Lecture 2 - Products and Quotients of Vector Spaces

Questions to ask Jeffrey Liese:

I'm confused on the idea of parallel. Aren't all affine subsets parallel, and vice versa?

(cont.) Products of Vector Spaces

Note from the example last time via:

!Year3/Spring2024/MATH406-LinearTree/LectureNotes/Lecture 1 - Intro to The Course#^0906d0

That we noticed the following thing:

Proposition

$dim (V_{1} \times \dots \times V_{n}) = \sum_{i = 1}^{n} dim (V_{i})$

Which is pretty clear from the example.

Connecting to Sums

If you recall from last lecture, we asked, given $U_{1}, . . ., U_{m}$ are subspaces of $V$ , how do we construct an element of their sum, $U_{1} + \dots + U_{m}$ ? The way we did that is we chose one vector from each subspace, and add them together:

u_{1} + \dots + u_{m}, u_{i} \in U_{i}

Choosing these vectors $u_{i}$ is like choosing a vector from the product $U_{1} \times \dots \times U_{m}$ .

Hence, we define the following:

Γ

Define $Γ : U_{1} \times \dots \times U_{m} \to U_{1} + \dots + U_{m}$ by:

Γ (u_{1}, u_{2}, . . ., u_{m}) = u_{1} + \dots + u_{m} \in V

then:

$Γ \in L (V)$
$Γ$ is surjective.

This is the formalism of the idea that we were looking at.

Notice that showing $Γ$ is linear is pretty trivial, like we did in Chapter 3 - Linear Maps. Showing surjectivity is pretty easy, as we did in Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^549aea.

One interesting thing is that we know $U_{1} + \dots + U_{m}$ is direct when the only mapping $\vec{0}$ from the product space to the zero vector of their sum, which is injective (since then $null (Γ) = {\vec{0}}$ ).

Γ

is injective when direct sum

$Γ$ is injective iff $U_{1} + \dots + U_{m}$ is direct.

Notice that that's insane! A direct sum implies that there's an isomorphism, which is $Γ$ ! But wait! ThErE's MoRe! If we apply FTOLM, we get this idea:

Sums of the dimension carry through

$\begin{aligned} dim (U_{1} \times \dots \times U_{m}) & = dim (range (Γ)) + dim (null (Γ)) \\ = dim (U_{1} + \dots + U_{m}) + dim (null (Γ)) \\ = dim (U_{1} + \dots + U_{m}) & iff Γ is injective \\ = \sum_{i = 1}^{n} dim (U_{i}) \end{aligned}$

Therefore:

Proposition

$U_{1} + \dots + U_{m}$ is a direct sum iff the dimension of the sum of these subspaces is the sum of their dimensions.

Quotients of Vector Spaces

If we had two vector spaces, we got a larger vector space. Now, with quotients we get a smaller vector space.

Let's consider $R^{3}$ and a plane in $R^{3}$ :

Note here that $P$ is not a subspace. But we can make a subspace if we add $v$ along with any vector in the plane $p \in P$ . Each point on $P$ can be expressed as $v + u$ where $u \in P$ , we'll denote as $U$ .

We also could slide $P$ down to the origin to make an actual subspace:

This gives rise to the following definition:

affine subset

Suppose $v \in V$ and $U$ is a subspace of $V$ . We define:

v + U = {v + u : u \in U}

is called an affine subset of $V$ (ie: a "shifted" subspace). In general, we say that $v + U$ is parallel to $U$ (always).

Notice also that the vector $v$ is very arbitrary. We could've chosen any vector $v$ and shift accordingly. Usually $v$ is called a "representative" of the affine subset. But they are not unique, while if we had $v_{1}, v_{2}$ as different representatives, then we still get the plane $P$ :

v_{1} + U = v_{2} + U

For example, considering $R^{3}$ , if ${(x, y, z) : a x + b y + c z = 0} = U$ as long as one of $a, b, c$ are no-zero. This makes a plane through the origin, which is a subspace. Looking at all affine subsets of $R^{3}$ parallel to $U$ are all planes that have the same normal vector $\vec{n} = ⟨ a, b, c ⟩$ . Namely:

v + U = {(x, y, z) : a x + b y + c z = d}

for any $d \in R$ .

For another example, looking at the set of solutions to $y^{″} + y = f (x)$ where $f (x) \neq 0$ . This is parallel to the subspace $U$ of solutions to $y^{″} + y = 0$ . This is because of the following. Suppose $y_{1}, y_{2}$ are solutions to $y^{″} + y = L (y)$ , so $L y = f$ have solutions $y_{1}, y_{2}$ . We know that $L y_{1} = f = L y_{2}$ so then $L y_{1} - L y_{2} = L (y_{1} - y_{2}) = 0$ , so then $y_{1} - y_{2} \in U$ since it's a solution to the homogeneous DE.

Notice then that:

y_{1} = (y_{1} - y_{2}) + y_{2} = y_{2} + \underset{\in U}{\underset{⏟}{(y_{1} - y_{2})}}

So given a specific solution $y_{p}$ to $L y = f$ , then any other solution $y$ could be written as:

y = (y - y_{p}) + y_{p}

So the set of solution to $L y = f$ is the affine subset:

y_{p} + U

Constructing the Quotient Space

Recall having all the parallel planes. We map each vector $v$ to it's related plane $v + U$ . Hence, each "plane" is really just a vector in disguise, and can be relabeled as such.

V / U

Given a subspace $U$ of vector space $V$ , we define the quotient space as:

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

this is the set of VECTORS.

We need two operations here.

Lemma

$V / U$ is a vector when vector addition and scalar multiplication are defined "representativewise":

\begin{aligned} (v + U) + (w + U) & = (v + w) + U \\ λ (v + U) & = (λ v) + U \end{aligned}