Linear Algebra/Comparing Set Descriptions/Solutions

SolutionsEdit

Problem 1

Decide if the vector is a member of the set.

  1.  ,  
  2.  ,  
  3.  ,  
  4.  ,  
  5.  ,  
  6.  ,  
Answer
  1. No.
  2. Yes.
  3. No.
  4. Yes.
  5. Yes; use Gauss' method to get   and  .
  6. No; use Gauss' method to conclude that there is no solution.
Problem 2

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

 
Answer

One easy thing to do is to double and triple the vector:

 
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.

Answer

Instead of showing all three equalities, we can show that the first equals the second, and that the second equals the third. Both equalities are easy, using the methods of this subsection.

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.

 
Answer

That system reduces like this:

 

showing that  ,   and  .

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  
Answer

For each item, we call the first set   and the other  .

  1. They are equal. To see that  , we must show that any element of the first set is in the second, that is, for any vector of the form
     
    there is an appropriate   such that
     
    Restated, given   we must find   so that this holds.
     
    That system reduces to
     
    That is,
     
    and so any vector in the form for   can be stated in the form needed for inclusion in  . For  , we look for   so that these equations hold.
     
    Rewrite that as
     
    and so
     
  2. These two are equal. To show that  , we check that for any   we can find an appropriate   so that these hold.
     
    Use Gauss' method
     
    to conclude that
     
    and so  . For  , solve
     
    with Gaussian reduction
     
    to get
     
    and so any member of   can be expressed in the form needed for  .
  3. These sets are equal. To prove that  , we must be able to solve
     
    for   and   in terms of  . Apply Gaussian reduction
     
    to conclude that any pair   where   will do. For instance,
     
    or
     
    Thus  . For  , we solve
     
    with Gauss' method
     
    to deduce that any vector in   is also in  .
     
  4. Neither set is a subset of the other. For   to hold we must be able to solve
     
    for   and   in terms of   and  . Gauss' method
     
    shows that we can only find an appropriate pair   when  . That is,
     
    has no expression of the form
     
    Having shown that   is not a subset of  , we know   so, strictly speaking, we need not go further. But we shall also show that   is not a subset of  . For   to hold, we must be able to solve
     
    for   and  . Apply row reduction
     
    to deduce that the only vectors from   that are also in   are of the form
     
    For instance,
     
    is in   but not in  .
  5. These sets are equal. First we change the parameters:
     
    Now, to show that  , we solve
     
    with Gauss' method
     
    to get that
     
    and so  . The proof that   involves solving
     
    with Gaussian reduction
     
    to conclude
     
    and so any vector in   is also in  .