7.3 Isomorphisms and Composition

1. 7.3 Isomorphisms and Composition

2. Definition T: VW, a LT, is an isomorphism if it is both onto and one-to-one. The vector spaces V and W are called isomorphic if there exists an isomorphism T: VW. (see examples on p. 362)

3. Theorem 1: If V and W are finite dimensional spaces, the following conditions are equivalent for a linear transformation T: VW: 1. T is an isomorphism 2. If {e1,…,en} is any basis of V, then {T(e1),…,T(en)} is a basis of W. 3. There exists a basis {e1,…,en} of V such that {T(e1),…,T(en)} is a basis of W.

4. Thm 1 -proof (1)(2): Given T is an isomorphism and that {e1,…,en} is any basis of V, show {T(e1),…,T(en)} is a basis of W: If t1T(e1)+…+tnT(en) = 0, then T[t1e1+…+tnen] = 0, so t1e1+…+tnen = 0 (since ker T =0 since it is one-to-one). t1=…=tn=0 (since the ei’s are LI). This tells us that {T(e1),…,T(en)} is LI -- just need to show that it spans W. Since T is onto, if w is any vector in W, w=T(v) for some v. v=t1e1+…+tnen. w=T(v) = t1T(e1)+…+tnT(en) so {T(e1),…,T(en)} spans W.

5. Thm 1 -proof (2)(3): Given that {e1,…,en} is any basis of V means that {T(e1),…,T(en)} is a basis of W, show that there exists a basis {e1,…,en} of V such that {T(e1),…,T(en)} is a basis of W. (Just need to show that V has a basis at all which it must since it is finite dimensional.)

6. Thm 1 -proof (3)(1): Given that there exists a basis {e1,…,en} of V such that {T(e1),…,T(en)} is a basis of W, show that T is an isomorphism (both one-to-one and onto) If T(v)=0, write v=v1e1+…+vnen, then 0=T(v)=v1T(e1)+…+vnT(en). So v1=…=vn=0 since {T(e1),…,T(en)} is LI. So v=0, so ker T=0 which tells us that T is one-to-one. Let w be any vector in W. w=w1T(e1)+…+wnT(en) since {T(e1),…,T(en)} spans W. w=T(w1e1+…+wnen). This tells us that we can create any w which means T is onto. 

7. Theorem 2 Two finite dimensional vector spaces V and W are isomorphic iff dim V=dim W. Proof:  Given T: VW is an isomorphism, and let {e1,…,en} be a basis of V, then {T(e1),…,T(en)} is a basis of W by thm 1. So dim W=n=dim V. : Given dim V=n=dimW, so let {e1,…,en} be a basis of V and let {f1,…,fn} be a basis of W. Then there exists a LT T: VW such that T(ei)=fi by thm 4 of 7.1. Then {T(e1),…,T(en)} is a basis of W. Therefore T is an isomorphism by thm 1. 

8. Example V denotes the space of all 2x2 symmetric matrices. Find an isomorphism T: P2V such that T(1) = I. Solution: We know that {1,x,x2} is a basis of P2, and we want a basis of V containing I. is LI, and is thus a basis (since dim V=3 from earlier in the year. Let T: P2V by taking T is an isomorphism by thm 1 (since we go from basis of P2 to basis of V. The action is:

9. Theorem 3 If dim V = dim W = n, a LT T: VW is an isomorphism if it is either one-to-one or onto. Proof: dim(ker T) + dim(im T) = n, so dim(ker T) = 0 iff dim(im T) = n. So T is one-to-one iff T is onto. 

10. Composition If T: VW and S: WU are LT’s, we can define a new function V U by first applying T and then S. Def: If this is the case, we define the composite ST: V U of T and S by ST(v) = S[T(v)] for all v in V The operation of forming the new function ST is called composition. (Just like with functions)

11. Example Let T: 3 2 and S: 2 4 be defined by T(x,y,z) = (x+y,y+z) and S(x,y)=(x-y,x+y,y,x). Show ST. ST(x,y,z)=S[T(x,y,z)] = S(x+y,y+z)=(x-z, x+2y+z, y+z, x+y) Work through example 6 on p. 366

12. Notation 1V: V  V is called the identity transformation for V. (It is an isomorphism)

13. Theorem 4 are LT’s. Then: 1. The composite ST is an LT. 2. T1V = T and 1WT = T 3. (RS)T = R(ST) Proof: (1) and (2) are homework. (3) {(RS)T}(v) = (RS){T(v)} = R{S[T(v)]}={R(ST)}(v).

14. Theorem 5 If V and W are finite dimensional vector spaces, then the following conditions are equivalent for LT T: VW. 1. T is an isomorphism 2. There exists a LT S: WV such that ST = 1V and TS = 1W Also, in this case, S is also an isomorphism and is uniquely determined by T: If w in W is written as w = T(v), then S(w)=v.

15. Theorem 5-proof (1)(2) Given T is an isomorphism, show that there exists a LT S: WV such that ST = 1V and TS = 1W B={e1,…,en} is a basis of V, then D={T(e1),…,T(en)} is a basis of W (by thm 1). So S: W V is defined by: S[T(ei)] = ei for each I which tells us that ST = 1V. Then apply T: T[S[T(ei)]]=T(ei), So TS = 1W.

16. Theorem 5-proof (2)(1) Given that there exists a LT S: WV such that ST = 1V and TS = 1W, show that T is an isomorphism. If T(v) = T(v1), then S[T(v)] = S[T(v1)]. Since ST = 1V, we have v = v1 so T is one-to-one Given w in W, TS = 1W means that w=T[S(w)] and so T is onto since we have a way of getting every w using T. S is uniquely determined by ST = 1V since this implies S[T(ei)]=ei (so S is the trans that takes D back to B) S is isomorphism since it carries the basis D to B. Given w in W, write w = r1T(e1)+…+rnT(en)=T(v) where v = r1e1+…+rnen Then S(w) = S(r1T(e1)+…+rnT(en))=v since S[T(ei)]=ei

17. Definition Given an isomorphism T: VW, the isomorphism S: W  V satisfying condition (2) of Thm 5 is called the inverse of T and is denoted by T-1. So T: VW and T-1: W  V are related by the fundamental identities: T-1[T(v)] = v for all v in V and T[T-1(w)] = w for all w in W

18. Example Go through examples 7,8,9 p. 368

19. Theorem 6 A, B matrices: 1. If TA=TB then A = B 2. 3. TA TB=TAB 4. TA has an inverse iff A is invertible, and then 5. im TA=col A, so rank TA=rank A