Here is midterm 2.
Here is midterm 2.
These questions are due Friday, April 9 at the beginning of lecture.
Here is quiz 7.
This set is due Friday, March 26.
Here is quiz 5 and here is quiz 6.
Problem 1 asks to show that the relation defined as follows is antisymmetric: Given a set the relation is defined on the subsets of by setting for iff where denotes set-theoretic difference of sets.
To show that is antisymmetric, we need to show that whenever are such that and then Suppose satisfy these assumptions. We need to show that they are equal as sets, i.e., that they have the same elements.
By definition, holds iff and holds iff Recall that if are sets, then It follows that iff every element of is also an element of and that iff every element of is also an element of But these two facts together mean precisely that and have the same elements, i.e., that as we needed to show.
Problem 2(a) of quiz 6 asks to consider the relation defined on by setting for iff and to show that is an equivalence relation. This means that is reflexive, symmetric, and transitive. Problem 2(a) of quiz 5 asks to show one of these properties.
To show that is reflexive, we need to show that for any we have that i.e., that But is certainly divisible by 5.
To show that is symmetric, we need to show that for any if it is the case that then it is also the case that Suppose then that This means that i.e., there is an integer such that But then showing that also i.e.,
To show that is transitive, we need to show that if and both and hold, then also holds. But if and then and But then it is certainly the case that Since this proves that, indeed or as we needed to show.
(If one feels the need to be somewhat more strict: That means that there is an integer such that Similarly, means that there is an integer such that But then showing that there is an integer such that namely, we can take )
Problem 2(b) asks to find all natural numbers such that where is as defined for problem 2(a).
That means exactly the same that Since and an integer is a multiple of 5 iff it ends in 5 or 0, we need to find all natural numbers such that ends in or Since and is even for all naturals we actually need to find all natural numbers such that ends in 8. For this, we only need to examine the last digit of the numbers and we find that these last digits form the sequence which is periodic, repeating itself each 4. This means that the numbers we are looking for are precisely i.e., the natural numbers of the form with
This set is due Friday, March 12.
It is common in algebraic settings to build new structures by taking quotients of old ones. This occurs in topology (building quotient spaces), in abstract algebra when building field extensions, or in the homomorphism theorems. Here we explore quotients in vector spaces.
First we briefly consider an example from differential equations.
Let be the space consisting of all continuously differentiable functions and let be the space of all continuous functions Let be the linear transformation
Show that is a real vector space of dimension 1, by showing that iff is a constant. This means, of course, that
Show that by finding a particular solution to the equation One way of doing this is by looking for such a function of the form for some constant Find the form of an arbitrary function such that by noting that if then
More generally, show that is surjective, by finding for any the explicit form of the solutions to the equation It may help you solve this equation if you first multiply both sides by
For another example, denote by the space of all matrices with real entries. Define a map by Show explicitly that has dimension 6 and that is surjective.
Now we abstract certain features of these examples to a general setting:
Suppose is a field and is a linear transformation between two -vector spaces and It is not necessary to assume that or are finite dimensional.
Let and let be any preimage, i.e., Show that the set of all preimages of is precisely
Define a relation in by setting iff Show that is an equivalence relation. Denote by the equivalence class of the vector
Let be the quotient of by i.e., the collection of equivalence classes of the relation We want to give the structure of an -vector space. In order to do this, we define and for all and Show that this is well-defined and satisfies the axioms of an -vector space. What is the usual name we give to the 0 vector of this space?
It is standard to denote by Define two functions and as follows: is given by Also, is given by Show that is well-defined and that both and are linear.
Show that that is a surjection, and that is an isomorphism between and In particular, any surjective image of a vector space by a linear map can be identified with a quotient of