Vector Space Isomorphism

S. F. Ellermeyer

Our goal here is to explain why two …nite–dimensional vector spaces, V and W , are isomorphic to each other if and only if they have the same dimension. In the process, we will also discuss the concept of an equivalence relation. This concept is important throughout mathematics.

1 What is an Equivalence Relation?

If X is some non–empty set of objects, then we can partition X into equiv- alence classes based on some criteria of our choosing. When we say that we are going to partition X, we mean that we are going to create disjoint subdivisions of X whose union gives us all of X. In the Venn diagram shown below, the big set X has been partitioned into three subsets, A; B; and C. In this diagram, we have A B = ;B C = ; and A C = and we also have X = A B C. Thus\X is the; disjoint\ union; of the\ subsets; A; B; and C. [ [

1 Here is an example of an equivalence relation: Suppose that X is a group of six students (Ann, Bob, Carrie, Doug, Evan, Frank) whose attributes are given in the table below. Name Gender Year Major Eye Color Ann Female Senior Computer Science Blue Bob Male Senior Mathematics Green Carrie Female Sophomore Biology Blue Doug Male Junior Chemistry Blue Evan Male Sophomore Computer Science Blue Frank Male Junior Computer Science Hazel We can partition X into a disjoint union of subsets (equivalence classes) by saying that any two students in the set will be considered to be equiv- alent to each other if they have the same gender. If we denote the gender equivalence relation by the symbol , then  Ann Carrie  and Bob Doug Evan Frank.    This equivalence relation partitions the set X into two disjoint subsets, which we might choose to call F and M, as shown in the following Venn diagram.

Note that F M = and that X = F M. Another way\ to partition; this group of[ students would be according to eye color. In this case, assuming that this equivalence relation is denoted by the symbol , we have  2 Ann Carrie Doug Evan    and Bob is an equivalence class by himself and Frank is in an equivalence class by himself. A Venn diagram of this equivalence relation is shown below.

In this case we have B G = ;B H = ; and G H = , and we also have X = B G H. \ ; \ ; \ ; In general,[ de…ning[ an equivalence relation, , on a given set, X, requires that we choose some criterion for which the following three conditions are satis…ed: 1. must be re‡exive. This means that x x for all x X.   2 2. must be symmetric. This means that if x y, then it must also be true that y x.   3. must be transitive. This means that if x y and y z, then x z.     Thus an equivalence relation is a relation which is re‡exive, symmetric, and transitive. We will now present some examples of equivalence relations that are more “mathematical”in nature than the examples given above.

1.1 Example: Row–Equivalence of Matrices Let X be the set of all 2 2 matrices with real entries. We will de…ne A B to mean that B can be obtained from A by performing a sequence of elementary row operations. To verify that is an equivalence relation, we must verify that is re‡exive, symmetric, and transitive.  3 1. To see that is re‡exive, let A be any 2 2 matrix. By performing   the row operation 1 R1 R1 on A, we end up with A itself. Thus A A.
!  2. To see that is re‡exive, suppose that A B. This means that B can be obtained from A by performing a sequence of elementary row operations. However, since all elementary row operations are reversible (meaning that they all can be “undone”), then A can also be obtained from B by performing a sequence of elementary row operations. There- fore B A.  3. To see that is transitive, suppose that A B and B C. This means that B can be obtained from A by performing a sequence of elementary row operations and that C can be obtained from B by performing a sequence of elementary row operations. But then clearly C can be obtained from A by performing a sequence of elementary row operations. Therefore A C.  1.2 Example: Equivalence of Integers Via Residues Let Z be the set of all integers and de…ne x y to mean that x and y have the same remainder when divided by 3. (Note that we are using the symbol rather than the symbol do denote this relation.) As an example to illustrate what we are doing here, notice that when we divide 16 by 3, the remainder is 1, and when we divide 58 by 3, the remainder is also 1. Thus 16 58. Let us now show that is re‡exive symmetric, and transitive (and hence that is an equivalence relation).  1. It is obvious that is re‡exive because if x is any integer, then x and x have the same remainder when divided by 3.

2. It is also obvious that is symmetric because if x and y have the same remainder when divided by 3, then y and x have the same remainder when divided by 3.

3. Transitivity is also obvious: If x and y have the same remainder when divided by 3 and y and z have the same remainder when divided by 3, then x and z have the same remainder when divided by 3.

4 Note that gives rise to exactly three equivalence classes. One of these is the set of integers that have remainder 0 when divided by 3; the other is the set of integers that have remainder 1 when divided by 3, and the third is the set of integers that have remainder 2 when divided by 3:

9 6 3 0 3 6 9


      


8 5 2 1 4 7 10


      


7 4 1 2 5 8 11 .


      


The above three equivalence classes are usually denoted, respectively, by 0; 1; and 2. These three subsets are then usually referred to as the integers modulo 3. Note that the sets 0; 1; and 2 are pairwise disjoint and that Z = 0 1 2. Thus these three sets form a partition of Z. Further study of the integers[ [ modulo n (where n is some given integer) is usually encountered in a course in Abstract Algebra (also called Modern Algebra). The integers modulo n are also often referred to as the ring of residues modulo n.

1.3 Example: An Equivalence Relation From Calculus Let X be the set of all continuous functions with domain [0; 1]. Recall from Calculus that all such functions, f, are integrable. This means that

1 f (x) dx Z0 is guaranteed to exist (and be equal to some real number). We can de…ne an equivalence relation, , on X by saying that f g if   1 1 f (x) dx = g (x) dx. Z0 Z0 2 2 For example, if f is the function f (x) = x and g is the function g (x) = 3 x, then f g. (Verify this for yourself.) It is not di¢ cult to see that is re‡exive, symmetric, and transitive and hence that is an equivalence relation. Certainly, determines in…nitely many equivalence classes (one corresponding to each real number).

5 1.4 Example: A Relation That is Not an Equivalence Relation Let Z be the set of all integers and consider the well–known relation . When we write x y, we mean that the number x is less than or equal to the number y. Why is not an equivalence relation on Z? It does satisfy two of the three requirements of an equivalence relation. Which one does it not satisfy?

2 Equivalence of Vector Spaces Via Isomor- phism

Now that we have introduced the concept of equivalence relations, we will embark on our main agenda which is to show that isomorphism is an equiv- alence relation on the set of all vector spaces. Recall that if V and W are vector spaces and T : V W is a linear transformation that is both one– to–one and onto W , then!T is called an isomorphism. Also recall that if V and W are vector spaces and there exists an isomorphism T : V W , then we say that V is isomorphic to W . ! With the above de…nitions in mind, let us take X to be the set of all vector spaces and let us de…ne V W to mean that V is isomorphic to W . We are going to prove that is an equivalence relation on X. The details needed to prove this fact will be established via three lemmas.

Lemma 1 Let V be any vector space and de…ne I : V V by I (v) = v for all v V . Then I is an isomorphism (which is called! the identity transformation).2

Proof. We want to prove that I is an isomorphism (meaning that I is a linear transformation that is both one–to–one and onto V ). To see that I is a linear transformation, let u and v V . Then 2 I (u + v) = u + v = I (u) + I (v) .

Next let u V and let c be a scalar. Then 2 I (cu) = cu = cI (u) .

This shows that I is a linear transformation.

6 To see that I maps V onto V , let v be any vector in V . Since I (v) = v, then v is in the range of I. Thus every vector in V is in the range of I which means that I maps V onto V . To see that I is one–to–one, let u and v be any two vectors in V with I (u) = I (v). Since I (u) = u and I (v) = v, then u = v. Thus any two di¤erent vectors in V are mapped by I to di¤erent vectors in V . This shows that I is one–to–one and completes the proof that I is an isomorphism. Before giving our next lemma, we must de…ne the concept of the inverse of a linear transformation. De…nition 2 Suppose that T : V W is an isomorphism. The inverse of !1 1 T is de…ned to be the mapping T : W V de…ned by T (w) = v where v is the unique vector in V such that T (!v) = w. Remark 3 We know that v, as given in the above de…nition, is unique be- cause T is one–to–one. In addition, we know that such a v exists for each 1 given w W because T maps V onto W . Thus the mapping T is well– de…ned and2 has domain W .

1 Remark 4 Observe that if v is any vector in V , then T (T (v)) = v. Also 1 observe that if w is any vector in W , then T (T (w)) = w. Lemma 5 If V and W are vector spaces and T : V W is an isomorphism, 1 ! then T : W V is also an isomorphism. ! 1 Proof. To see that T is a linear transformation, let w1 and w2 be vectors 1 in W . Then T (w1) = v1 where v1 is the unique vector in V such that 1 T (v1) = w1 and T (w2) = v2 where v2 is the unique vector in V such that T (v2) = w2. Also, since T is a linear transformation, then

T (v1 + v2) = T (v1) + T (v2) = w1 + w2 and since T is one–to–one, then it must be the case that v1 + v2 is the only vector in V that is mapped by T to the vector w1 + w2. Thus 1 1 T (w1 + w2) = v1 + v2 (by de…nition of T ). We now see that 1 1 1 T (w1 + w2) = v1 + v2 = T (w1) + T (w2) .

1 Next let w be a vector in W and let c be a scalar. Then T (w) = v where v is the unique vector in V such that T (v) = w. Also, since T is a linear transformation, then T (cv) = cT (v) = cw

7 and since T is one–to–one, then cv is the only vector in V such that is mapped 1 by T onto the vector cw. Therefore T (cw) = cv and we now observe that

1 1 T (cw) = cv = cT (w) .

1 This completes the proof that T is a linear transformation. 1 To see that T maps W onto V , let v be any vector in V and suppose 1 that T (v) = w. Then T (w) = v which means that v is in the range of 1 1 T . Therefore T maps W onto V . 1 To see that T is one–to–one, let w1 and w2 be two vectors in W and 1 1 suppose that T (w1) = T (w2). (We need to show that w1 = w2.) Since T is a linear transformation, then

1 1 1 1 T (0V ) = T T (w1) T (w2) = T T (w1) T T (w2) = w1 w2. However, since T is a linear transformation,
then
it also must be
true that

T (0V ) = 0W .

Since T is one–to–one, we conclude that w1 w2 = 0W and hence that 1 w1 = w2. This completes the proof that T is one–to–one. Before stating our third lemma, we recall the meaning of the composition of two functions.

De…nition 6 Suppose that V , W , and Y are vector spaces and suppose that T : V W and that S : W Y . The composition, S T , is the mapping S T :!V Y de…ned by !   ! S T (v) = S (T (v)) for all v V .  2 Lemma 7 Suppose that V , W , and Y are vector spaces and suppose that T : V W and S : W Y are both isomorphisms. Then S T : V Y is also an! isomorphism. !  !

Proof. To see that S T is a linear transformation, let u and v be vectors in V . Since S and T are both linear transformations, we see that S T (u + v) = S (T (u + v))  = S (T (u) + T (v)) = S (T (u)) + S (T (v)) = S T (u) + S T (v) .  

8 In addition, if u is a vector in V and c is a scalar, then

S T (cu) = S (T (cu))  = S (cT (u)) = cS (T (u)) = cS T (u)  showing that S T is a linear transformation. To see that S T maps V onto Y , let y be any vector in Y . Since S maps W onto Y , then there is some vector w W such that S (w) = y. In addition, since T maps V onto W , then there is2 some vector v V such that T (v) = w. Thus we have that 2

S T (v) = S (T (v)) = S (w) = y  and this shows that y is in the range of S T . Therefore S T maps V onto Y .   To see that S T is one–to–one, let v1 and v2 be vectors in V and  suppose that S T (v1) = S T (v2). Then S (T (v1)) = S (T (v2)). Since   S is one–to–one, it must then be true that T (v1) = T (v2). However, T is also one–to–one, so it must be true that v1 = v2. This shows that S T is one–to–one and completes our proof that S T is an isomorphism.  We now combine Lemmas 1, 5, and 7 to obtain the following main result.

Theorem 8 Let X be the set of all vector spaces and de…ne the relation  on X by V W if and only if V is isomorphic to W .  Then is an equivalence relation on X.  Proof. We must show that is re‡exive, symmetric, and transitive.  1. Let V be any vector space. By Lemma 1, the identity transformation, I : V V is an isomorphism. Thus V V . This shows that is re‡exive.!  

2. Let V and W be vector spaces and suppose that V W . Then there 1 exists an isomorphism T : V W . By Lemma 5, T : W V is also an isomorphism. Thus W !V . This shows that is symmetric.!  

9 3. Let V , W , and Y be vector spaces and suppose that V W and W Y . Then there exists an isomorphism T : V W and there also exists an isomorphism S : W Y . By Lemma 7,! the composition S T : V Y is also an isomorphism.! Therefore V Y . This shows that is transitive.!  

3 Equivalence of Finite–Dimensional Vector Spaces

Theorem 8 applies to the set of all vector spaces. The theorem tells us that the set of all vector spaces can be decomposed into equivalence classes that consist of vector spaces that are isomorphic to each other. There is a simpler equivalent criterion by which to obtain this decomposition for the set of all …nite–dimensional vector spaces.

Theorem 9 Two …nite–dimensional vector spaces, V and W , are isomor- phic to each other if and only if they have the same dimension.

Proof. Suppose that dim (V ) = dim (W ) = n. Then V has a basis

B = v1; v2;:::; vn f g and W has a basis C = w1; w2;:::; wn . f g To show that V must be isomorphic to W , we de…ne T : V W as follows: ! For any v V , there exist unique scalars a1; a2; : : : ; an such that 2

v = a1v1 + a2v2 + + anvn.


We de…ne T (v) = a1w1 + a2w2 + + anwn.


It can be seen that T is a linear transformation that is both one–to–one and onto W . Thus V is isomorphic to W . Now, to prove the converse statement, suppose that V is isomorphic to W and suppose that dim (V ) = n. Then there exists an isomorphism T :

10 V W and there also exists a basis, B = v1; v2;:::; vn , for V . We claim ! f g that C = T (v1) ;T (v2) ;:::;T (vn) is a basis for W . To provef that C is linearly independent,g consider the equation

c1T (v1) + c2T (v2) + + cnT (vn) = 0W .


Since T is a linear transformation, this equation is equivalent to

T (c1v1 + c2v2 + + cnvn) = 0W .

Because T (0V ) = 0W and because T is one–to–one, it must be the case that

c1v1 + c2v2 + + cnvn = 0V .

Because B is linearly independent, it must then be the case that c1 = c2 = = cn = 0. This shows that C is linearly independent.


To prove that C spans W , let w W . Since T maps V onto W , there exists a vector v V such that T (v2) = w. Also, since B is a basis for V 2 (and hence spans V ) then there are scalars, a1; a2; : : : ; an; such that

v = a1v1 + a2v2 + + anvn.


Now observe that

w = T (v) = a1T (v1) + a2T (v2) + + anT (vn)


which shows that w Span(C). This proves that W = Span(C). Since W has a basis2 containing exactly n vectors, then dim (W ) = n. Therefore dim (V ) = dim (W ).

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