Abstract Vector Spaces, Linear Transformations, and Their Coordinate Representations

Transcription

1 Abstract Vector Spaces, Linear Transformations, and Their Coordinate Representations Contents 1 Vector Spaces Definitions Basics Linear Combinations, Spans, Bases, and Dimensions Sums and Products of Vector Spaces and Subspaces Basic Vector Space Theory Linear Transformations Definitions Linear Transformations, or Vector Space Homomorphisms Basic Properties of Linear Transformations Monomorphisms and Isomorphisms Rank-Nullity Theorem Matrix Representations of Linear Transformations Definitions Matrix Representations of Linear Transformations Change of Coordinates Definitions Change of Coordinate Maps and Matrices Vector Spaces 1.1 Definitions Basics A vector space (linear space) V over a field F is a set V on which the operations addition, + : V V V, and left F -action or scalar multiplication, : F V V, satisfy: for all x, y, z V and a, b, 1 F VS1 x + y = y + x VS2 (x + y) + z = x + (y + z) VS3 0 V such that x + 0 = x VS4 x V, y V such that x + y = 0 VS5 1x = x VS6 (ab)x = a(bx) VS7 a(x + y) = ax + ay VS8 (a + b)x = ax + ay (V, +) is an abelian group If V is a vector space over F, then a subset W V is called a subspace of V if W is a vector space over the same field F and with addition and scalar multiplication + W W and F W. 1

2 1.1.2 Linear Combinations, Spans, Bases, and Dimensions If S V, v 1,..., v n S and a 1,..., a n F, then a linear combination of v 1,..., v n is the finite sum a 1 v a n v n (1.1) which is a vector in V. The a i F are called the coefficients of the linear combination. If a 1 = = a n = 0, then the linear combination is said to be trivial. In particular, considering the special case of 0 in V, the zero vector, we note that 0 may always be represented as a linear combination of any vectors u 1,..., u n V, 0u u n }{{} 0 F = 0 }{{} 0 V This representation is called the trivial representation of 0 by u 1,..., u n. If, on the other hand, there are vectors u 1,..., u n V and scalars a 1,..., a n F such that a 1 u a n u n = 0 where at least one a i 0 F, then that linear combination is called a nontrivial representation of 0. Using linear combinations we can generate subspaces, as follows. If S is a subset of V, then the span of S is given by The span of is by definition span(s) := {v V v is a linear combination of vectors in S} (1.2) span( ) := {0} (1.3) In this case, S V is said to generate or span V, and to be a generating or spanning set. If span(s) = V, then S said to be a generating set or a spanning set for V. A nonempty subset S of a vector space V is said to be linearly independent if, taking any finite number of distinct vectors u 1,..., u n S, we have for all a 1,..., a n F that a 1 u 1 + a 2 u a n u n = 0 = a 1 = = a n = 0 That is S is linearly independent if the only representation of 0 V by vectors in S is the trivial one. In this case, the vectors u 1,..., u n themselves are also said to be linearly independent. Otherwise, if there is at least one nontrivial representation of 0 by vectors in S, then S is said to be linearly dependent. Given S V, a nonzero vector v S is said to be an essentially unique linear combination of the vectors in S if, up to order of terms, there is one and only one way to express v as a linear combination of u 1,..., u n S. That is, if there are a 1,..., a n, b 1,..., b m F \{0} and distinct u 1,..., u n S and distinct v 1,..., v m S distinct, then, re-indexing the b i s if necessary, } { } v = a 1 u a n u n ai = b i = m = n and i = 1,..., n = b 1 v b m v m u i = v i A subset β V is called a (Hamel) basis if it is linearly independent and span(β) = V. We also say that the vectors of β form a basis for V. Equivalently, as explained in Theorem 1.13 below, β is a basis if every nonzero vector v V is an essentially unique linear combination of vectors in β. In the context of inner product spaces of inifinite dimension, there is a difference between a vector space basis, the Hamel basis of V, and an orthonormal basis for V, the Hilbert basis for V, because though the two always exist, they are not always equal unless dim(v ) <. 2

3 The dimension of a vector space V is the cardinality of any basis for V, and is denoted dim(v ). V finite-dimensional if it is the zero vector space {0} or if it has a basis of finite cardinality. Otherwise, if it s basis has infinite cardinality, it is called infinite-dimensional. In the former case, dim(v ) = β = n < for some n N, and V is said to be n-dimensional, while in the latter case, dim(v ) = β = κ, where κ is a cardinal number, and V is said to be κ-dimensional. If V is finite-dimensional, say of dimension n, then an ordered basis for V a finite sequence or n-tuple (v 1,..., v n ) of linearly independent vectors v 1,..., v n V such that {v 1,..., v n } is a basis for V. If V is infinite-dimensional but with a countable basis, then an ordered basis is a sequence (v n ) n N such that the set {v n n N} is a basis for V. The term standard ordered basis is applied only to the ordered bases for F n, (F ω ) 0, (F B ) 0 and F [x] and its subsets F n [x]. For F n, the standard ordered basis is given by (e 1, e 2,..., e n ) where e 1 = (1, 0,..., 0), e 2 = (0, 1, 0,..., 0),..., e n = (0,..., 0, 1). Since F is any field, 1 represents the multiplicative identity and 0 the additive identity. For F = R, the standard ordered basis is just as above with the usual 0 and 1. If F = C, then 1 = (1, 0) and 0 = (0, 0), so the ith standard basis vector for C n looks like this: For (F ω ) 0 the standard ordered basis is given by e i = ( ith term (0, 0),..., (1, 0),..., (0, 0) ) (e n ) n N, where e n (k) = δ nk and generally for (F B ) 0 the standard ordered basis is given by: { 1, if x = b (e b ) b B, where e b (x) = 0, if x b When the term standard ordered basis is applied to the vector space F n [x], the subspace of F [x] of polynomials of degree less than or equal to n, it refers the basis For F [x], the standard ordered basis is given by (1, x, x 2,..., x n ) (1, x, x 2,... ) Sums and Products of Vector Spaces and Subspaces If S 1,..., S k P (V ) are subsets or subspaces of a vector space V, then their (internal) sum is given by k S S k = S i := {v v k v i S i for 1 i k} (1.4) i=1 More generally, if {S i i I} P (V ) is any collection of subsets or subspaces of V, then their sum is defined to be { S i = v v n si } S i and n N (1.5) i I i I If S is a subspace of V and v V, then the sum {v} + S = {v + s s S} is called a coset of S and is usually denoted v + S (1.6) 3

4 We are of course interested in the (internal) sum of subspaces of a vector space. In this case, we may have the special condition V = k i=1 S i and S j i j S i = {0}, j = 1,..., k (1.7) When this happens, V is said to be the (internal) direct sum of S 1,..., S k, and this is symbolically denoted k V = S 1 S 2 S k = S i = {s s k s i S i } (1.8) i=1 In the infinite-dimensional case we proceed similarly. If F = {S i i I} is a family of subsets of V such that V = S i and S j S i = {0}, j I (1.9) i I i j then V is a direct sum of the S i F, denoted V = F = { S i = s s n s j } S i, and s ji S i if s ji 0 i I i I (1.10) We can also define the (external) sum of distinct vector spaces which do not lie inside a larger vector space: if V 1,..., V n are vector spaces over the same field F, then their external direct sum is the cartesian product V 1 V n, with addition and scalar multiplication defined componentwise. And we denote the sum, confusingly, by the same notation: V 1 V 2 V n := {(v 1, v 2,..., v n ) v i V i, i = 1,..., n} (1.11) In the infinite-dimensional case, we have two types of external direct sum, one where there is no restriction on the sequences, the other where we only allow sequences with finite support. To distinguish these two cases we use the terms direct product and external direct product, respectively: let F = {V i i I} be a family of vector spaces over the same field F. Then their direct product is given by } V i := {v = (v i ) i I vi V i (1.12) i I = { f : I } V i f(i) V i i I (1.13) The infinite-dimensional external direct sum is then defined: { } V i := v = (v i ) i I vi V i and v has finite support i I = { f : I } V i f(i) Vi and f has finite support i I (1.14) (1.15) When all the V i are equal, there is some special notation: V n (finite-dimensional) (1.16) V I := V (infinite-dimensional) (1.17) i I (V I ) 0 := V (1.18) i I 4

5 Example 1.1 Common examples of vector spaces are the sequence space F n, F ω, F B, (F ω ) 0 and (F B ) 0 in a field F over that field, i.e. the various types of cartesian products of F equipped with addition and scalar multiplication operations defined componentwise (ω = N and B is any set, and where (F ω ) 0 and (F B ) 0 denote all functions f : ω F, respectively f : B F, with finite support). Some typical ones are Q n R n C n Q ω R ω C ω Q B R B C B (Q ω ) 0 (R ω ) 0 (C ω ) 0 (Q B ) 0 (R B ) 0 (C B ) 0 each over Q or R or C, respectively. 5

6 1.2 Basic Vector Space Theory Theorem 1.2 following hold: If V is a vector space over a field F, then for all x, y, z V and a, b F the (1) x + y = z + y = x = z (Cancellation Law) (2) 0 V is unique. (3) x + y = x + z = 0 = y = z (Uniqueness of the Additive Inverse) (4) 0x = 0 (5) ( a)x = (ax) = a( x) (6) The finite external direct sum V 1 V n is a vector space over F if each V i is a vector space over F and if we define + and componentwise by (v 1,..., v n ) + (u 1,..., u n ) := (v 1 + u 1,..., v n + u n ) c(v 1,..., v n ) := (cv 1,..., cv n ) More generally, if F = {V i i I} is a family of vector spaces over the same field F, then the direct product i I V i and external direct sum i I V i = (V I ) 0 are vector spaces over F, with + and given by v + w = ( f : I ) ( V i + g : I ) V i i I i I ( cv = c f : I ) V i i I := := ( cf : I ) V i i I ( f + g : I ) V i i I Proof: (1) x = x+0 = x+(y +( y)) = (x+y)+( y) = (z +y)+( y) = z +(y +( y)) = z +0 = z. (2) 0 1, 0 2 = 0 1 = = = 0 2. (3)(a) x + y 1 = x + y 2 = 0 = y 1 = y 2 by the Cancellation Law. (b) y 1 = y = y 1 + (x + y 2 ) = (y 1 + x) + y 2 = 0 + y 2 = y = y 2. (4) 0x + 0x = (0 + 0)x = 0x = 0x + 0 V = 0x = 0 V by the Cancellation Law. (5) ax + ( (ax)) = 0, and by 3 (ax) is unique. But note ax + ( a)x = (a + ( a))x = 0x = 0, hence ( a)x = (ax), and also a( x) = a(( 1)x) = (a( 1))x = ( a)x, which we already know equals (ax). (6) All 8 vector space properties are verified componentwise, since that is how addition and scalar multiplication are defined. But the components are vectors in a vector space, so all 8 properties will be satisfied. Theorem 1.3 Let V be a vector space over F. A subset S of V is a subspace iff for all a, b F and x, y S we have one of the following equivalent conditions: (1) ax + by S (2) x + y S and ax S (3) x + ay S Proof: If S satisfies any of the above implications, then it is a subspace because vector space properties VS1-2 and VS5-8 follow from the fact that S V, while VS3 follows because we can choose a = b = 0 and 0 V is unique (Theorem 1.2), and VS4 follows from VS3, in the first case by choosing b = a and x = y, in the second case by choosing y = ( 1)x, and in the third case by choosing a = 1 and y = x. Conversely, if S is a vector space, then (1)-(3) follow by definition. 6

7 Theorem 1.4 Let V be a vector space. If C = {S i i K} is a collection of subspaces of V, then S i, and S i (1.19) S i i K i K i K are subspaces. Proof: (1) If C = {S i i K} is a collection of subspaces, then none of the S i are empty by the previous theorem, so if a, b F and v, w i K S i, then v = s i1 + + s in i K S i and w = s j1 + + s jm i K S i such that s ik S ik for all i k K. Consequently av + bw = a(s i1 + + s in ) + b(s j1 + + s jn ) = as i1 + + as in + bs j1 + + bs jn S i i K because as ik S ik for all i k K because the S ik are subspaces. Hence by the previous theorem i K S i is a subspace of V. The direct sum case is a subcase of this general case. (2) If a, b F and u, v i K S i, then u, v S i for all i K, whence au + bv S i for all i K, or au + bv i K S i, and i K S i is a subspace by the previous theorem. Theorem 1.5 T S. If S and T are subspaces of a vector space V, then S T is a subspace iff S T or Proof: If S T is a subspace, then for all a, b F and x, y S T we have ax + by S T. Suppose neither S T nor S T holds, and take x S\T and y T \S. Then, x + y is neither in S nor in T (for if it were in S, say, then y = (x + y) + ( x) S, a contradiction), contradicting the fact that S T is a subspace. Conversely, suppose, say, S T, and choose any a, b F and x, y S T. If x, y S, then ax + by S S T, while if x, y T, then ax + by T S T. If x S and y T, then x T as well and ax + by T S T. Thus in all cases ax + by S T, and S T is a subspace. Theorem 1.6 If V is a vector space, then for all S V, span(s) is a subspace of V. Moreover, if T is a subspace of V and S T, then span(s) T. Proof: If S =, span( ) = {0}, which is a subspace of V ; if T is a subspace, then T and {0} T. If S, the any x, y S have the form x = a 1 v a n v n and y = b 1 w b m w m for some a 1,... a n, b 1,..., b n F and v 1,..., v n, w 1,..., w n S. Then, if a, b F we have ax + by = a(a 1 v a n v n ) + b(b 1 w b m w m ) = (aa 1 )v (aa n )v n + (bb 1 )w (bb m )w m span(s) so span(s) is a subspace of V. Now, suppose T is a subspace of V and S T. If w span(s), then w = a 1 w a k w n for some w 1,..., w n S and a 1,..., a n F. Since S T, w 1,..., w k T, whence w = c 1 w c k w k T. Corollary 1.7 If V is a vector space, then S V is a subspace of V iff span(s) = S. Proof: If S is a subspace of V, then by the theorem S S = span(s) S. But if s S, then s = 1s span(s), so S span(s), and therefore S = span(s). Conversely, if S = span(s), then S is a subspace of V by the theorem. 7

8 Corollary 1.8 If V is a vector space and S, T V, then the following hold: (1) S T V = span(s) span(t ) (2) S T V and span(s) = V = span(t ) = V (3) span(s T ) = span(s) + span(t ) (4) span(s T ) span(s) span(t ) Proof: (1) and (2) are immediate, so we only need to prove 3 and 4: (3) If v span(s T ), then v 1,..., v m S, u 1,..., u n T and a 1,..., a m, b 1,..., b n F such that v = v 1 a v m a m + b 1 u b n u n Note, however, v 1 a v m a m span(s) and b 1 u b n u n span(t ), so that v span(s) + span(t ). Thus span(s T ) span(s) + span(t ). Conversely, if v = s + t span(s) + span(t ), then s span(s) and t span(t ), so that by 1, since S S T and T S T, we must have span(s) span(s T ) and span(t ) span(s T ). Consequently, s, t span(s T ), and since span(s T ) is a subspace, v = s + t span(s T ). That is span(s) + span(t ) span(s T ). Thus, span(s) + span(t ) = span(s T ). (4) By Theorem 1.6 span(s T ), span(s) and span(t ) are subspaces, and by Theorem 1.4 span(s) span(t ) is a subspace. Now, consider x span(s T ). There exist vectors v 1,..., v n S T and scalars a 1,..., a n F such that x = a 1 v a n v n. But since v 1,..., v n belong to both S and T, x span(s) and x span(t ), so that x span(s) span(t ). It is not in general true, however, that span(s) span(t ) span(s T ). For example, if S = {e 1, e 2 } R 2 and T = {e 1, (1, 1)}, then span(s) span(t ) = R 2 R 2 = R 2, but span(s T ) = span({e 1 }) = R, and R 2 R. Theorem 1.9 If V is a vector space and S T V, then (1) S linearly dependent = T linearly dependent (2) T linearly independent = S linearly independent Proof: (1) If S is linearly dependent, there are scalars a 1,..., a m F, not all zero, and vectors v 1,..., v m S such that a 1 v a m v m = 0. If S T, then v 1,..., v m are also in T, and have the same property, namely a 1 v a m v m = 0. Thus, T is linearly dependent. (2) If T is linearly independent, then if v 1,..., v n S, a 1,..., a n F and a 1 v a n v n = 0, since S T implies v 1,..., v n T, we must have a 1 = a 2 = = a n = 0, and S is linearly independent. Theorem 1.10 If V is a vector space, then S V is linearly dependent iff S = {0} or there exist distinct vectors v, u 1,..., u n S such that v is a linear combination of u 1,..., u n. Proof: If S is linearly dependent, then S consists of at least one element by definition. Suppose S has only one element, v. Then, either v = 0 or v 0. If v 0, consider a F. Then, av = 0 = a = 0, which means S is linearly independent, contrary to assumption. Therefore, v = 0, and so S = {0}. If S contains more than one element and v S, then since S is linearly dependent, there exist distinct u 1,..., u n S and a 1,..., a n F, not all zero such that a 1 u 1 + a n u n = 0 If v = 0, we re done. If v 0, then, since {u 1,..., u n } is linearly dependent, so is {u 1,..., u n, v}, by Theorem 1.9, so that there are b 1,..., b n+1 F, not all zero, such that b 1 u b n u n + b n+1 v = 0 8

9 Choose b n+1 0, for if we let b n+1 = 0 we have gained nothing. Then, ( v = b ) ( 1 u b ) n u n b n+1 b n+1 and v is a linear combination of distinct vectors in S. Conversely, if S = {0}, clearly S is linearly dependent, for we can choose a, b F both nonzero and have a0 + b0 = 0. Otherwise, if there are distinct u 1,..., u n, v S and a 1,..., a n F such that v = a 1 u a n u n, then obviously 0 = 1v + a 1 u a n u n, and since 1 0, S is linearly dependent. Corollary 1.11 If S = {u 1,..., u n } V, then S is linearly dependent iff u 1 = 0 or for some 1 k < n we have u k+1 span(u 1,..., u k ). Theorem 1.12 If V is a vector space and u, v V are distinct, then {u, v} is linearly dependent iff u or v is a multiple of the other. Proof: If {u, v} is linearly dependent, there are scalars a, b F not both zero such that au+bv = 0. If a = 0, then b 0, and so au + bv = bv = 0 = v = 0 = u 0, and so v = 0u. If neither a nor b is equal to zero, then au + bv = 0 = u = a 1 bv. Conversely, suppose, say, u = av. If a = 0, then u = 0 and v can be any vector other than v, since by hypothesis u v, whence au + 0v = 0 for any a F, including nonzero a, so that {u, v} is linearly dependent. Otherwise, if a 0, we have u = av = 1u + ( a)v = 0, and both 1 0 and a 0, so that {u, v} is linearly dependent. Theorem 1.13 (Properties of a Basis) If V is a vector space, then the following are equivalent, and any S V satisfying any of them is a basis: (1) S is linearly dependent and V = span(s). (2) (Essentially Unique Representation) Ever nonzero vector v V is an essentially unique linear combination of vectors in S. (3) No vector in S is a linear combination of other vectors in S. (4) S is a minimal spanning set, that is V = span(s) but no proper subset of S spans V. (5) S is a maximal linearly independent set, that is S is linearly independent but any proper superset is not linearly independent. Proof: (1)= (2): If S is linearly independent and v S is a nonzero vector such that for distinct s i and t i in S v = a 1 s a n s n = b 1 t b m s m then grouping any s ij and t ij that are equal, we have 0 = (a i1 b i1 )s i1 + + (a ik b ik )s ik + a ik+1 s ik a in s in + b ik+1 t ik b im t im implies a ik+1 = = a in = b ik+1 = = b im = 0, so n = m = k, and for j = 1,..., k we have a ij = b ij, whence v is essentially uniquely represented as a linear combination of vectors in S. 9

10 (1)= (2): The contrapositive of Theorem (3)= (1): If 3 holds and a 1 s a n s n = 0, where the s i are distinct and a 1 0, then n > 1 and s 1 = 1 a 1 (a 2 s a n s n ), which violates 3. Thus 3 implies 1. (1) (4): If S is a linearly independent spanning set and T is a proper subset of S that also spans V, then any vectors in S\T would have to be linear combinations of the vectors in T, violating 3 for S (3 is equivalent to 1, as we just showed). Thus S is a minimal spanning set. Conversely, if S is a minimal spanning set, then it must be linearly independent, for otherwise there would be some s S that is a linear combination of other vectors in S, which would mean S\{s} also spans V, contradicting minimality. Thus 4 implies 1. (1) (5): If 1 holds, so S is linearly independent and spans V, but S is not maximal, then v V \S such that S {v} is also linearly independent. But v / span(s) by 3, contradiction. Therefore S is maximal and 1 implies 5. Conversely, if S is a maximal linearly independent set, then S must span V, for otherwise there is a v V \S that isn t a linear combination of vectors in S, implying S {v} is linearly independent proper superset, violating maximality. Therefore span(s) = V, and 5 implies 1. Corollary 1.14 If V is a vector space, then S = {v 1,..., v n } V is a basis for V iff V = span(v 1 ) span(v 2 ) span(v n ) (1.20) The previous theorem did not say that any such basis exists for a given vector space. It merely gave the defining characteristics of a basis, should we ever come across one. The next theorem assures us that all vector spaces have a basis a consequence of Zorns lemma. Theorem 1.15 (Existence of a Basis) If V is a nonzero vector space and I S V with I linearly independent and V = span(s), then B basis for V for which More specifically, (1) Any nonzero vector space has a basis. (2) Any linearly independent set in V is contained in a basis. (3) Any spanning set in V contains a basis. I B S (1.21) Proof: Let A P (V ) be the set of all linearly independent sets L such that I L S. Then A is non-empty because I I S, so I A. Now, if C = {I k k K} A is a chain in A, that is a totally ordered (under set inclusion) subset of A, then the union U = C = is linearly independent and satisfies I U S, that is U A. But by Zorn s lemma every chain has a maximal element B, so that B A maximal which is linearly independent. But of course such a B is a basis for V = span(s), for if any s S is not a linear combination of elements in B, then B {s} is linearly independent and B B {s}, contradicting the maximality of B. Therefore S span(b), and so V = span(s) span(b) V, or V = span(b). This shows that (1) there is a basis B for V, (2) any linearly independent set I has I B for this B, and (3) any spanning set S has B S for this B. i K I i 10

11 Corollary 1.16 (All Subpaces Have Complements) If S is a subspace of a vector space V, then there exists a subspace T of V such that V = S T (1.22) Proof: If S = V, then let T = {0}. Otherwise, if V \S, then since V has a basis B and B is maximal, we must have S = span(b S) and B (V \S). That is, there is a nonzero subspace T = span(b (V \S)), and by Corollary 1.8 V = span(b) = span((b S) (B (V \S))) = span(b S) + span(b (V \S)) = S + T But S T = {0} because of the essentially unique representation of vectors in V as linear combinations of vectors in B. Hence, V = S T, and T is a complement of S. Theorem 1.17 (Replacement Theorem) If V is a vector space such that V = span(s) for some S V with S = n, and if B V is linearly independent with B = m, then (1) m n (2) C S with C = n m such that V = span(b C) Proof: The proof is by induction on m, beginning with m = 0. For this m, B =, so taking C = S, we have B C = S = S, which generates V. Now suppose the theorem holds for some m 0. For m+1, let B = {v 1,..., v m+1 } V be linearly independent. By Corollary 1.8, B = {v 1,..., v m } is linearly independent, and so by the induction hypothesis m n and C = {u 1,..., u n m } S such that {v 1,..., v n } C generates V. This means a 1,..., a m, b 1,..., b n m F such that v m+1 = a 1 v a m v m + b 1 u b n m u n m (1.23) Note that if n = m, C =, so that v m+1 span(b ), contradicting the assumption that B is linearly independent. Therefore, n > m, or n m + 1. Moreover, some b i, say b 1, is nonzero, for otherwise we have v m+1 = a 1 v a m v m, leading again to B being linearly dependent in contradiction to assumption. Solving (1.23) for u 1, we get u 1 = ( b 1 1 a 1)v ( b 1 1 a m)v m + b 1 1 v m+1 + ( b 1 1 b 2)u ( b 1 1 b n m)u n m Let C = {u 2,..., u n m }. Then u 1 span(b C), and since v 1,..., v m, u 2,..., u n m B C, they are also in span(b C), whence B C span(b C) Now, since span(b C ) = V by the induction hypothesis, we have by Corollary 1.8 that span(b C) = V. So we have B linearly independent with B = m + 1 vectors, m + 1 n, span(b C) = V, and C S with C = (n m) 1 = n (m + 1) vectors, so the theorem holds for m + 1. Therefore the theorem is true for all m N by induction. Corollary 1.18 If V is a vector space with a finite spanning set, then every basis for V contains the same number of vectors. Proof: Suppose S is a finite spanning set for V, and let β and γ be two bases for V with γ > β = k, so that some subset T γ contains exactly k + 1 vectors. Since T is linearly independent and β generates V, the replacement theorem implies that k + 1 k, a contradiction. Therefore γ is finite and γ k. Reversing the roles of β and γ shows that k γ, so that γ = β = k. 11

12 Corollary 1.19 If V is a finite-dimensional vector space with dim(v ) = n, then the following hold: (1) Any finite spanning set for V contains at least n vectors, and a spanning set for V that contains exactly n vectors is a basis for V. (2) Any linearly independent subset of V that contains exactly n vectors is a basis for V. (3) Every linearly independent subset of V can be extended to a basis for V. Proof: Let B be a basis for V. (1) Let S V be a finite spanning set for V. By Theorem 1.15, B S that is a basis for V, and by Corollary 1.18 B = n vectors. Therefore S n, and S = n = S = B, so that S is a basis for V. (2) Let I V be linearly independent with I = n. By the replacement theorem T B with T = n n = 0 vectors such that V = span(i T ) = span(i ) = span(i). Since I is also linearly independent, it is a basis for V. (3) If I V is linearly independent with I = m vectors, then the replacement theorem asserts that H B containing exactly n m vectors such that V = span(i H). Now, I H n, so that by 1 I H = n, and so I H is a basis for V extended from I. Theorem 1.20 Any two bases for a vector space V have the same cardinality. Proof: If V is finite-dimensional, then the previous corollary applies, so we need only consider bases that are infinite sets. Let B = {b i i I} ba a basis for V and let C be any other basis for V. Then all c C can be written as finite linear combinations of vectors in B, where all the coefficients are nonzero. That is, if U c = {1,..., n c }, we have n c c = a i b i = a i b i i=1 i U c for unique a 1,..., a nc F. But because C is a basis, we must have I = c C U c, for if c C I, then all c C can be expressed by a proper subset B of B, so that V = span(b ), which is contradiction of the minimality of B as a spanning set. Now, for all c C we have U c < ℵ 0, which implies that B = I = U c ℵ0 C = C c C Reversing the roles of B and C gives C B, so by the Schröder-Bernstein theorem we have B = C. Corollary 1.21 If V is a vector space and S is a subspace of V : (1) dim(s) dim(v ) (2) dim(s) = dim(v ) < = S = V (3) V is infinite-dimensional iff it contains an infinite linearly independent subset. Theorem 1.22 Let V be a vector space. (1) If B is a basis for V and B = B 1 B 2, where B 1 B 2 =, then V = span(b 1 ) span(b 2 ). (2) If V = S T and B 1 is a basis for S and B 2 is a basis for T, then B 1 B 2 = and B = B 1 B 2 is a basis for V. Proof: (1) If B 1 B 2 = and B = B 1 B 2 is a basis for V, then 0 / B 1 B 2. But, if a nonzero vector v span(b 1 ) span(b 2 ), then B 1 B 2, a contradiction. Hence {0} = span(b 1 ) span(b 2 ). Moreover, since B 1 B 2 is a basis for V, and since it is also a basis for span(b 1 ) + span(b 2 ), we 12

13 must have V = span(b 1 ) + span(b 2 ), and hence V = span(b 1 ) span(b 2 ). (2) If V = S T, then S T = {0}, and since 0 / B 1 B 2, we have B 1 B 2 =. Also, since all v V = S T have the form a 1 u a m u m + b 1 v b n v n for u 1,..., u m B 1 and v 1,..., v n B 2, v span(b 1 B 2 ), so B 1 B 2 is a basis for V by Theorem Theorem 1.23 If S and T are subspaces of a vector space V, then dim(s) + dim(t ) = dim(s + T ) + dim(s T ) (1.24) As a consequence, V = S T iff dim(v ) = dim(s) + dim(t ). Proof: If B = {b i i I} is a basis for S T, then we can extend this to a basis A B for S and to another basis B C for T, where A = {a j j J}, C = {c k k K} and A B = and B C =. Then A B C is a basis for S + T : clearly span(a B C) = S + T, so we need only verify that A B C is linearly independent. To that end, suppose not, suppose c 1,..., c n F \{0} and v 1,..., v n A B C such that c 1 v 1 + c 2 v c n v n = 0 Then some v i A C since by our construction A B and B C are linearly independent. Isolating the vectors in A, say v 1,..., v k, on one side of the equality shows that there is a nonzero vector, x = span(a) {}}{ a 1 v a k v k = a k+1 v k a n v n }{{} span(b C) = x span(a) span(b C) = span(a) T S T since span(a) S. Consequently x = 0, and therefore a 1 = = a n = 0 since A and B C are linearly independent sets, a contradiction. Thus A B C is linearly independent and hence a basis for S + T. Moreover, dim(s) + dim(t ) = A B + B C Of course if S T = {0}, then dim(s T ) = 0, and = A + B + B + C = A B C + B = dim(s + T ) + dim(s T ) dim(s) + dim(t ) = dim(s T ) = dim(v ) while if dim(v ) = dim(s) + dim(t ), then dim(s T ) = 0, so S T = {0}, and so V = S T. Corollary 1.24 If S and T are subspaces of a vector space V, then (1) dim(s + T ) dim(s) + dim(t ) (2) dim(s + T ) max{dim(s), dim(t )}, if S and T are finite-dimensional. 13

14 Theorem 1.25 (Direct Sums) If F = {S i i I} is a family of subspaces of a vector space V such that V = i I S i, then the following statements are equivalent: (1) V = i I S i. (2) 0 V has a unique expression as a sum of vectors each from different S i, namely as a sum of zeros: for any distinct j 1,..., j n I, we have v j1 + v j2 + + v jn = 0 and v ji S ji for each i = v j1 = = v jn = 0 (3) Each v V has a unique expression as a sum of distinct v ji S ji \{0}, v = v j1 + v j2 + + v jn (4) If γ i is a basis for S i, then γ = i I γ i is a basis for V. If V is finite-dimensional, then this may be restated in terms of ordered bases γ i and γ. Proof: (1) = (2): Suppose 1 is true, that is V = i I S i, and choose v 1,..., v n V such that v i S ki and v v k = 0. Then for any j, v j = i j v i i j S i However, v j S j, whence v j W j i j S i = {0} so that v j = 0. This is true for all j = 1,..., k, so v 1 = = v k = 0, proving 2. (2) = (3): Suppose 2 is true and let v V = i I S i be given by v = u u n = w w m for some u i S ki and w j S kj. Then, grouping terms from the same subspaces 0 = v v = (u i1 w i1 ) + + (u ik w ik ) + u ik u in w ik+1 w im { u i1 = w i1,, u ik = w ik = and u ik+1 = = u in = w ik+1 = = w im = 0 proving uniqueness. (3) = (4): Suppose each vector v V = n i I S i can be uniquely written as v = v v k, for v i S ji. For each i I let γ i be an ordered basis for S i, so that since V = n ( i I S i, we must k ) have that V = span(γ) = span i=1 γ i, so we only need to show that γ is linearly independent. To that end, let v 11, v 12,..., v 1n1 γ i1 v m1, v m2,..., v mnm γ im for any bases γ ij for S ij F, then let a ij F for i = 1,..., m and j = 1,..., n i, and let w i = ni j=1 a ijv ij. Then, suppose w w m = a ij v ij = 0 i,j. 14

15 since 0 m i=1 S j i and 0 = , by the assumed uniqueness of expression of v V by w i S i we must have w i = 0 for all i, and since all the γ i are linearly independent, we must have a ij = 0 for all i, j. (4) = (1): if 4 is true, then there exist ordered bases γ i the S i such that γ = i I γ i is an ordered basis for V. By Corollary 1.8 and Theorem 1.22, V = span(γ) = span( γ i ) = span(γ i ) = i I i I i I S i Choose any j I and suppose for a nonzero vector v V we have v S j i j S i. Then, v S j = span(γ j ) span( i j γ i ) = i j S i which means v is a nontrivial linear combination of elements in γ i and elements in i j γ i, so that v can be expressed as a linear combination of i I γ i in more than one way, which contradicts uniqueness of representation of vectors in V in terms of a basis for V, Theorem Consequently, any such v must be 0. That is, S j i j S i = {0} and the sum is direct, i.e. V = i I S i, proving 1. 15

16 2 Linear Transformations 2.1 Definitions Linear Transformations, or Vector Space Homomorphisms If V and W are vector spaces over the same field F, a function T : V W is called a linear transformation or vector space homomorphism if it preserves the vector space structure, that is if for all vectors x, y V and all scalars a, b F we have T (x + y) = T (x) + T (y) T (ax) = at (x) or equivalently T (ax + by) = at (x) + bt (y) (2.1) The set of all linear transformations from V to W is denoted L(V, W ) or Hom(V, W ) (2.2) Linear transformation, like group and ring homomorphisms, come in different types: 1. linear operator (vector space endomorphism) on V if V is a vector space over F, a linear operator, or vector space endomorphism, is a linear transformation T L(V, V ). The set of all linear operators, or vector space endomorphisms, is denoted L(V ) or End(V ) (2.3) 2. monomorphism (embedding) a 1-1 or injective linear transformation. 3. epimorphism an onto or surjective linear transformation. 4. isomorphism (invertible linear transformation) from V to W a bijective (and therefore invertible) linear transformation T L(V, W ), i.e. a monomorphism which is also an epimorphism. If such an isomorphism exists, V is said to be isomorphic to W, and this relationship is denoted V = W (2.4) The set of all linear isomorphomorphisms from V to W is denoted by analogy with the general linear group. GL(V, W ) (2.5) 5. automorphism a bijective linear operator. The set of all automorphisms of V is denoted Aut(V ) or GL(V ) (2.6) Some common and important linear transformations are: 1. The identity transformation, I V : V V, given by I V (x) = x for all x V. It s linearity is obvious: I V (ax + by) = ax + by = ai V (x) + bi V (y). 2. The zero transformation, T 0 : V W, given by T 0 (x) = 0 for all x V. Sometimes this is denoted by 0 if the context is clear, like the zero polynomial 0 F [x]. This is also clearly a linear transformation. 3. An idempotent operator T L(V ) satisfies T 2 = T. 16

17 A crucial concept, which will allow us to classify and so distinguish operators, is that of similarity of operators. Two linear operators T, U L(V ) are said to be similar, denoted if there exists an automorphism φ Aut(V ) such that Similarity is an equivalence relation on L(V ): 1. T T, because T = I T I 1 T U (2.7) T = φ U φ 1 (2.8) 2. T U implies T = φ U φ 1 for some automorphism φ L(V ), so that U = φ 1 T φ, or U = ψ T ψ 1, where ψ = φ 1 is an automorphism. Thus U T. 3. S T and T U implies S = φ T φ 1 and T = ψ U ψ 1 where φ and ψ are automorphisms, so S = φ T φ 1 = φ (ψ U ψ 1 ) φ 1 = (φ ψ) U(φ ψ) 1 and since φ ψ is an automorphism of V, we have S U. This partitions L(V ) into equivalence classes. As we will see, each equivalence class corresponds to a single matrix representation, depending only on the choice of basis for V, which maybe different for each operator. The kernel or null space of a linear tranformation T L(V, W ) is the pre-image of 0 W under T : ker(t ) = T 1 (0) (2.9) i.e. ker(t ) = {x V T (x) = 0}. As we will show below, ker(t ) is a subspace of V. The range or image of a linear tranformation T L(V, W ) is the range of T as a set function: it is variously denoted by R(T ), T (V ) or im(t ), R(T ) = im(t ) = T (V ) := {T (x) x V } (2.10) As we show below, R(T ) is a subspace of W. The rank of a linear transformation T L(V, W ) is the dimension of its range: rank(t ) := dim ( R(T ) ) (2.11) The nullity of T L(V, W ) is the dimension of its kernel: null(t ) := dim ( ker(t ) ) (2.12) Given an operator T L(V ), a subspace S of V is said to be T -invariant if v S = T (v) S, that is if T (S) S. The restriction of T to a T -invariant subspace S is denoted T S, and of course T S L(S) is a linear operator on S. In the case that V = R S decomposes as a direct sum of T -invariant subspaces R and S, we say the pair (R, S) reduces T, or is a reducing pair for T, and we say that T is the direct sum of operators T R and T S, denoting this by T = T R T S (2.13) The expression T = τ σ L(V ) means there are subspaces R and S of V for which (R, S) reduces T and τ = T R and σ = T S. If V is a direct sum of subspaces V = R S, then the function π R,S : V V given by where r R and s S, is called the projection onto R along S. π R,S (v) = π R,S (r + s) = r (2.14) 17

18 Example 2.1 The projection on the y-axis along the x-axis is the map π Ry,R x : R 2 R 2 given by π Ry,R x (x) = y. Example 2.2 The projection on the y-axis along the line L = {(s, s) s R} is the map π Ry,L : R 2 R 2 given by π Ry,L(x) = π Ry,L(s, s + y) = y. Any projection π R,S is linear, i.e. π R,S L(V ): π R,S (au + bv) = π R,S (a(r 1 + s 1 ) + b(r 2 + s 2 )) = π R,S ((ar 1 + br 2 ) + (as 1 + bs 2 )) = ar 1 + br 2 = aπ R,S (u) + bπ R,S (v) And clearly R = im(π R,S ) and S = ker(π R,S ) (2.15) Two projections ρ, σ L(V ) are said to be orthogonal, denoted if ρ σ (2.16) ρ σ = σ ρ = 0 T 0 (2.17) This is of course by analogy with orthogonal vectors v w, for which we have v, w = w, v = 0. We use projections to define a resolution of the identity, a sum of the form π π k = I L(V ), where the π i are pairwise orthogonal projections, that is π i π j if i j. This occurs when V = S 1 S 2 S k, π i = π Si, j i Sj, and π i π j for i j. Resolutions of the identity will be important in establishing the spectral resolution of a linear operator. 18

19 2.2 Basic Properties of Linear Transformations Theorem 2.3 If V and W are vector spaces over the same field F and T L(V, W ), then ker(t ) is a subspace of V and im(t ) is a subspace of W. Moreover, if T L(V ), then {0}, ker(t ), im(t ) and V are T -invariant. Proof: (1) If x, y ker(t ) and a, b F, we have T (ax + by) = at (x) + bt (y) = a0 + b0 = 0 so ax + by ker(t ), and the statement follows by Theorem 1.3. If x, y im(t ) and a, b F, then u, v V such that T (u) = x and T (v) = y, whence ax + by = at (u) + bt (v) = T (au + bv) im(t ). (2) Clearly {0} and V are T -invariant, the first because T (0) = 0, the second because the codomain of T is V. If x im(t ), then of course x V, so that T (x) im(t ), while if x ker(t ), then T (x) = 0 ker(t ). Theorem 2.4 If V and W are vector spaces and B = {v i i I} is a basis for V, then for any T L(V, W ) we have im(t ) = span(t (B)) (2.18) Proof: T (v i ) im(t ) for each i I by definition, and because im(t ) is a subspace, we have span(t (B)) im(t ). Moreover, span(t (B)) is a subspace by Theorem 1.6. For the reverse inclusion, choose any w im(t )j. Then v V such that w = T (v), so that by Theorem 1.13!a 1,..., a n F such that v = a 1 v a n v n, where by v i we mean v i v ji. Consequently, w = T (v) = T (a 1 v a n v n ) = a 1 T (v 1 ) + + a n T (v n ) span(t (B)) Theorem 2.5 (A Linear Transformation Is Defined by Its Action on a Basis) Let V and W be vector spaces over the same field F. If B = {v i i I} is a basis for V, then we can define a unique linear transformation T L(V, W ) by arbitrarily specifying the values w i = T (v i ) W for each i I and extending T by linearity, i.e. specifying that for all a 1,..., a n F our T satisfy T (a 1 v a n v n ) = a 1 T (v 1 ) + + a n T (v n ) (2.19) = a 1 w a n w n Proof: By Theorem 1.13: v V, a 1,..., a n F such that v = a 1 v a n v n. Defining T W V by (2.19) we have the following results: (1) T L(V, W ): For any u, v V, there exist unique a 1,..., a n, b 1,..., b m F and u 1,..., u n, v 1,..., v m B such that u = a 1 u a n u n v = b 1 v b m v m If w 1,..., w n, z 1,..., z m W are the vectors for which we have specified T (u) = a 1 T (u 1 ) + + a n T (u n ) = a 1 w a n w n T (v) = b 1 T (v 1 ) + + b m T (v m ) = b 1 z b m z m then, for all a, b F we have T (au + bv) = T ( a(a 1 u a n u n ) + b(b 1 v b m v m ) ) = aa 1 T (u 1 ) + + aa n T (u n ) + bb 1 T (v 1 ) + + bb m T (v m ) = at (a 1 u a n u n ) + bt (b 1 v b m v m ) = at (u) + bt (u) 19

20 (2) T is unique: suppose U L(V, W ) such that U(v i ) = w i = T (v i ) for all i I. Then for any v V we have v = n i=1 a iv i for unique a i F, so that U(v) = n a i U(v i ) = i=1 n a i w i = T (v) i=1 and U = T. Theorem 2.6 If V, W, Z are vector spaces over the same field F and T L(V, W ) and U L(W, Z), then U T L(V, Z). Proof: If v, w V and a, b F, then (U T )(ac + bw) = U(T (ac + bw)) = U(aT (v) + bt (w)) = au(t (v)) + bu(t (w)) = a(u T )(x) + b(u T )(y) so U T L(V, Z). Theorem 2.7 If V and W are vector spaces over F, then L(V, W ) is a vector space over F under ordinary addition and scalar multiplication of functions. Proof: First, note that if T, U L(V, W ) and a, b, c, d F, then v, w V (at + bu)(cv + dw) = at (cv + dw) + bu(cv + dw) = act (v) + adt (w) + bcu(v) + bdu(w) = c ( at (v) + bu(v) ) + d ( at (w) + bu(w) ) = c(at + bu)(v) + d(at + bu)(w) so (at + bu) L(V, W ), and L(V, W ) is closed under addition and scalar multiplication. Moreover, L(V, W ) satisfies VS1-8 because W is a vector space, and addition and scalar multiplication in L(V, W ) are defined pointwise for values which any functions S, T, U L(V, W ) take in W. 20

21 2.3 Monomorphisms and Isomorphisms Theorem 2.8 If V, W, Z are vector spaces over the same field F and T GL(V, W ) and U GL(W, Z) are isomorphisms, then (1) T 1 L(W, V ) (2) (T 1 ) 1 = T, and hence T 1 GL(W, V ). (3) (U T ) 1 = T 1 U 1 GL(Z, V ). Proof: (1) If w, z W and a, b F, since T is bijective!u, v V such that T (u) = w and T (v) = z. Hence T 1 (w) = u and T 1 (z) = v, so that T 1 (aw + bz) = T 1( at (u) + bt (v) ) = T 1( T (au + bv) ) = au + bv = at 1 (w) + bt 1 (z) (2) T 1 T = I V and T T 1 = I W shows that T is the inverse of T 1, that is T = (T 1 ) 1. (3) First, any composition of bijections is a bijection, so U T is bijective. Moreover, and (U T ) (U T ) 1 = I Z = U U 1 = U I W U 1 (U T ) 1 (U T ) = I V = T 1 T = T 1 I W T = U (T T 1 ) U 1 = (U T ) (T 1 U 1 ) = T 1 (U 1 U) T = (U 1 T 1 ) (U T ) so by the cancellation law and the fact that L(V, Z) is a vector space, we must have (U T ) 1 = T 1 U 1 L(Z, V ) and by (1) it s an isomorphism. Alternatively, we can get (3) from (2) via Theorem 2.5: U T : V Z, so (U T ) 1 : Z V and of course T 1 U 1 : Z V Let α, β and γ be bases for V, W and Z, respectively, such that T (α) = β and U(β) = U(T (α)) = (U T )(α) = γ, so that (U T ) 1 (γ) = α which is possible by Theorem But we also have U 1 (γ) = β and T 1 (β) = α so that (T 1 U 1 )(γ) = α By Theorem 2.5 (uniqueness) (U T ) 1 = T 1 U 1 and by (1) it s an isomorphism. Theorem 2.9 If V and W are vector spaces over the same field F and T L(V, W ), then T is injective iff ker(t ) = {0}. Proof: If T is 1-1 and x ker(t ), then T (x) = 0 = T (0), so that x = 0, whence ker(t ) = {0}. Conversely, if ker(t ) = {0} and T (x) = T (y), then, 0 = T (x) T (y) = T (x y) = x y = 0 or x = y, and so T is

22 Theorem 2.10 Let V and W be vector spaces over the same field and S V. Then for any T L(V, W ), (1) T is injective iff it carries linearly independent sets into linearly independent sets. (2) T is injective implies S V is linearly independent iff T (S) W is linearly independent. Proof: (1) If T is 1-1 and S V is linearly independent, then for any v 1,..., v n S we have 0 V = a 1 v a n v n = a 1 = a n = 0, so 0 W = T (0 V ) = T (a 1 v a n v n ) = a 1 T (v 1 ) + + a n T (v n ) whence {T (v 1 ),..., T (v n )} is linearly independent, and since this is true for all v i S, T (S) is linearly independent, so T carries linearly independent sets into linearly independent sets. Conversely, if T carries linearly independent sets into linearly independent sets, let B be a basis for V and suppose T (u) = T (v) for some u, v V. Since u = a 1 u a n u n and v = b 1 v b m v m for unique a i, b i F and u 1,..., u n, v 1,..., v m B, we have 0 = T (u) T (v) = T (u v) = T ( (a i1 b j1 )v i1 + + (a ik b jk )v ik + a ik+1 u ik a in u in +b ik+1 v ik b im v im ) = (a i1 b j1 )T (v i1 ) + + (a ik b jk )T (v ik ) +a ik+1 T (u ik+1 ) + + a in T (u in ) + b ik+1 T (v ik+1 ) + + b im T (v im ) so that a ik+1 = = a in = b ik+1 = = b im = 0, which means m = n = k and therefore a i1 = b j1,..., a ik = b jk. Consequently u = a i1 v i1 + + a ik v ik = v whence T is 1-1. (2) If T is 1-1 and S V is linearly independent, then by (1) T (S) is linearly independent. Conversely, if T is 1-1 and T (S) is linearly independent, then for K = {v 1,..., v k } S V we have that T (0) = 0 = a 1 T (v 1 ) + + a k T (v k ) = T (a 1 v a k v k ) implying simultaneously that a 1 v a k v k = 0 and a 1 = = a k = 0, so K is linearly independent. But this is true for all such K S, so S is linearly independent. Theorem 2.11 If V and W are vector spaces over the same field F and T GL(V, W ), then for any S V we have (1) V = span(s) iff W = span(t (S)). (2) S is linearly independent in V iff T (S) is linearly independent in W. (3) S is a basis for V iff T (S) is a basis for W. Proof: (1) V = span(s) W = im(t ) = T GL(V,W ) {}}{ T (span(s)) = span(t (S)). (2) S linearly independent means for any s 1,..., s n S we have n i=1 a is i = 0 for all a i = 0, which implies ( n ) T a i s i = i=1 n a i T (s i ) = 0 = T (0) i=1 T GL(V,W ) = n a i s i = 0 i=1 S lin. indep. = a 1 = = a n = 0 22

23 so T (S) is linearly independent, since this is true for all s i S. Conversely, if T (S) is linearly independent we have for any T (s 1 ),..., T (s n ) T (S) n ( n ) T GL(V,W ) n 0 = a i T (s i ) = T a i s i = T (0) = a i s i = 0 i=1 so S is linearly independent. (1), (2) i=1 i=1 T (S) lin. indep. = a 1 = = a n = 0 (3) S basis for V T (S) linearly independent in W and W = span(t (S)) = T (S) is a basis for W. We could also say that since T 1 is an isomorphism by Theorem 2.8, we have that since T (S) is a basis for W, S = T 1 (T (S)) is a basis for V. Theorem 2.12 (Isomorphisms Preserve Bases) If V and W are vector spaces over the same field F and B is a basis for V, then T GL(V, W ) iff T (B) is a basis for W. Proof: If T GL(V, W ) is an isomorphism, then T is bijective, so by Theorem 2.11 T (B) is a basis for W. Conversely, if T (B) is a basis for W, then u V,!a 1,..., a n F and v 1,..., v n B such that u = a 1 v a n v n. Therefore, 0 = T (u) = T (a 1 v a n v n ) = a 1 T (v 1 ) + + a n T (v n ) = a 1 = = a n = 0 so ker(t ) = {0}, whence by Theorem 2.9 T is injective. But since span(t (B)) = W, we have for all w W that!a 1,... a n F such that w = a 1 T (v 1 ) + + a n T (v n ) = T (a 1 v a n v n ), so u = a 1 v a n v n V such that w = T (u), and W im(t ). We obviously have im(t ) W, and so T is surjective. Therefore it is bijective and an isomorphism. Theorem 2.13 (Isomorphisms Preserve Dimension) If V and W are vector spaces over the same field F, then V = W dim(v ) = dim(w ) (2.20) Proof: V = W = T GL(V, W ), so B basis for V = T (B) basis for W, and dim(v ) = B = T (B) = dim(w ). Conversely, if dim(v ) = B = C = dim(w ), where C is a basis for W, then T : B C bijective. Extending T to V by linearity defines a unique T L(V, W ) by Theorem 2.5 and T is an isomorphism because it is surjective, im(t ) = W, and injective, ker(t ) = {0}, so V = W. Corollary 2.14 If V and W are vector spaces over the same field and T L(V, W ), then dim(v ) < dim(w ) = T cannot be onto and dim(v ) > dim(w ) = T cannot be 1-1. Corollary 2.15 If V is an n-dimensional vector space over F, where n N, then V = F n (2.21) If κ is any cardinal number, B is a set of cardinality κ and V is a κ-dimensional vector space over F, then V = (F B ) 0 (2.22) Proof: As a result of the fact that dim(f n ) = n and dim ( (F B ) 0 ) = κ, by the previous theorem. Corollary 2.16 If V and W are vector spaces over the same field F and T GL(V, W ), then for any subspace S of V we have that T (S) is a subspace of W and dim(s) = dim(t (S)). Proof: If S is a subspace of V, then x, y S and a, b F, we have ax+by S, and T (x), T (y) S and T (ax+by) = at (x)+bt (y) T (S) imply that S is a subspace of W, while dim(s) = dim(t (S)) follows from the theorem. 23

24 2.4 Rank-Nullity Theorem Lemma 2.17 If V and W are vector spaces over the same field F and T L(V, W ), then any complement of the kernel of T is isomorphic to the range of T, that is where ker(t ) c is any complement of ker(t ). V = ker(t ) ker(t ) c = ker(t ) c = im(t ) (2.23) Proof: By Theorem 1.23 dim(v ) = dim ( ker(t ) ) + dim ( ker(t ) c). Let T c = T ker(t ) c : ker(t ) c im(t ) and note that T is injective by Theorem 2.9 because ker(t c ) = ker(t ) ker(t c ) = {0} Moreover T c is surjective because T c (V ) = im(t ): first we obviously have T c (V ) im(t ), while for the reverse inclusion suppose v im(t ). Then by Theorem 1.25!s ker(t ) and t ker(t ) c such that v = s + t, so that T (v) = T (s + t) = T (s) + 0 T (t) = T (s) = T c (s) T c (V ) so im(t ) = T c (V ) and T c is surjective, so that T c is an isomorphism, whence ker(t ) c = im(t ) Theorem 2.18 (Rank-Nullity Theorem) If V and W are vector spaces over the same field F and T L(V, W ), then dim(v ) = rank(t ) + null(t ) (2.24) Proof: By the lemma and we have ker(t ) c = im(t ), which by the previous theorem implies dim ( ker(t ) c) = dim ( im(t ) ) = rank(t ), while by Theorem 1.23 V = ker(t ) ker(t ) c implies which completes the proof. dim(v ) = dim ( ker(t ) ) + dim ( ker(t ) c) = dim ( ker(t ) ) + dim ( im(t ) ) = null(t ) + rank(t ) Corollary 2.19 Let V and W are vector spaces over the same field F and T L(V, W ). If dim(v ) = dim(w ), then the following are equivalent: (1) T is injective. (2) T is surjective. (3) rank(t ) = dim(v ) Proof: By the Rank-Nullity Theorem, rank(t ) + null(t ) = dim(v ), and by Theorem 2.9 we have T is 1-1 which completes the proof. Thm 2.9 ker(t ) = {0} null(t ) = 0 R-N Thm dim(im(t )) = rank(t ) = dim(v ) assump. = dim(w ) Thm im(t ) = W. T is onto 24