Linear Algebra/Rangespace and Nullspace/Solutions

SolutionsEdit

This exercise is recommended for all readers.
Problem 1

Let   be given by  . Which of these are in the nullspace? Which are in the rangespace?

  1.  
  2.  
  3.  
  4.  
  5.  
Answer

First, to answer whether a polynomial is in the nullspace, we have to consider it as a member of the domain  . To answer whether it is in the rangespace, we consider it as a member of the codomain  . That is, for  , the question of whether it is in the rangespace is sensible but the question of whether it is in the nullspace is not because it is not even in the domain.

  1. The polynomial   is not in the nullspace because   is not the zero polynomial in  . The polynomial   is in the rangespace because   is mapped by   to  .
  2. The answer to both questions is, "Yes, because  ." The polynomial   is in the nullspace because it is mapped by   to the zero polynomial in  . The polynomial   is in the rangespace because it is the image, under  , of  .
  3. The polynomial   is not in the nullspace because   is not the zero polynomial in  . The polynomial   is not in the rangespace because there is no member of the domain that when multiplied by   gives the constant polynomial  .
  4. The polynomial   is not in the nullspace because  . The polynomial   is in the rangespace because it is the image of  .
  5. The polynomial   is not in the nullspace because  . The polynomial   is not in the rangespace because of the constant term.
This exercise is recommended for all readers.
Problem 2

Find the nullspace, nullity, rangespace, and rank of each map.

  1.   given by
     
  2.   given by
     
  3.   given by
     
  4. the zero map  
Answer
  1. The nullspace is
     
    while the rangespace is
     
    and so the nullity is one and the rank is one.
  2. The nullspace is this.
     
    The rangespace
     
    is all of   (we can get any real number by taking   to be   and taking   to be the desired number). Thus, the nullity is three and the rank is one.
  3. The nullspace is
     
    while the rangespace is  . Thus, the nullity is two and the rank is two.
  4. The nullspace is all of   so the nullity is three. The rangespace is the trivial subspace of   so the rank is zero.
This exercise is recommended for all readers.
Problem 3

Find the nullity of each map.

  1.   of rank five
  2.   of rank one
  3.  , an onto map
  4.  , onto
Answer

For each, use the result that the rank plus the nullity equals the dimension of the domain.

  1.  
  2.  
  3.  
  4.  
This exercise is recommended for all readers.
Problem 4

What is the nullspace of the differentiation transformation  ? What is the nullspace of the second derivative, as a transformation of  ? The  -th derivative?

Answer

Because

 

we have this.

 

In the same way,

 

for  .

Problem 5

Example 2.7 restates the first condition in the definition of homomorphism as "the shadow of a sum is the sum of the shadows". Restate the second condition in the same style.

Answer

The shadow of a scalar multiple is the scalar multiple of the shadow.

Problem 6

For the homomorphism   given by   find these.

  1.  
  2.  
  3.  
Answer
  1. Setting   gives   and   and  , so the nullspace is  .
  2. Setting   gives that  , and  , and  . Taking   as a parameter, and renaming it   gives this set description  .
  3. This set is empty because the range of   includes only those polynomials with a   term.
This exercise is recommended for all readers.
Problem 7

For the map   given by

 

sketch these inverse image sets:  ,  , and  .

Answer

All inverse images are lines with slope  .

 

This exercise is recommended for all readers.
Problem 8

Each of these transformations of   is nonsingular. Find the inverse function of each.

  1.  
  2.  
  3.  
  4.  
Answer

These are the inverses.

  1.  
  2.  
  3.  
  4.  

For instance, for the second one, the map given in the question sends   and then the inverse above sends  . So this map is actually self-inverse.

Problem 9

Describe the nullspace and rangespace of a transformation given by  .

Answer

For any vector space  , the nullspace

 

is trivial, while the rangespace

 

is all of  , because every vector   is twice some other vector, specifically, it is twice  . (Thus, this transformation is actually an automorphism.)

Problem 10

List all pairs   that are possible for linear maps from   to  .

Answer

Because the rank plus the nullity equals the dimension of the domain (here, five), and the rank is at most three, the possible pairs are:  ,  ,  , and  . Coming up with linear maps that show that each pair is indeed possible is easy.

Problem 11

Does the differentiation map   have an inverse?

Answer

No (unless   is trivial), because the two polynomials   and   have the same derivative; a map must be one-to-one to have an inverse.

This exercise is recommended for all readers.
Problem 12

Find the nullity of the map   given by

 
Answer

The nullspace is this.

 
 

Thus the nullity is  .

Problem 13
  1. Prove that a homomorphism is onto if and only if its rank equals the dimension of its codomain.
  2. Conclude that a homomorphism between vector spaces with the same dimension is one-to-one if and only if it is onto.
Answer
  1. One direction is obvious: if the homomorphism is onto then its range is the codomain and so its rank equals the dimension of its codomain. For the other direction assume that the map's rank equals the dimension of the codomain. Then the map's range is a subspace of the codomain, and has dimension equal to the dimension of the codomain. Therefore, the map's range must equal the codomain, and the map is onto. (The "therefore" is because there is a linearly independent subset of the range that is of size equal to the dimension of the codomain, but any such linearly independent subset of the codomain must be a basis for the codomain, and so the range equals the codomain.)
  2. By Theorem 2.21, a homomorphism is one-to-one if and only if its nullity is zero. Because rank plus nullity equals the dimension of the domain, it follows that a homomorphism is one-to-one if and only if its rank equals the dimension of its domain. But this domain and codomain have the same dimension, so the map is one-to-one if and only if it is onto.
Problem 14

Show that a linear map is nonsingular if and only if it preserves linear independence.

Answer

We are proving that   is nonsingular if and only if for every linearly independent subset   of   the subset   of   is linearly independent.

One half is easy— by Theorem 2.21, if   is singular then its nullspace is nontrivial (contains more than just the zero vector). So, where   is in that nullspace, the singleton set   is independent while its image   is not.

For the other half, assume that   is nonsingular and so by Theorem 2.21 has a trivial nullspace. Then for any  , the relation

 

implies the relation  . Hence, if a subset of   is independent then so is its image in  .

Remark. The statement is that a linear map is nonsingular if and only if it preserves independence for all sets (that is, if a set is independent then its image is also independent). A singular map may well preserve some independent sets. An example is this singular map from   to  .

 

Linear independence is preserved for this set

 

and (in a somewhat more tricky example) also for this set

 

(recall that in a set, repeated elements do not appear twice). However, there are sets whose independence is not preserved under this map;

 

and so not all sets have independence preserved.

Problem 15

Corollary 2.17 says that for there to be an onto homomorphism from a vector space   to a vector space  , it is necessary that the dimension of   be less than or equal to the dimension of  . Prove that this condition is also sufficient; use Theorem 1.9 to show that if the dimension of   is less than or equal to the dimension of  , then there is a homomorphism from   to   that is onto.

Answer

(We use the notation from Theorem 1.9.) Fix a basis   for   and a basis   for  . If the dimension   of   is less than or equal to the dimension   of   then the theorem gives a linear map from   to   determined in this way.

 

We need only to verify that this map is onto.

Any member of   can be written as a linear combination of basis elements  . This vector is the image, under the map described above, of  . Thus the map is onto.

Problem 16

Let   be a homomorphism, but not the zero homomorphism. Prove that if   is a basis for the nullspace and if   is not in the nullspace then   is a basis for the entire domain  .

Answer

By assumption,   is not the zero map and so a vector   exists that is not in the nullspace. Note that   is a basis for  , because it is a size one linearly independent subset of  . Consequently   is onto, as for any   we have   for some scalar  , and so  .

Thus the rank of   is one. Because the nullity is given as  , the dimension of the domain of   (the vector space  ) is  . We can finish by showing   is linearly independent, as it is a size   subset of a dimension   space. Because   is linearly independent we need only show that   is not a linear combination of the other vectors. But   would give   and applying   to both sides would give a contradiction.

This exercise is recommended for all readers.
Problem 17

Recall that the nullspace is a subset of the domain and the rangespace is a subset of the codomain. Are they necessarily distinct? Is there a homomorphism that has a nontrivial intersection of its nullspace and its rangespace?

Answer

Yes. For the transformation of   given by

 

we have this.

 

Remark. We will see more of this in the fifth chapter.

Problem 18

Prove that the image of a span equals the span of the images. That is, where   is linear, prove that if   is a subset of   then   equals  . This generalizes Lemma 2.1 since it shows that if   is any subspace of   then its image   is a subspace of  , because the span of the set   is  .

Answer

This is a simple calculation.

 
This exercise is recommended for all readers.
Problem 19
  1. Prove that for any linear map   and any  , the set   has the form
     
    for   with   (if   is not onto then this set may be empty). Such a set is a coset of   and is denoted  .
  2. Consider the map   given by
     
    for some scalars  ,  ,  , and  . Prove that   is linear.
  3. Conclude from the prior two items that for any linear system of the form
     
    the solution set can be written (the vectors are members of  )
     
    where   is a particular solution of that linear system (if there is no particular solution then the above set is empty).
  4. Show that this map   is linear
     
    for any scalars  , ...,  . Extend the conclusion made in the prior item.
  5. Show that the  -th derivative map is a linear transformation of   for each  . Prove that this map is a linear transformation of that space
     
    for any scalars  , ...,  . Draw a conclusion as above.
Answer
  1. We will show that the two sets are equal   by mutual inclusion. For the   direction, just note that   equals  , and so any member of the first set is a member of the second. For the   direction, consider  . Because   is linear,   implies that  . We can write   as  , and then we have that  , as desired, because  .
  2. This check is routine.
  3. This is immediate.
  4. For the linearity check, briefly, where   are scalars and   have components   and  , we have this.
     
    The appropriate conclusion is that  .
  5. Each power of the derivative is linear because of the rules
     
    from calculus. Thus the given map is a linear transformation of the space because any linear combination of linear maps is also a linear map by Lemma 1.16. The appropriate conclusion is  , where the associated homogeneous differential equation has a constant of  .
Problem 20

Prove that for any transformation   that is rank one, the map given by composing the operator with itself   satisfies   for some real number  .

Answer

Because the rank of   is one, the rangespace of   is a one-dimensional set. Taking   as a basis (for some appropriate  ), we have that for every  , the image   is a multiple of this basis vector— associated with each   there is a scalar   such that  . Apply   to both sides of that equation and take   to be  

 

to get the desired conclusion.

Problem 21

Show that for any space   of dimension  , the dual space

 

is isomorphic to  . It is often denoted  . Conclude that  .

Answer

Fix a basis   for  . We shall prove that this map

 

is an isomorphism from   to  .

To see that   is one-to-one, assume that   and   are members of   such that  . Then

 

and consequently,  , etc. But a homomorphism is determined by its action on a basis, so  , and therefore   is one-to-one.

To see that   is onto, consider

 

for  . This function   from   to  

 

is easily seen to be linear, and to be mapped by   to the given vector in  , so   is onto.

The map   also preserves structure: where

 

we have

 

so  .

Problem 22

Show that any linear map is the sum of maps of rank one.

Answer

Let   be linear and fix a basis   for  . Consider these   maps from   to  

 

for any  . Clearly   is the sum of the  's. We need only check that each   is linear: where   we have  .

Problem 23

Is "is homomorphic to" an equivalence relation? (Hint: the difficulty is to decide on an appropriate meaning for the quoted phrase.)

Answer

Either yes (trivially) or no (nearly trivially).

If   "is homomorphic to"   is taken to mean there is a homomorphism from   into (but not necessarily onto)  , then every space is homomorphic to every other space as a zero map always exists.

If   "is homomorphic to"   is taken to mean there is an onto homomorphism from   to   then the relation is not an equivalence. For instance, there is an onto homomorphism from   to   (projection is one) but no homomorphism from   onto   by Corollary 2.17, so the relation is not reflexive.[1]

Problem 24

Show that the rangespaces and nullspaces of powers of linear maps   form descending

 

and ascending

 

chains. Also show that if   is such that   then all following rangespaces are equal:  . Similarly, if   then  .

Answer

That they form the chains is obvious. For the rest, we show here that   implies that  . Induction then applies.

Assume that  . Then   is the same map, with the same domain, as  . Thus it has the same range:  .

  1. More information on equivalence relations is in the appendix.