Linear Algebra/Definition and Examples of Isomorphisms/Solutions

SolutionsEdit

This exercise is recommended for all readers.
Problem 1

Verify, using Example 1.4 as a model, that the two correspondences given before the definition are isomorphisms.

  1. Example 1.1
  2. Example 1.2
Answer
  1. Call the map  .
     
    It is one-to-one because if   sends two members of the domain to the same image, that is, if  , then the definition of   gives that
     
    and since column vectors are equal only if they have equal components, we have that   and that  . Thus, if   maps two row vectors from the domain to the same column vector then the two row vectors are equal:  . To show that   is onto we must show that any member of the codomain   is the image under   of some row vector. That's easy;
     
    is  . The computation for preservation of addition is this.
     
    The computation for preservation of scalar multiplication is similar.
     
  2. Denote the map from Example 1.2 by  . To show that it is one-to-one, assume that  . Then by the definition of the function,
     
    and so   and   and  . Thus  , and consequently   is one-to-one. The function   is onto because there is a polynomial sent to
     
    by  , namely,  . As for structure, this shows that   preserves addition
     
    and this shows
     
    that it preserves scalar multiplication.
This exercise is recommended for all readers.
Problem 2

For the map   given by

 

Find the image of each of these elements of the domain.

  1.  
  2.  
  3.  

Show that this map is an isomorphism.

Answer

These are the images.

  1.  
  2.  
  3.  

To prove that   is one-to-one, assume that it maps two linear polynomials to the same image  . Then

 

and so, since column vectors are equal only when their components are equal,   and  . That shows that the two linear polynomials are equal, and so   is one-to-one.

To show that   is onto, note that this member of the codomain

 

is the image of this member of the domain  .

To check that   preserves structure, we can use item 2 of Lemma 1.9.

 
Problem 3

Show that the natural map   from Example 1.5 is an isomorphism.

Answer

To verify it is one-to-one, assume that  . Then   by the definition of  . Members of   are equal only when they have the same coefficients, so this implies that   and   and  . Therefore   implies that  , and so   is one-to-one.

To verify that it is onto, consider an arbitrary member of the codomain   and observe that it is indeed the image of a member of the domain, namely, it is  . (For instance,  .)

The computation checking that   preserves addition is this.

 

The check that   preserves scalar multiplication is this.

 
This exercise is recommended for all readers.
Problem 4

Decide whether each map is an isomorphism (if it is an isomorphism then prove it and if it isn't then state a condition that it fails to satisfy).

  1.   given by
     
  2.   given by
     
  3.   given by
     
  4.   given by
     
Answer
  1. No; this map is not one-to-one. In particular, the matrix of all zeroes is mapped to the same image as the matrix of all ones.
  2. Yes, this is an isomorphism. It is one-to-one:
     
    gives that  , and that  , and that  , and that  . It is onto, since this shows
     
    that any four-tall vector is the image of a   matrix. Finally, it preserves combinations
     
    and so item 2 of Lemma 1.9 shows that it preserves structure.
  3. Yes, it is an isomorphism. To show that it is one-to-one, we suppose that two members of the domain have the same image under  .
     
    This gives, by the definition of  , that   and then the fact that polynomials are equal only when their coefficients are equal gives a set of linear equations
     
    that has only the solution  ,  ,  , and  . To show that   is onto, we note that   is the image under   of this matrix.
     
    We can check that   preserves structure by using item 2 of Lemma 1.9.
     
  4. No, this map does not preserve structure. For instance, it does not send the zero matrix to the zero polynomial.
Problem 5

Show that the map   given by   is one-to-one and onto.Is it an isomorphism?

Answer

It is one-to-one and onto, a correspondence, because it has an inverse (namely,  ). However, it is not an isomorphism. For instance,  .

This exercise is recommended for all readers.
Problem 6

Refer to Example 1.1. Produce two more isomorphisms (of course, that they satisfy the conditions in the definition of isomorphism must be verified).

Answer

Many maps are possible. Here are two.

 

The verifications are straightforward adaptations of the others above.

Problem 7

Refer to Example 1.2. Produce two more isomorphisms (and verify that they satisfy the conditions).

Answer

Here are two.

 

Verification is straightforward (for the second, to show that it is onto, note that

 

is the image of  ).

This exercise is recommended for all readers.
Problem 8

Show that, although   is not itself a subspace of  , it is isomorphic to the  -plane subspace of  .

Answer

The space   is not a subspace of   because it is not a subset of  . The two-tall vectors in   are not members of  .

The natural isomorphism   (called the injection map) is this.

 

This map is one-to-one because

 

which in turn implies that   and  , and therefore the initial two two-tall vectors are equal.

Because

 

this map is onto the  -plane.

To show that this map preserves structure, we will use item 2 of Lemma 1.9 and show

 
 

that it preserves combinations of two vectors.

Problem 9

Find two isomorphisms between   and  .

Answer

Here are two:

 

Verification that each is an isomorphism is easy.

This exercise is recommended for all readers.
Problem 10

For what   is   isomorphic to  ?

Answer

When   is the product  , here is an isomorphism.

 

Checking that this is an isomorphism is easy.

Problem 11

For what   is   isomorphic to  ?

Answer

If   then  . (If we take   and   to be trivial vector spaces, then the relationship extends one dimension lower.) The natural isomorphism between them is this.

 

Checking that it is an isomorphism is straightforward.

Problem 12

Prove that the map in Example 1.7, from   to   given by  , is a vector space isomorphism.

Answer

This is the map, expanded.

 

To finish checking that it is an isomorphism, we apply item 2 of Lemma 1.9 and show that it preserves linear combinations of two polynomials. Briefly, the check goes like this.

 
 
 
Problem 13

Why, in Lemma 1.8, must there be a  ? That is, why must   be nonempty?

Answer

No vector space has the empty set underlying it. We can take   to be the zero vector.

Problem 14

Are any two trivial spaces isomorphic?

Answer

Yes; where the two spaces are   and  , the map sending   to   is clearly one-to-one and onto, and also preserves what little structure there is.

Problem 15

In the proof of Lemma 1.9, what about the zero-summands case (that is, if   is zero)?

Answer

A linear combination of   vectors adds to the zero vector and so Lemma 1.8 shows that the three statements are equivalent in this case.

Problem 16

Show that any isomorphism   has the form   for some nonzero real number  .

Answer

Consider the basis   for   and let   be  . For any   we have that   and so  's action is multiplication by  . Note that   or else the map is not one-to-one. (Incidentally, any such map   is an isomorphism, as is easy to check.)

This exercise is recommended for all readers.
Problem 17

These prove that isomorphism is an equivalence relation.

  1. Show that the identity map   is an isomorphism. Thus, any vector space is isomorphic to itself.
  2. Show that if   is an isomorphism then so is its inverse  . Thus, if   is isomorphic to   then also   is isomorphic to  .
  3. Show that a composition of isomorphisms is an isomorphism: if   is an isomorphism and   is an isomorphism then so also is  . Thus, if   is isomorphic to   and   is isomorphic to  , then also   is isomorphic to  .
Answer

In each item, following item 2 of Lemma 1.9, we show that the map preserves structure by showing that the it preserves linear combinations of two members of the domain.

  1. The identity map is clearly one-to-one and onto. For linear combinations the check is easy.
     
  2. The inverse of a correspondence is also a correspondence (as stated in the appendix), so we need only check that the inverse preserves linear combinations. Assume that   (so  ) and assume that  .
     
  3. The composition of two correspondences is a correspondence (as stated in the appendix), so we need only check that the composition map preserves linear combinations.
     
Problem 18

Suppose that   preserves structure. Show that   is one-to-one if and only if the unique member of   mapped by   to   is  .

Answer

One direction is easy: by definition, if   is one-to-one then for any   at most one   has  , and so in particular, at most one member of   is mapped to  . The proof of Lemma 1.8 does not use the fact that the map is a correspondence and therefore shows that any structure-preserving map   sends   to  .

For the other direction, assume that the only member of   that is mapped to   is  . To show that   is one-to-one assume that  . Then   and so  . Consequently  , so  , and so   is one-to-one.

Problem 19

Suppose that   is an isomorphism. Prove that the set   is linearly dependent if and only if the set of images   is linearly dependent.

Answer

We will prove something stronger— not only is the existence of a dependence preserved by isomorphism, but each instance of a dependence is preserved, that is,

 
 

The   direction of this statement holds by item 3 of Lemma 1.9. The   direction holds by regrouping

 

and applying the fact that   is one-to-one, and so for the two vectors   and   to be mapped to the same image by  , they must be equal.

This exercise is recommended for all readers.
Problem 20

Show that each type of map from Example 1.6 is an automorphism.

  1. Dilation   by a nonzero scalar  .
  2. Rotation   through an angle  .
  3. Reflection   over a line through the origin.

Hint. For the second and third items, polar coordinates are useful.

Answer
  1. This map is one-to-one because if   then by definition of the map,   and so  , as   is nonzero. This map is onto as any   is the image of   (again, note that   is nonzero). (Another way to see that this map is a correspondence is to observe that it has an inverse: the inverse of   is  .) To finish, note that this map preserves linear combinations
     
    and therefore is an isomorphism.
  2. As in the prior item, we can show that the map   is a correspondence by noting that it has an inverse,  . That the map preserves structure is geometrically easy to see. For instance, adding two vectors and then rotating them has the same effect as rotating first and then adding. For an algebraic argument, consider polar coordinates: the map   sends the vector with endpoint   to the vector with endpoint  . Then the familiar trigonometric formulas   and   show how to express the map's action in the usual rectangular coordinate system.
     
    Now the calculation for preservation of addition is routine.
     
    The calculation for preservation of scalar multiplication is similar.
  3. This map is a correspondence because it has an inverse (namely, itself). As in the last item, that the reflection map preserves structure is geometrically easy to see: adding vectors and then reflecting gives the same result as reflecting first and then adding, for instance. For an algebraic proof, suppose that the line   has slope   (the case of a line with undefined slope can be done as a separate, but easy, case). We can follow the hint and use polar coordinates: where the line   forms an angle of   with the  -axis, the action of   is to send the vector with endpoint   to the one with endpoint  .

     

    To convert to rectangular coordinates, we will use some trigonometric formulas, as we did in the prior item. First observe that   and   can be determined from the slope   of the line. This picture

     

    gives that   and  . Now,

     

    and thus the first component of the image vector is this.

     

    A similar calculation shows that the second component of the image vector is this.

     

    With this algebraic description of the action of  


     


    checking that it preserves structure is routine.

Problem 21

Produce an automorphism of   other than the identity map, and other than a shift map  .

Answer

First, the map   doesn't count because it is a version of  . Here is a correct answer (many others are also correct):  . Verification that this is an isomorphism is straightforward.

Problem 22
  1. Show that a function   is an automorphism if and only if it has the form   for some  .
  2. Let   be an automorphism of   such that  . Find  .
  3. Show that a function   is an automorphism if and only if it has the form
     
    for some   with  . Hint. Exercises in prior subsections have shown that
     
    if and only if  .
  4. Let   be an automorphism of   with
     
    Find
     
Answer
  1. For the "only if" half, let   to be an isomorphism. Consider the basis  . Designate   by  . Then for any   we have that  , and so  's action is multiplication by  . To finish this half, just note that   or else   would not be one-to-one. For the "if" half we only have to check that such a map is an isomorphism when  . To check that it is one-to-one, assume that   so that   and divide by the nonzero factor   to conclude that  . To check that it is onto, note that any   is the image of   (again,  ). Finally, to check that such a map preserves combinations of two members of the domain, we have this.
     
  2. By the prior item,  's action is  . Thus  .
  3. For the "only if" half, assume that   is an automorphism. Consider the standard basis   for  . Let
     
    Then the action of   on any vector is determined by by its action on the two basis vectors.
     
    To finish this half, note that if  , that is, if   is a multiple of  , then   is not one-to-one. For "if" we must check that the map is an isomorphism, under the condition that  . The structure-preservation check is easy; we will here show that   is a correspondence. For the argument that the map is one-to-one, assume this.
     
    Then, because  , the resulting system
     
    has a unique solution, namely the trivial one   and   (this follows from the hint). The argument that this map is onto is closely related— this system
     
    has a solution for any   and   if and only if this set
     
    spans  , i.e., if and only if this set is a basis (because it is a two-element subset of  ), i.e., if and only if  .
  4.  
Problem 23

Refer to Lemma 1.8 and Lemma 1.9. Find two more things preserved by isomorphism.

Answer

There are many answers; two are linear independence and subspaces.

To show that if a set   is linearly independent then its image   is also linearly independent, consider a linear relationship among members of the image set.

 

Because this map is an isomorphism, it is one-to-one. So   maps only one vector from the domain to the zero vector in the range, that is,   equals the zero vector (in the domain, of course). But, if   is linearly independent then all of the  's are zero, and so   is linearly independent also. (Remark. There is a small point about this argument that is worth mention. In a set, repeats collapse, that is, strictly speaking, this is a one-element set:  , because the things listed as in it are the same thing. Observe, however, the use of the subscript   in the above argument. In moving from the domain set   to the image set  , there is no collapsing, because the image set does not have repeats, because the isomorphism   is one-to-one.)

To show that if   is an isomorphism and if   is a subspace of the domain   then the set of image vectors   is a subspace of  , we need only show that it is closed under linear combinations of two of its members (it is nonempty because it contains the image of the zero vector). We have

 

and   is a member of   because of the closure of a subspace under combinations. Hence the combination of   and   is a member of  .

Problem 24

We show that isomorphisms can be tailored to fit in that, sometimes, given vectors in the domain and in the range we can produce an isomorphism associating those vectors.

  1. Let   be a basis for   so that any   has a unique representation as  , which we denote in this way.
     
    Show that the   operation is a function from   to   (this entails showing that with every domain vector   there is an associated image vector in  , and further, that with every domain vector   there is at most one associated image vector).
  2. Show that this   function is one-to-one and onto.
  3. Show that it preserves structure.
  4. Produce an isomorphism from   to   that fits these specifications.
     
Answer
  1. The association
     
    is a function if every member   of the domain is associated with at least one member of the codomain, and if every member   of the domain is associated with at most one member of the codomain. The first condition holds because the basis   spans the domain— every   can be written as at least one linear combination of  's. The second condition holds because the basis   is linearly independent— every member   of the domain can be written as at most one linear combination of the  's.
  2. For the one-to-one argument, if  , that is, if   then
     
    and so   and   and  , which gives the conclusion that  . Therefore this map is one-to-one. For onto, we can just note that
     
    equals  , and so any member of the codomain   is the image of some member of the domain  .
  3. This map respects addition and scalar multiplication because it respects combinations of two members of the domain (that is, we are using item 2 of Lemma 1.9): where   and  , we have this.
     
  4. Use any basis   for   whose first two members are   and  , say  .
Problem 25

Prove that a space is  -dimensional if and only if it is isomorphic to  . Hint. Fix a basis   for the space and consider the map sending a vector over to its representation with respect to  .

Answer

See the next subsection.

Problem 26

(Requires the subsection on Combining Subspaces, which is optional.) Let   and   be vector spaces. Define a new vector space, consisting of the set   along with these operations.

 

This is a vector space, the external direct sum of   and  .

  1. Check that it is a vector space.
  2. Find a basis for, and the dimension of, the external direct sum  .
  3. What is the relationship among  ,  , and  ?
  4. Suppose that   and   are subspaces of a vector space   such that   (in this case we say that   is the internal direct sum of   and  ). Show that the map   given by
     
    is an isomorphism. Thus if the internal direct sum is defined then the internal and external direct sums are isomorphic.
Answer
  1. Most of the conditions in the definition of a vector space are routine. We here sketch the verification of part 1 of that definition. For closure of  , note that because   and   are closed, we have that   and   and so  . Commutativity of addition in   follows from commutativity of addition in   and  .
     
    The check for associativity of addition is similar. The zero element is   and the additive inverse of   is  . The checks for the second part of the definition of a vector space are also straightforward.
  2. This is a basis
     
    because there is one and only one way to represent any member of   with respect to this set; here is an example.