Linear Algebra/Comparing Set Descriptions

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This subsection is optional. Later material will not require the work here.

Comparing Set DescriptionsEdit

A set can be described in many different ways. Here are two different descriptions of a single set:

 

For instance, this set contains

 

(take   and  ) but does not contain

 

(the first component gives   but that clashes with the third component, similarly the first component gives   but the third component gives something different). Here is a third description of the same set:

 

We need to decide when two descriptions are describing the same set. More pragmatically stated, how can a person tell when an answer to a homework question describes the same set as the one described in the back of the book?

Set EqualityEdit

Sets are equal if and only if they have the same members. A common way to show that two sets,   and  , are equal is to show mutual inclusion: any member of   is also in  , and any member of   is also in  .[1]

Example 4.1

To show that

 

equals

 

we show first that   and then that  .

For the first half we must check that any vector from   is also in  . We first consider two examples to use them as models for the general argument. If we make up a member of   by trying   and  , then to show that it is in   we need   and   such that

 

that is, this relation holds between   and  .

 

Similarly, if we try   and  , then to show that the resulting member of   is in   we need   and   such that

 

that is, this holds.

 

In the general case, to show that any vector from   is a member of   we must show that for any   and   there are appropriate   and  . We follow the pattern of the examples; fix

 

and look for   and   such that

 

that is, this is true.

 

Applying Gauss' method

 

gives   and  . This shows that for any choice of   and   there are appropriate   and  . We conclude any member of   is a member of   because it can be rewritten in this way:

 

For the other inclusion,  , we want to do the opposite. We want to show that for any choice of   and   there are appropriate   and  . So fix   and   and solve for   and  :

 

shows that   and  . Thus any vector from  

 

is also of the right form for  

 
Example 4.2

Of course, sometimes sets are not equal. The method of the prior example will help us see the relationship between the two sets. These

 

are not equal sets. While   is a subset of  , it is a proper subset of   because   is not a subset of  .

To see that, observe first that given a vector from   we can express it in the form for  — if we fix   and  , we can solve for appropriate  ,  , and  :

 

shows that that any

 

can be expressed as a member of   with  ,  , and  :

 

Thus  .

But, for the other direction, the reduction resulting from fixing  ,  , and   and looking for   and  

 

shows that the only vectors

 

representable in the form

 

are those where  . For instance,

 

is in   but not in  .

ExercisesEdit

Problem 1

Decide if the vector is a member of the set.

  1.  ,  
  2.  ,  
  3.  ,  
  4.  ,  
  5.  ,  
  6.  ,  
Problem 2

Produce two descriptions of this set that are different than this one.

 
This exercise is recommended for all readers.
Problem 3

Show that the three descriptions given at the start of this subsection all describe the same set.

This exercise is recommended for all readers.
Problem 4

Show that these sets are equal

 

and that both describe the solution set of this system.

 
This exercise is recommended for all readers.
Problem 5

Decide if the sets are equal.

  1.   and  
  2.   and  
  3.   and  
  4.   and  
  5.   and  

Solutions

FootnotesEdit

  1. More information on set equality is in the appendix.
Linear Algebra
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