3
$\begingroup$

I am reading the textbook Linear Algebra Done Right Chapter 3 section E on Products and Quotients of Vectors Spaces.

It tried to prove the dimension of a quotient space is equal to $\text{dim }V/U = \text{dim }V -\text{dim }U$.

Before that it defines the quotient map $\pi$ as follow:

Suppose $U$ is a subspace of $V$. The quotient map $\pi$ is the linear map $\pi:V \to V/U$ defined by $$\pi(v) = v+U$$ for $v \in V$.

I can understand that the range of $\pi$ is $v+U$ which is $V/U$ according to the definition of $V/U$.

But I don't understand why the null space of $\pi$ is $U$. The book said it is due to the proof like this following:

Suppose $U$ is a subspace of $V$ and $v,w \in V$. Then the following are equivalent: $$v-w \in U$$ $$v+U=w+U$$ $$(v+U) \cap (w+U) \neq \emptyset$$

$\endgroup$
1
  • $\begingroup$ $v+U$ is rather the image of $v$ by $\pi$, instead of the range of $\pi$. If you take $u\in U$, then $\pi(u)=u+U$, but $u+U=\{u+v:\ v\in U\}=U$, since $U$ is a vector subspace. Therefore $\pi(u)=U=0+U$. Likewise if $\pi(v)=U$, then $v+U=U$, which implies that $v=v+0\in U$. $\endgroup$
    – logarithm
    May 5, 2019 at 4:32

5 Answers 5

6
$\begingroup$

The definition of $\ker\pi$ is

\begin{align} \ker \pi &= \{ x: \pi(x) = 0_{V/U} \}\\ &= \{x : x + U = 0_{V/U}\}\\ &= \{x :x + U = U \}\\ &= U \end{align}

$\endgroup$
9
  • $\begingroup$ If I take $u \in U$, then $\pi(u) = u + U$. How can I see that $\pi(u) = 0$ from that? $\endgroup$
    – JOHN
    May 5, 2019 at 4:56
  • $\begingroup$ @JOHN, $u \in U$. It is closed under addition. Hence $u + U = U.$ $\endgroup$
    – IAmNoOne
    May 5, 2019 at 5:48
  • $\begingroup$ so $\pi(u) = U$. But I still cannot see how $\pi(u) = U$ becomes $\pi(u) = 0$ $\endgroup$
    – JOHN
    May 5, 2019 at 5:59
  • $\begingroup$ @JOHN $0_{V/U} = U$ $\endgroup$
    – IAmNoOne
    May 5, 2019 at 6:00
  • $\begingroup$ After a month, I still a bit confused on why the zero vector of quotient space is U. the line $0_{V/U}=U$ seems like given which I don't understand $\endgroup$
    – JOHN
    May 27, 2019 at 0:16
1
$\begingroup$

Here's another way of thinking about it. Suppose that $\nu' \in$ $null\ \pi$. Then, we can see that:

$\pi(\nu') = \nu' + U = 0 + U$

*Because $0 + U$ is the additive identity of $V/U$.

And according to the additional proof that you provided, we see that $\nu'- 0 = \nu' \in U$.

So we see that whenever we map $\nu' \in U$ through $\pi$, we will always get $0 + U = U$ because the affine set of $\{\nu' + u : \nu',u \in U\} = \{0 + (\nu' + u)\}$. Since taking any vector from $U$ will map to the null space of $\pi$, we can see now that $null \ \pi = U$.

$\endgroup$
0
$\begingroup$

The null space of $\pi$ is the set of $v\in V$ such that $\pi(v)=0\in V/U$. Now using what you wrote, convince yourself that the zero element in the vector space $V/U$ is $U$.

$\endgroup$
0
$\begingroup$

Not the answer but intuition to understand why $U_{0} (= U)$ should be the Nullspace:

The $0$ of any vector space is an element that gives you back what you added it to. In the case of a quotient space, $U$ is only such 'element' as:

\begin{align} v + U + U_{0} = v + U \end{align} as proved in Question 15 of Exercise 1.C

Note: Both $U$ and $U_{0}$ are one and the same.

$\endgroup$
0
$\begingroup$

@IAmNoOne has already given a good explanation. I want to add some details:

First, note that the additive identity of $V/U$ is $0+U$, which equals U. In other words, $0_{V/U} = U$.

Second, let $\pi(v) = 0_{V/U}$, solve $v$:

$$\because \pi(v) = v + U, v\in V$$

$$\therefore v+U=0_{V/U}=U$$ which can be written as:$$v+U = 0_V+U$$ From 3.85: $$v-0_V=v\in U$$ $$\therefore \ker \pi\subset U$$ Apparently, $\forall u \in U, \pi(u) = 0_{V/U}$ $$\therefore U\subset \ker \pi$$ $$\therefore \ker \pi = U$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.