinite dimensional. In particular Ker ΦX 6= {0}.
87 Suppose V is a finite dimensional k-vector space, and X ∈ Link(V ). Prove that there exists a unique polynomial M ∈ k[t] with the following properties: (i) M(X) = 0. (ii) M is monic, in the sense that the leading coefficient is 1. (iii) If P ∈ k[t] is any polynomial with P (X) = 0, then P is divisible by M Hint: One (conceptual) method of proving this result is to quote the property of k[t] of being a principal ideal domain. Then Ker ΦX , being an ideal, must be presented as Ker ΦX = M · k[t], for a unique monic polynomial M. Here is the sketch of a “direct” proof (which in fact traces the steps used in proving that k[t] is a pid, in this particular case). Choose a (non-zero) polynomial N ∈ k[t], of minimal degree, such that N(X) = 0. Replace N with M = λN, so that M is monic (we will still have M(X) = 0). Suppose now P is a polynomial with P (x) = 0. Divide P by M with remainder, so that we have P = MQ+R, for some polynomials Q and R, with deg R < deg M. Using the fact that M(X) = 0, we get R(X) = M(X)Q(X) + R(X) = P (X) = 0. By minimality, this forces R = 0, thus proving property (iii). The uniqueness is a direct application of (iii).
Definition. Suppose V is a finite dimensional k-vector space, and X : V → V is a linear map. The polynomial M, described in the preceding exercise, is called the minimal polynomial of X, and will be denoted by MX .
88* Give an infinite dimensional counter example to exercise ??. Hint: Take V = k[t] and define the linear map X : k[t] 3 P (t) 7−→ tP (t) ∈ k[t]. Prove that there is no polynomial M such that M(X) = 0.
89 Let V be a finite dimensional vector space, and fix some X ∈ Lin(V ). Prove that the following are equivalent: • X is a scalar multiple of the identity, i.e. there exists some α ∈ k such that X = αI. • The minimal polynomial MX (t) is of degree one.
Definition. Suppose V is a vector space. A linear map X : V → V is said to be nilpotent, if there exists some integer n ≥ 1, such that Xn = 0.
90 Let V be a finite dimensional vector space, and let X ∈ Lin(V ) be nilpotent. p Prove that the minimal polynomial is of the form MX (t) = t , for some p ≤ dim V . In particular, we have Xdim V = 0. n k Hint: Fix n such that X = 0 Consider the linear subspaces W0 = {0} and Wp = Ker(X ), p k ≥ 1. It is clear that MX (t) = t , where p is the smallest index for which Wp = V . Prove that 3. THE MINIMAL POLYNOMIAL 21 for every k ∈ {1, . . . , p} we have the strict inclusion Wk−1 ( Wk. (Use the Descent Lemma ??) Use then (recursively) the dimension formula for quotient spaces (2), to get p X dim V = dim(Wk/Wk−1). k=1 Since each term in the sum is ≥ 1, this will force dim V ≥ p.
91 Let V be a vector space, and let M,N : V → V be nilpotent linear maps. If M and N commute, then prove that M + N is again nilpotent. If X : V → V is an arbitrary linear map, which commutes with N, then prove that XN is nilpotent.
92* Suppose V is finite dimensional, X ∈ Lin(V ), and W is an invariant linear subspace, i.e. with the property T (W ) ⊂ W . Let T : V/X → V/X be the unique linear map which makes the digram quotient map V −−−−−−−−→ V/W Xy yT quotient map V −−−−−−−−→ V/W commutative (see exercise ??). Let us also consider the linear map S = X W : W → W . Prove that
(i) The minimal polynomial MX (t) of X divides the product of minimal poly- nomials MS(t)MT (t). (ii) The polynomial MX (t) is divisble by both minimal polynomials MS(t) and MT (t), hence MX (t) is divisible by the lowest common multiple (lcm) of MS(t) and MT (t).
Conclude that, if MS(t) and MT (t) are relatively prime, then
MX (t) = MS(t)MT (t).
Hint: Observe first that, for any polynomial P ∈ k[t], one has a commutative diagram quotient map V −−−−−−−−−→ V/W P (X)y yP (T ) quotient map V −−−−−−−−−→ V/W and the equality P (S) = P (X) W . In particular, since MX (X) = 0, we immediately get MX (S) = 0 and MX (T ) = 0, thus proving (ii). To prove (i), we only need to show that MS (X)MT (X) = 0. Start with an arbitrary element v ∈ V . Since MT (T ) = 0, using the above diagram we get the equality q [MT (X)](v) = 0, in V/W, which means that the element w = [MT (X)](v) belongs to W . But then, using MS (S) = 0, we get 0 = [MS (S)](w) = [MS (X)](w) = [MS (X)] [MT (X)](v) = [MS (X)MT (X)](v).
93 The following example shows that, in general the minimal polynomial MX (t) can differ from both the product MS(t)MT (t) and lcm(MS(t),MT (t)). Consider the space V = k4 and the linear map 4 4 X : k 3 (α1, α2, α3, α4) 7−→ (0, α1, 0, α2) ∈ k . 22 1. VECTOR SPACES AND LINEAR MAPS
Take the linear subspace W = {(0, 0, λ, µ): λ, µ ∈ k} ⊂ V . Check the equalities: 2 4 MS(t) = t, MT (t) = t , and MX (t) = t . Hint: Prove that S = 0, T 2 = 0, but T 6= 0, and X4 = 0, but X3 6= 0.
94* Suppose V is finite dimensional, X ∈ Lin(V ), and W1,...,Wn are invariant subspaces, such that W1 + ··· + Wn = V . For each k = 1, . . . , n we take Mk(t) to be the minimal polynomial of X Wk. Prove that the minimal polynomial of X is given as MX (t) = lcm(M1(t),...,Mk(t)).
Hint: Use induction on k. In fact, it suffices to prove the case k = 2. As in the previ- ous exercise, prove that MX (t) is divisible by lcm(M1(t),...,Mk(t)). Conversely, if we de- note lcm(M1(t),...,Mk(t)) simply by P (t), we know that for every k, we have a factorization P (t) = Qk(t)Mk(t). In particular, for every w ∈ Wk, we get [P (X)](w) = [Qk(X)] [Mk(X)](w) = [Qk(X)] [Mk(X )](w) = 0. Wk
Now we have P (X) = 0, for all k, which gives P (X) = 0, thus proving that P (t) is divisible Wk by MX (t).
95* Let V be a finite dimensional k-vector space, and let X : V → V be a linear map. Prove the following Theorem 1.3.2. (Abstract Spectral Decomposition) Assume the minimal poly- nomial is decomposed as
(7) MX (t) = M1(t) · M2(t) ····· Mn(t), such that for any j, k ∈ {1, . . . , n} with j 6= k, we have gcd(Mj,Mk) = 1. Define the polynomials MX (t) Pj(t) = , j = 1, . . . , n. Mj(t) Since gcd(P1,P2,...,Pn) = 1, we know there exist polynomials Q1,...,Qn ∈ k[t], such that
(8) Q1(t)P1(t) + Q2(t)P2(t) + ··· + Qn(t)Pn(t) = 1. Define the linear maps
Ej = Qj(X)Pj(X) ∈ Link(V ), j = 1, 2, . . . , n. 2 (i) The linear maps Ej, j = 1, . . . , n are idempotents, i.e. Ej = Ej. (ii) For any j, k ∈ {1, . . . , n} with j 6= k, we have EjEk = 0. (iii) E1 + E2 + ··· + En = I. (iv) For each j ∈ {1, . . . , n} we have the equality Ran Ej = Ker[Mj(X)]. (v) For every j, the minimal polynomial of the restriction X is precisely Ran Ej Mj(t).
Sketch of proof: By construction, we have (iii), so in fact conditions (i) and (ii) are equivalent. To prove (ii) start with j 6= k. Using the fact that Pj Pk is divisible by all the Mi’s it follows that Pj Pk is divisible by MX , so we get Pj (X)Pk(X) = 0. Using polynomial functional calculus, we get Ej Ek = Qj (X)Pj (X)Qk(X)Pk(X) = Qj (X)Qk(X)Pj (X)Pk(X) = 0. To prove (iv) we first observe that since Mj (t)Pj (t) = MX (t), we have (again by functional calculus)
Mj (X)Ej = Mj (X)Qj (X)Pj (X) = Qj (X)Mj (X)Pj (X) = Qj (X)MX (X) = 0, 3. THE MINIMAL POLYNOMIAL 23
which proves the inclusion Ran Ej ⊂ Ker[Mj (X)]. Conversely, since Pk(t) is divisible by Mj (t) for all k 6= j, we see that Pk(X) = 0, which forces Ek = 0, for all k 6= j. Ker[Mj (X)] Ker[Mj (X)]
Using (iii) this gives Ej = Id , thus proving the other inclusion. Ker[Mj (X)] Ker[Mj (X)] To prove (v), denote for simplicity the subspaces Ran Ej by Wj . By (iii) we know that W1 + ··· + Wn = V . By exercise ?? we know that 0 0 (9) MX = lcm(M1 ,...,Mn), 0 where M is the minimal polynomial of X . By (iii) we know that Mj (X ) = 0, hence Mj (t) j Wj Wj 0 0 0 is divisble by Mj (t). In particular we will also have gcd(Mj ,Mk ) = 1, for all j 6= k, so by (9) we will get 0 0 0 MX (t) = M1 (t)M2 (t) ····· Mn(t). 0 This forces Mj = Mj , for all j
Definition. Use the hypothesis of the above Theorem. The idempotents E1,...,En are called the spectral idempotents of X, associated with the decom- position (7). The subspaces Ran Ej = Ker[Mj(X)], j = 1, . . . , n are called the spectral subspaces of X, associated with the decomposition (7). Notice that
(i) V = Ker[M1(X)] + Ker[M2(X)] + ··· + Ker[Mn(X)]. n (ii) Ker[Mj(X)] j=1 is a linearly independent family of linear subspaces. n In other words, Ker[Mj(X)] j=1 is a direct sum decomposition of V .
96 Prove the properties (i) and (ii) above.
97 The above definition to carries a slight ambiguity, in the sense that the spectral idempotents Ej are defined using a particular choice of the polynomials Qj for which (8) holds. Prove that if we choose another sequence of polynomials, Q1,..., Qn, such that
Q1(t)P1(t) + Q2(t)P2(t) + ··· + Qn(t)Pn(t) = 1, then we have the equalities
Qj(X)Pj(X) = Ej, for all j = 1, . . . , n. Therefore the spectral idempotents are un-ambiguously defined.
Hint: Define Ej = Qj (X)Pj (X), j = 1, . . . , n. Use the Theorem to show Ran Ej = Ran Ej , and then use the Theorem again to show that Ej = Ej , for all j. Another approach is to look at the spectral decomposition
V = Ker[M1(X)] + Ker[M2(X)] + ··· + Ker[Mn(X)], which depends only on the factorization (7). For every element v ∈ V , there are unique elements wj ∈ Ker[Mj (X)], j = 1, . . . , n, such that v = w1 + ··· + wn. Then Ej (v) = wj .
98 Prove the following Theorem 1.3.3. (Uniqueness of Spectral Decomposition) Let V be a finite dimensional vector space, let X : V → V be a linear map, and let W1,...,Wn be a collection of invariant linear subspaces. For each j = 1, . . . , n, define Mj(t) to be the minimal polynomial of X ∈ Lin(Wj). Assume that: Wj
(i) W1 + ··· + Wn = V . (ii) For every j, k ∈ {1, . . . , n} with j 6= k, we have gcd(Mj,Mk) = 1. 24 1. VECTOR SPACES AND LINEAR MAPS
Then the minimal polynomial decomposes as MX (t) = M1(t) ····· Mn(t), and moreover, the Wj’s are precisely the spectral subspaces associated with this decom- position, i.e. Wj = Ker[Mj(X)], for all j = 1, . . . , n.
Hint: The decomposition of MX (t) has already been discussed (see exercise ??). It is clear n that Wj ⊂ Ker[Mj (X)]. On the one hand, (Wj )j=1 is a linearly independent family of linear subspaces, so
dim V = dim(W1 + ··· + Wn) = dim W1 + ··· + dim Wn.
On the other hand, we have dim Wj ≤ dim(Ker[Mj (X)]), and we also have
dim V = dim(Ker[M1(X)]) + ··· + dim(Ker[Mn(X)]).
By finiteness, this forces dim Wj = dim(Ker[Mj (X)]), so we must have Wj = Ker[Mj (X)].
99* Work again under the hypothesis of the Abstract Spectral Decomposition Theorem. The following is an alternative construction of the spectral idempotents. Suppose R1,...,Rn are polynomials such that, for each j ∈ {1, . . . , n}, we have:
(i) Rj is divisible by Mk, for all k ∈ {1, . . . , n}, k 6= j; (ii) 1 − Rj is divisible by Mj.
(See the hint on why such polynomials exist.) Prove that Rj(X) = Ej, for all j.
Hint: The simplest example of the existence of R1,...,Rn is produced by taking Rj (t) = Qj (t)Pj (t), j = 1, . . . , n. Define Fj = Rj (X). On the one hand, j, k ∈ {1, . . . , m} with j 6= k, the polynomial Rj Rk is divisible by both Pj and Pk, hence it is divisible by gcd(Pj ,Pk) = MX . In particular we get
Fj Fk = Rj (X)Rk(X) = (Rj Rk)(X) = 0.
On the other hand, for a fixed j, it is clear that Rk is divisible by Mj , for all k 6= j. Therefore, if we consider the polynomial R(t) = R1(t) + ··· + Rn(t), we see that R − Rj is divisible by Mj . But so is 1 − Rj . This means that 1 − R is divisble by Mj . Since this is true for all j, it follows that 1 − R is divisible by gcd(M1,...,Mn) = MX . This forces R(X) = 1, which means that 2 F1 + ··· + Fn = I, so we must have Fj = Fj . n Now we have a linearly independent family of linear subspaces Ran Fj j=1, with
(10) Ran Fn + ··· + Ran Fn = V.
Notice that Mj Rj is divisible by MX , so Mj (X)Fj = Mj (X)Rj (X) = 0. This means that Ran Fj ⊂ Ker[Mj (X)]. Because of (10), this forces Ran Fj = Ker[Mj (X)] = Ran Ej , which in turn forces Fj = Ej .
100 Suppose V and W are k-vector spaces, and S : V → W is a linear isomor- phism. Prove that the map −1 Link(V ) 3 X 7−→ SXS ∈ Link(W ) is an isomorphism of k-algebras. Use this fact tow prove that, for a linear map X ∈ Link(V ) one has P (SXS−1) = SP (X)S−1, for all P ∈ k[t].
101 Suppose V , W and S are as above, and V (hence also W ) finite dimensional. Prove that for any X ∈ Lin(V ) we have the equality of minimal polynomials MX (t) = MSXS−1 (t). 4. SPECTRUM 25
4. Spectrum Convention. Throughout this section we restrict ourselves to a particular choice of the field k = C, the field of complex numbers. All vector spaces are assumed to be over C. This choice of the field is justified by the fact that the arithmetic in C[t] is particularly nice, as shown by the following: Theorem 1.4.1. (Fundamental Theorem of Algebra) For every non-zero poly- nomial P ∈ C[t] there exists a collection of different numbers λ1, . . . , λn ∈ C, and integers m1, . . . , mk ≥ 1, such that
m1 mn P (t) = (t − λ1) ····· (t − λn) .
The numbers λ1, . . . , λn are called the roots of P . They are precisely the soultions (in C) of the equation P (λ) = 0. The numbers m1, . . . , mn are called the multiplicities. More precisely, mk is called the multiplicity of the root λk. Remark that
deg P = m1 + ··· + mn. Definitions. Let V be a vector space, and X : V → V be a linear map. The spectrum of X is defined to be the set
Spec(X) = {λ ∈ C : MX (λ) = 0}. Suppose λ ∈ Spec(X). The multiplicity of λ, as a root of the minimal polynomial, is called the spectral multiplicity of λ, and is denoted by mX (λ). The linear subspace
N(X, λ) = Ker[(X − λI)mX (λ), is called the spectral subspace of λ. The number
dX (λ) = dim N(X, λ) is called the spectral dimension of λ. According to the Fundamental Theorem of Algebra, if we list the spectrum as Spec(X) = {λ1, . . . , λn}, with the λ’s different, we have a decomposition
mX (λ1) mX (λn) (11) MX (t) = (t − λ1) ····· (t − λn) . When we apply the Abstract Spectral Decomposition Theorem for the polynomials mX (λj ) Mj(t) = (t − λj) , j = 1, . . . , n, we get the following Theorem 1.4.2. (Spectral Decomposition Theorem) Suppose V is a finite dimensional vector space, and X : V → V is a linear map. List the spectrum as Spec(X) = {λ1, . . . , λn}, with all λ’s different.
(i) The spectral subspaces N(X, λj), j = 1, . . . , n, form a linearly independent family of linear subspaces. (ii) N(X, λ1) + ··· + N(X, λn) = V . If for every j ∈ {1, . . . , n} we define Xj = X ∈ Lin N(X, λj) , then N(X,λj )
mX (λj ) (iii) the minimal polynomial of Xj is (t−λj) ; in particular the spectrum is a singleton Spec(Xj) = {λj}. 26 1. VECTOR SPACES AND LINEAR MAPS
Definitions. Suppose V is a finite dimensional vector space, and X : V → V is a linear map. The family N(X, λ) λ∈Spec(X), which by Theorem ?? is a direct sum decomposition of V , is called the spectral decomposition of X. The spectral idempotents of X are the ones described in the Abstract Spectral Decomposition Theorem (exercise ??), corresponding to (11). They are be denoted by EX (λ), λ ∈ Spec(X). These idempotents have the following properties (which uniquely characterize them):
(i) EX (λ)EX (µ) = 0, for all λ, µ ∈ Spec(X) with λ 6= µ; (ii) Ran EX (λ) = N(X, λ), for all λ ∈ Spec(X). P As a consequence, we also get λ∈Spec(X) EX (λ) = I.
102 Let V be a finite dimensional vector space, and let X : V → V be a linear map. Prove that, for every λ ∈ Spec(X), we have the inequality mX (λ) ≤ dX (λ).