## 403/503 – Homework 4

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 $V$ be the space $C^1({\mathbb R})$ consisting of all continuously differentiable functions $f:{\mathbb R}\to{\mathbb R}$ and let $W$ be the space $C({\mathbb R})$ of all continuous functions $g:{\mathbb R}\to{\mathbb R}.$ Let $T:V\to W$ be the linear transformation $Tf=f'-2f.$

Show that ${\rm null}(T)$ is a real vector space of dimension 1, by showing that $Tf=0$ iff $e^{-2x}f$ is a constant. This means, of course, that ${\rm null}(T)={\rm span}(e^{2x})=\{\alpha e^{2x}\mid \alpha\in{\mathbb R}\}.$

Show that $e^x\in{\rm ran}(T)$ by finding a particular solution $f_0$ to the equation $Tf=e^x.$ One way of doing this is by looking for such a function $f_0$ of the form $ae^x$ for some constant $a.$ Find the form of an arbitrary function $f$ such that $Tf=e^x$ by noting that if $Tf=Tf_0$ then $T(f-f_0)=0.$

More generally, show that $T$ is surjective, by finding for any $g\in C({\mathbb R})$ the explicit form of the solutions $f$ to the equation $Tf=g.$ It may help you solve this equation if you first multiply both sides by $e^{-2x}.$

For another example, denote by ${\mathbb R}^{3\times 3}$ the space of all $3\times 3$ matrices with real entries. Define a map $T:{\mathbb R}^{3\times 3}\to{\mathbb R}^3$ by $TA=A\begin{pmatrix}1\\1\\1\end{pmatrix}.$ Show explicitly that ${\rm null}(T)$ has dimension 6 and that $T$ is surjective.

Now we abstract certain features of these examples to a general setting:

Suppose ${\mathbb F}$ is a field and $T:V\to W$ is a linear transformation between two ${\mathbb F}$-vector spaces $V$ and $W.$ It is not necessary to assume that $V$ or $W$ are finite dimensional.

Let $w\in{\rm ran}(T),$ and let $v_0\in V$ be any preimage, i.e., $Tv_0=w.$ Show that the set $T^{-1}\{w\}$ of all preimages of $w$ is precisely $v_0+{\rm null}(T)=\{v_0+v\mid Tv=0\}.$

Define a relation $\sim$ in $V$ by setting $v_1\sim v_2$ iff $Tv_1=Tv_2.$ Show that $\sim$ is an equivalence relation. Denote by ${}[v]$ the equivalence class of the vector $v\in V.$

Let $U=V/\sim$ be the quotient of $V$ by $\sim,$ i.e., the collection of equivalence classes of the relation $\sim.$ We want to give $U$ the structure of an ${\mathbb F}$-vector space. In order to do this, we define ${}[v_1]+[v_2]=[v_1+v_2]$ and $a[v]=[av]$ for all $v_1,v_2,v\in V$ and $a\in{\mathbb F}.$ Show that this is well-defined and satisfies the axioms of an ${\mathbb F}$-vector space. What is the usual name we give to the 0 vector of this space?

It is standard to denote $V/\sim$ by $V/{\rm null}(T).$ Define two functions $\pi$ and $\phi$ as follows: $\pi:V\to V/{\rm null}(T)$ is given by $\pi(v)=[v].$ Also, $\phi:V/{\rm null}(T)\to W$ is given by $\phi([v])=Tv.$ Show that $\phi$ is well-defined and that both $\pi$ and $\phi$ are linear.

Show that $T=\phi\pi,$ that $\pi$ is a surjection, and that $\phi$ is an isomorphism between $V/{\rm null}(T)$ and ${\rm ran}(T).$ In particular, any surjective image of a vector space $V$ by a linear map can be identified with a quotient of $V.$

### 6 Responses to 403/503 – Homework 4

1. Nick Davidson says:

In the second example, where T maps the real 3×3 matrices to R^3, you ask us to show explicitly that the dimension of null(T) is 6. By explicitly, do you mean that you actually want us to find a basis for null(T), or would it suffice to reference an appropriate theorem?

Thanks,
Nick

2. I would prefer an actual basis, I’ve been seen some difficulties carrying out explicit computations in the previous homework sets.

3. Amy Griffin says:

When we define an equivalence relation, what is required to show that it is an equivalence relation? The book doesn’t discuss them, so from google I found three properties: a~a, a~b then b~a, and finally a~b and b~c then a~c. Is this all that’s required to show it is one?

Thanks, Amy

• Yes, that’s all that is needed. To be precise: You need to show that $v\sim v$ holds for all vectors $v\in V$. (This is called the reflexive property.) Similarly, you need to show that, for any vectors $v,w,u\in V$, if it is the case that $v\sim w$, then also $w\sim v$ (symmetry), and that if it is the case that $v\sim w$ and $w\sim u$, then also $v\sim u$ (transitivity).

4. Summer says:

I had a question referring to what I have called Show #7: “Show that this is well-defined and satisfies the axioms of an ${\mathbb F}$-vector space.”
I’m not sure what you mean by “show that this is well-defined”? What exactly do I need to show? I tried to look this up and I found things that said well defined meant the mapping was one-to-one, is this what we are talking about here?
Thanks,
Summer

• Hi Summer,
We are defining a map on a quotient. Whenever we do this, we typically run into the following problem:

I am defining ${}[v_1]+[v_2]=[v_1+v_2].$ However, if $v_1\sim v_3$ and $v_2\sim v_4,$ then ${}[v_1]=[v_3]$ and ${}[v_2]=[v_4],$ right? This is potentially a problem, because we are defining ${}[v_3]+[v_4]=[v_3+v_4].$ But we had already said that ${}[v_3]+[v_4]=[v_1]+[v_2]=[v_1+v_2].$ So, there is the risk that the definition of sum we gave makes no sense: Which one is it: ${}[v_1+v_2]$ or ${}[v_3+v_4]$?

The answer ought to be that both expressions are the same, i.e., that no matter what elements of ${}[v_1]$ and ${}[v_2]$ are chosen, when we add them, we always land in the same equivalence class. This is usually expressed by saying that the definition of sum that we gave is independent of the specific representatives $v_1$ and $v_2$ that we choose in order to compute $v_1+v_2,$ from which we then obtain ${}[v_1+v_2].$

Then you have to argue that, similarly, if $v_1\sim v_2$ then $av_1\sim av_2$ for any $a\in{\mathbb F},$ so the definition of scalar multiplication we are giving is also independent of the specific representatives of the equivalence classes that we choose in order to compute it.

(The point is that when somebody hands you two equivalence classes ${}[v_1]$ and ${}[v_2]$ and asks you to add them, you are not given $v_1$ and $v_2$, only their classes. You have to pick an element of the first class, and one of the second, and then you add these 2 elements, and form the class of the result that you obtain, and that’s your answer. Of course, you want that your answer does not change if you pick different elements of the first and second classes to begin with.)