Linear Algebra/Comparing Set Descriptions

Linear Algebra Comparing Set Descriptions General = Particular + Homogeneous Automation This subsection is optional. Later material will not require the work here. Comparing Set Descriptions 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 Equality Sets are equal if and only if they have the same members. mutual inclusionA 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 .More information on set equality is in the appendix. 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 . Exercises Problem 1 Decide if the vector is a member of the set. <li> , <li> , <li> , <li> , <li> , <li> , Problem 2 Produce two descriptions of this set that are different than this one. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. Problem 4 Show that these sets are equal and that both describe the solution set of this system. Problem 5 Decide if the sets are equal. <li> and <li> and <li> and <li> and <li> and /Solutions/ Footnotes Linear Algebra Comparing Set Descriptions General = Particular + Homogeneous Automation