Notes:Advanced Linear Algebra - Roman/Chapter 1

Overview

The chapter starts on page 35.

Notes

I am not doing this chapter in order, if a title referring to page [ilmath]n[/ilmath] occurs above some content, that content came after or on page [ilmath]n[/ilmath]. Content starts close to the nearest page marking above it. Just because something occurs under page [ilmath]n[/ilmath] does not mean it is on page [ilmath]n[/ilmath].

Page 39

Definition: Sum

Let [ilmath]S[/ilmath], [ilmath]T[/ilmath] be subspaces of a vector space [ilmath]V[/ilmath]. Then the sum, [ilmath]S+T[/ilmath] is defined by:

• [ilmath]S+T:=\{u+v\ \vert\ u\in S\wedge v\in T\}[/ilmath]

Let [ilmath]\{S_\alpha\}_{\alpha\in I} [/ilmath] be an arbitrary collection of subspaces of a vector space [ilmath]V[/ilmath]. Then the sum of the collection is:

• The set of all finite sums of vectors from the union [math]\bigcup_{\alpha\in I} S_\alpha[/math]
  • [math]\sum_{\alpha\in I}S_\alpha:=\left\{s_1+\cdots+s_n\left\vert\forall i\in\{1,\ldots,n\}\subset\mathbb{N}\left[s_i\in\bigcup_{\alpha\in I}S_\alpha\right]\right.\right\}[/math]

It is "easy to see" that, under set inclusion as the partial ordering:

• [ilmath]S+T=\text{lub}(\{S,T\})[/ilmath] and more generally:
• [math]\sum_{\alpha\in I}S_\alpha=\text{lub}(\{S_\alpha\ \vert\ \alpha\in I\})[/math]

We're now at the bottom of page 39 and there's some notes about when a poset is a lattice and a complete lattice

Theorem 4.3: The set of all subspaces of a vector space is a complete lattice under set inclusion

Let [ilmath]\mathcal{S}(V)[/ilmath] denote the set of all vector subspaces of a vector space [ilmath](V,\mathcal{K})[/ilmath]. Then [ilmath]\mathcal{S}(V)[/ilmath] is a complete lattice under the partial order relation of set inclusion, with smallest element [ilmath]\{0\} [/ilmath] (the zero vector) and largest element [ilmath]V[/ilmath] itself. The meet and join are defined as follows:

• Meet: [math]\text{glb}(\{S_\alpha\vert\alpha\in I\})=\bigcap_{\alpha\in I}S_\alpha[/math]
• Join: [math]\text{lub}(\{S_\alpha\vert\alpha\in I\})=\sum_{\alpha\in I}S_\alpha[/math]

Page 40: Direct sums

Definition:External Direct Sums

Let [ilmath]V_1,\ldots,V_n[/ilmath] be vector spaces over a field, [ilmath]\mathcal{K} [/ilmath], the external direct sum of [ilmath]V_1,\ldots,V_n[/ilmath], [ilmath]V[/ilmath], is denoted by:

• [ilmath]V=V_1\boxplus V_2\boxplus\cdots\boxplus V_{n-1}\boxplus V_n[/ilmath]
  • [ilmath]V=\{(v_1,\ldots,v_n)\ \vert\ \forall i\in\{1,\ldots,n\}[v_i\in V_i]\}[/ilmath]
  • The vector operations on [ilmath]V[/ilmath] are component-wise addition, and multiplication by scalar multiplying each component by that scalar.

For example [ilmath]\mathcal{K}^n:=\underbrace{\mathcal{K}\boxplus\mathcal{K}\boxplus\cdots\boxplus\mathcal{K}\boxplus\mathcal{K} }_{n\text{ copies of }\mathcal{K} }[/ilmath]

Definition:Direct product

Let [ilmath]\mathcal{F}:=\{V_\alpha\}_{\alpha\in I}[/ilmath] be an arbitrary collection of vector spaces over a field, [ilmath]\mathcal{K} [/ilmath]. The direct product of [ilmath]\mathcal{F} [/ilmath] is defined as:

• [math]\prod_{\alpha\in I}V_\alpha:=\left.\left\{f:I\rightarrow\bigcup_{\alpha\in I}V_\alpha\right\vert f(\alpha)\in V_\alpha\right\}[/math] - note that the functions, [ilmath]f[/ilmath] are just generalisations of tuples to those of arbitrary (perhaps uncountable) length.

He then adds: "thought of as a vector subspace of the vector space of all functions [ilmath]:I\rightarrow\bigcup_{\alpha\in I}V_\alpha [/ilmath]" - Caution:which I do not get. I don't think the union of some vector spaces is actually a vector space! However if we take this to have component-wise addition and multiplication by scalar to be multiplication of each component by that scalar I am happy that this is a vector space. It's a class product (category theory)

Definition: External direct sum

First we need to define "support".

• Let [ilmath]\mathcal{F}:=\{V_\alpha\vert\alpha\in I\}[/ilmath] be an arbitrary family of vector spaces over a field [ilmath]\mathcal{K} [/ilmath].
  • We define the support of a function, [ilmath]f:I\rightarrow\bigcup_{\alpha\in I}V_\alpha[/ilmath] to be the set:
    • [ilmath]\text{Supp}(f):=\{\alpha\in I\ \vert\ f(\alpha)\ne 0_{V_\alpha}\}[/ilmath]
  • A function, [ilmath]f[/ilmath], has finite support if [ilmath]f(\alpha)=0[/ilmath] for all but a finite number of [ilmath]\alpha\in I[/ilmath], that is if the cardinality of the support is finite (or [ilmath]\in\mathbb{N} [/ilmath]).

The external direct sum of the family [ilmath]\mathcal{F} [/ilmath] is the vector space:

• [math]\bigoplus^\text{ext}_{\alpha\in I}V_i=\left.\left\{f:I\rightarrow\bigcup_{\alpha\in I}V_\alpha\right\vert f(\alpha)\in V_\alpha\wedge \vert\text{Supp}(f)\vert\in\mathbb{N}\right\}[/math]

Suppose that [ilmath]V_\alpha=V[/ilmath] for all [ilmath]\alpha\in I[/ilmath]. Then, if we denote by [ilmath]V^I[/ilmath] the set of all functions from [ilmath]I[/ilmath] to [ilmath]V[/ilmath], and [ilmath](V^I)_0[/ilmath] all those functions with finite support as defined above, then:

• [math]\prod_{\alpha\in I}V=V^I[/math] and [math]\bigoplus^\text{ext}_{\alpha\in I}V=(V^I)_0[/math] and are the same for a finite family (when [ilmath]I[/ilmath] is a finite set).
  • Caution:I am not convinced of this, however considering functions as tuples and doing a bit of the footwork to show it is true seems easy and believable. I should investigate this later

Definition: Internal direct sum

A vector space [ilmath](V,\mathcal{K})[/ilmath] is the (internal) direct sum of a family, [ilmath]\mathcal{F}:=\{S_\alpha\}_{\alpha\in I}[/ilmath] of subspaces of [ilmath]V[/ilmath], written:

• [math]V=\bigoplus\mathcal{F}[/math] or [math]V=\bigoplus_{\alpha\in I}V_\alpha[/math] if the following hold:
  1. Join of the family: that [ilmath]V[/ilmath] is the join of the family [ilmath]\mathcal{F} [/ilmath], or [ilmath]V=\sum_{\alpha\in I}S_\alpha[/ilmath]
  2. Independence of the family: that for each [ilmath]\alpha\in I[/ilmath] we have:
     • [math]S_\alpha\cap\left(\sum_{\beta\in(I-\{\alpha\})}S_\beta\right)=\{0\}[/math]

In this case each [ilmath]S_\alpha[/ilmath] is called a direct summand of [ilmath]V[/ilmath]. If [ilmath]\mathcal{F} [/ilmath] is a finite family then the direct sum is often written:

• [ilmath]V=S_1\oplus\cdots\oplus S_n[/ilmath]

and lastly, if [ilmath]V=S\oplus T[/ilmath] then [ilmath]T[/ilmath] is called a complement of [ilmath]S[/ilmath] in [ilmath]V[/ilmath] (and probably: [ilmath]S[/ilmath] is a complement of [ilmath]T[/ilmath] in [ilmath]V[/ilmath] I'd imagine!)

• Caution:Caveat: Notice that condition 2 is stronger than saying that each [ilmath]S_\alpha[/ilmath] are pairwise disjoint (not counting the identity, so the intersection is only the identity) this isn't hard to see, if you take, for example:
  • The 3 axies in [ilmath]\mathbb{R}^3[/ilmath] as subspaces and another line through the origin that isn't one of the axies. These are all pairwise disjoint (except for the 0 vector), but the line, cap the sum of the axies is the line! Not the set containing only the identity. Caveat means "beware" so... beware!

Terminology note

If [ilmath]S[/ilmath] and [ilmath]T[/ilmath] are subspaces of [ilmath]V[/ilmath] then we may always say the sum [ilmath]S+T[/ilmath] exists, but to say the direct sum of [ilmath]S\oplus T[/ilmath] exists is to imply that [ilmath]S\cap T=\{0\}[/ilmath]. So the direct sum does not always exist!

Later we will find out that internal and external sums are essentially equivalent (isomorphic) and thus direct sum will often just be used without qualification.

Alec's note: Direct seems to mean "unique". I wish someone would define it! Actually theorem 1.5 lists equivalent statements to condition 2 (for a family of subspaces of [ilmath]V[/ilmath]) and that shows that it is equivalent to every non-zero vector in [ilmath]V[/ilmath] has a unique (up to order of the terms) expression of non-zero vectors from distinct subspaces in [ilmath]\mathcal{F} [/ilmath]. Direct seems to mean like "the sum of the dimensions of the subspaces is the dimension of the result" - kind of - for infinite cases this phrasing obviously doens't work.

Theorem 1.4: Any subspace of a vector space has a complement

Let [ilmath]S[/ilmath] be a vector subspace of [ilmath]V[/ilmath], then there exists a complement of [ilmath]S[/ilmath] in [ilmath]V[/ilmath] (so there is a [ilmath]T[/ilmath] such that [ilmath]V=S\oplus T[/ilmath])

• Caveat: the complement need not be unique! For example in [ilmath]\mathbb{R}^2[/ilmath] the [ilmath]x[/ilmath]-axis and any other line through the origin that isn't the [ilmath]x[/ilmath]-axis is a complement to the [ilmath]x[/ilmath]-axis!