Key Sequence

• \(\mathbb{F}^{n}\) not being a field kinda sucks, so we made an object called a “vector space” which essentially does everything a field does except without necessitating a multiplicative inverse
• Formally, a vector space is closed over addition and have a scalar multiplication. Its addition is commutative, both addition and scalar multiplication is associative, and distributivity holds. There is an additive identity, additive inverse, and multiplicative identity.
• We defined something called \(\mathbb{F}^{S}\), which is the set of functions from a set \(S\) to \(\mathbb{F}\). Turns out, \(\mathbb{F}^{S}\) is a Vector Space Over \(\mathbb{F}\) and we can secretly treat \(\mathbb{F}^{n}\) and \(\mathbb{F}^{\infty}\) as special cases of \(\mathbb{F}^{s}\).
• We established that identity and inverse are unique additively in vector spaces.
• Lastly, we proved some expressions we already know: \(0v=0\), \(-1v=-v\).

New Definitions

• addition and scalar multiplication
• vector space and vectors
• vector space “over” fields
• \(V\) denotes a vector space over \(\mathbb{F}\)
• \(-v\) is defined as the additive inverse of \(v \in V\)

Results and Their Proofs

• \(\mathbb{F}^{\infty}\) is a Vector Space over \(\mathbb{F}\)
• \(\mathbb{F}^{S}\) is a Vector Space Over \(\mathbb{F}\)
• All vector spaces \(\mathbb{F}^{n}\) and \(\mathbb{F}^{\infty}\) are just special cases \(\mathbb{F}^{S}\): you can think about those as a mapping from coordinates \((1,2,3, \dots )\) to their actual values in the “vector”
• additive identity is unique in a vector space
• additive inverse is unique in a vector space
• \(0v=0\), both ways (for zero scalars and vectors)
• \(-1v=-v\)

Questions for Jana

• The way Axler presented the idea of “over” is a tad weird; is it really only scalar multiplication which hinders vector spaces without \(\mathbb{F}\)? In other words, do the sets that form vector spaces, apart from the \(\lambda\) used for scalar multiplication, need anything to do with the \(\mathbb{F}\) they are “over”? The name of the field and what its over do not have to be the same—“vector space \(\mathbb{C}^2\) over \(\{0,1\}\)” is a perfectly valid statement
• If lists have finite length \(n\), then what are the elements of \(\mathbb{F}^{\infty}\) called? “we could think about \(\mathbb{F}^{\infty}\), but we aren’t gonna.”
• Why is \(1v=v\) an axiom, whereas we say that some \(0\) exists? because we know 1 already, and you can follow the behavor of scalar multiplication
• what’s that thing called again in proofs where you just steal the property of a constituent element?: inherits

Interesting Factoids

• The simplest vector space is \(\{0\}\)

Key Sequence

• we defined subspace and how to check for them
• we want to operate on subsets, so we defined the sum of subsets
• we saw that the sum of subspaces are the smallest containing subspace
• and finally, we defined direct sums and how to prove them

New Definitions

• subspace
• sum of subsets
• direct sum

Results and Their Proofs

• checking for subspaces
  • simplified check for subspace
  • sum of subspaces is the smallest subspace with both subspaces
• creating direct sums
  • a sum of subsets is a direct sum IFF there is only one way to write \(0\)
  • a sum of subsets is only a direct sum IFF their intersection is the set containing \(0\)

Questions for Jana

• Does the additive identity have be the same between different subspaces of the same vector space? yes, otherwise the larger vector space has two additive identities.
• Does the addition and multiplication operations in a subspace have to be the same as its constituent vector space? by definition
• Why are direct sums defined on sub-spaces and not sum of subsets? because the union is usually not a subspace so we use sums and keep it in subspaces

3: Show that the set of differential real-valued functions \(f\) on the interval \((-4,4)\) such that \(f’(-1)=3f(2)\) is a subspace of \(\mathbb{R}^{(-4,4)}\)

4: Suppose \(b \in R\). Show that the set of continuous real-valued functions \(f\) on the interval \([0,1]\) such that \(\int_{0}^{1}f=b\) is a subspace of \(\mathbb{R}^{[0,1]}\) IFF \(b=0\)

Additive Identity:

assume \(\int_{0}^{1}f=b\) is a subspace

Key Sequence

• we defined the combination of a list of vectors as a linear combination and defined set of all linear combination of vectors to be called a span
• we defined the idea of a finite-dimensional vector space vis a vi spanning
• we took a god-forsaken divergence into polynomials that will surely not come back and bite us in chapter 4
• we defined linear independence + linear dependence and, from those definition, proved the actual usecase of these concepts which is the Linear Dependence Lemma
• we apply the Linear Dependence Lemma to show that length of linearly-independent list \(\leq\) length of spanning list as well as that finite-dimensional vector spaces make finite subspaces. Both of these proofs work by making linearly independent lists—the former by taking a spanning list and making it smaller and smaller, and the latter by taking a linearly independent list and making it bigger and bigger

New Definitions

• linear combination
• span + “spans”
• finite-dimensional vector space
  • infinite-demensional vector space
  • finite-dimensional subspaces
• polynomial
  • \(\mathcal{P}(\mathbb{F})\)
  • \(\mathcal{P}_{m}(\mathbb{F})\)
  • degree of a polynomial \(\deg p\)
• linear independence and linear dependence
• Linear Dependence Lemma

Results and Their Proofs

• span is the smallest subspace containing all vectors in the list
• \(\mathcal{P}(\mathbb{F})\) is a vector space over \(\mathbb{F}\)
• the world famous Linear Dependence Lemma and its fun issue
• length of linearly-independent list \(\leq\) length of spanning list
• subspaces of inite-dimensional vector spaces is finite dimensional

Questions for Jana

• obviously polynomials are non-linear structures; under what conditions make them nice to work with in linear algebra?
• what is the “obvious way” to change Linear Dependence Lemma’s part \(b\) to make \(v_1=0\) work?
• for the finite-dimensional subspaces proof, though we know that the process terminates, how do we know that it terminates at a spanning list of \(U\) and not just a linearly independent list in \(U\)?
• direct sum and linear independence related; how exactly?

Interesting Factoids

I just ate an entire Chinese new-year worth of food while typing this up. That’s worth something right

Key Sequence

• we defined basis of a vector space—a linearly independent spanning list of that vector space—and shown that to be a basis one has to be able to write a write an unique spanning list
• we show that you can chop a spanning list of a space down to a basis or build a linearly independent list up to a basis
• because of this, you can make a spanning list of finite-dimensional vector spaces and chop it down to a basis: so every finite-dimensional vector space has a basis
• lastly, we can use the fact that you can grow list to basis to show that every subspace of \(V\) is a part of a direct sum equaling to \(V\)

New Definitions

basis and criteria for basis