Pre-conditions and post-conditions for specify logical formula; this is the basis of verification systems.

Key sequence

1. In this chapter, we defined complex numbers, their definition, their closeness under addition and multiplication, and their properties
2. These properties make them a field: namely, they have, associativity, commutativity, identities, inverses, and distribution.
3. notably, they are different from a group by having 1) two operations 2) additionally, commutativity and distributivity. We then defined \(\mathbb{F}^n\), defined addition, additive inverse, and zero.
4. These combined (with some algebra) shows that \(\mathbb{F}^n\) under addition is a commutative group.
5. Lastly, we show that there is this magical thing called scalar multiplication in \(\mathbb{F}^n\) and that its associative, distributive, and has an identity. Technically scalar multiplication in \(\mathbb{F}^n\) commutes too but extremely wonkily so we don’t really think about it.

New Definitions

• complex number
  • addition and multiplication of complex numbers
  • subtraction and division of complex numbers
• field: \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\)
  • power
• list
• \(\mathbb{F}^n\): F^n
  • coordinate
  • addition in \(\mathbb{F}^n\)
  • additive inverse of \(\mathbb{F}^n\)
  • \(0\): zero
  • scalar multiplication in \(\mathbb{F}^n\)

Results and Their Proofs

• properties of complex arithmetic
  • commutativity
  • associativity
  • identities
  • additive inverse
  • multiplicative inverse
  • distributive property
• properties of \(\mathbb{F}^n\)
  • addition in \(\mathbb{F}^n\) is associative
  • addition in \(\mathbb{F}^n\) is commutative
  • addition in \(\mathbb{F}^n\) has an identity (zero)
  • addition in \(\mathbb{F}^n\) has an inverse
  • scalar multiplication in \(\mathbb{F}^n\) is associative
  • scalar multiplication in \(\mathbb{F}^n\) has an identity (one)
  • scalar multiplication in \(\mathbb{F}^n\) is distributive

Question for Jana

• No demonstration in exercises or book that scalar multiplication is commutative, why?

Interesting Factoids

• You can take a field, look at an operation, and take that (minus the other op’s identity), and call it a group
• (groups (vector spaces (fields )))

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