Linear Algebra/Comparing Set Descriptions/Solutions

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Solutions[edit]

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 .