Lecture 3 - Quotient Spaces

Recall we define:

V / U = {v + U : v \in V}

where $U$ is a subspace of $V$ . This is the quotient space.

If we have two affine subsets $v + U$ and $w + U$ , then:

(v + U) + (w + U) = (v + w) + U

and scalar multiplication over $α \in F$ is similarly:

α (v + U) = (α v) + U

We claimed that these operations allowed $V / U$ to be a vector space, which is verifiable, albeit time consuming. However, we know that it's possible that $v + U = w + U$ without having $v = w$ . Namely, see Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^a9c447's proof for this problem.

So what if we have $v + U = w + U$ ? If that's the case, then $v \in v + U$ and similarly $w \in w + U$ . If they're equal, then clearly $v \in w + U$ and $w \in v + U$ . As such, then $v = w + u$ where $u \in U$ is some vector. Thus:

v - w = u \in U

Two affine subsets parallel to

U

are equal or disjoint

Suppose $U$ is a subspace of $V$ and $v, w \in V$ . Then the following are equivalent:

$v - w \in U$
$v + U = w + U$
$(v + U) \cap (w + U) \neq \emptyset$

This tells us that affine subsets are either equal, or disjoint. Think of the plane example. If you have two parallel planes, they either don't intersect, or are equal to each other.

The proof of this is pretty easy as shown in the book:

Proof
First suppose (1) holds, so $v - w \in U$ . If any $u \in U$ then:

To show our operations are well defined, we did the proof in the book via Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^6ab226. We want to make sure that $(v + U) + (w + U) = (\hat{v} + U) + (\hat{w} + U)$ , and similar for the scalar multiplication case.

We defined the quotient map:

quotient map, $π$

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

Recall that the zero vector for $v + U$ is $U$ itself. Namely, if you think of the line of planes in $R^{3}$ that are parallel to $U$ , the only vector that would have:

0 (v + U) = (0 v) + U = 0 + U = U = \vec{0}

As such, then $V / {\vec{0}}$ is isomorphic to $V$ since their dimensions are the same via:

Dimension of a quotient space

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$ . Then:
$dim (V / U) = dim (V) - dim (U)$

Now suppose that $T \in L (V, W)$ . Suppose $T v_{1} = T v_{2}$ . Then:

T (v_{1} - v_{2}) = 0 \Rightarrow v_{1} - v_{2} \in null (T)

So then, treating $null (T)$ like a $U$ :

v_{1} + null (T) = v_{2} + null (T)

So then any vector in an affine subset, parallel to $null (T)$ , has the same image under $T$ .

\tilde{T}

Suppose $T \in L (V, W)$ . Define $\tilde{T} : V / (null (T)) \to W$ by:
$\tilde{T} (v + null (T)) = T v$

This map sends pieces to outputs, rather than vectors in $V$ to corresponding outputs. Think of a plane through the origin $U$ . Say $U = null (T)$ for some transformation $T$ . All the other planes $v + null (T)$ don't get mapped to zero, and only the ones that do are $0 + null (T)$ . Essentially, the collection of all things mapped to $0$ , ie: $null (T)$ , is isolated from the other $v$ 's, and thus forces $\tilde{T}$ to be injective (as any $v$ has to get mapped to it's counterpart $T v$ , as there's no other parts as we isolated the nullspace).

Duality

We defined:

Definition

A linear functional on $V$ is an element of $L (V, F)$ .

Notice that $F$ is a vector space, like $F^{1}$ .