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

Key Sequence

• Because Length of Basis Doesn’t Depend on Basis, we defined dimension as the same, shared length of basis in a vector space
• We shown that lists of the right length (i.e. dim that space) that is either spanning or linearly independent must be a basis—“half is good enough” theorems
• we also shown that \(dim(U_1+U_2) = dim(U_1)+dim(U_2) - dim(U_1 \cap U_2)\): dimension of sums

New Definitions

• dimension

Results and Their Proofs

• Length of Basis Doesn’t Depend on Basis
• lists of right length are basis
  • linearly independent list of length dim V are a basis of V
  • spanning list of length of dim V are a basis of V
• dimension of sums

Questions for Jana

• Example 2.41: why is it that \(\dim U \neq 4\)? We only know that \(\dim \mathcal{P}_{3}(\mathbb{R}) = 4\), and \(\dim U \leq 4\). Is it because \(U\) (i.e. basis of \(U\) doesn’t span the polynomial) is strictly a subset of \(\mathcal{P}_{3}(\mathbb{R})\), so there must be some extension needed? because we know that \(U\) isn’t all of \(\mathcal{P}_{3}\).

Interesting Factoids

OMGOMGOMG its Linear Maps time! “One of the key definitions in linear algebra.”

Key Sequence

• We define these new-fangled functions called Linear Maps, which obey \(T(u+v) = Tu+Tv\) and \(T(\lambda v) = \lambda Tv\)
• We show that the set of all linear maps between two vector spaces \(V,W\) is denoted \(\mathcal{L}(V,W)\); and, in fact, by defining addition and scalar multiplication of Linear Maps in the way you’d expect, \(\mathcal{L}(V,W)\) is a vector space!
  • this also means that we can use effectively the \(0v=0\) proof to show that linear maps take \(0\) to \(0\)
• we show that Linear Maps can be defined uniquely by where it takes the basis of a vector space; in fact, there exists a Linear Map to take the basis anywhere you want to go!
• though this doesn’t usually make sense, we call the “composition” operation on Linear Maps their “product” and show that this product is associative, distributive, and has an identity

New Definitions

• Linear Map — additivity (adding “distributes”) and homogeneity (scalar multiplication “factors”)
  • \(\mathcal{L}(V,W)\)
  • any polynomial map from Fn to Fm is a linear map
  • addition and scalar multiplication on \(\mathcal{L}(V,W)\); and, as a bonus, \(\mathcal{L}(V,W)\) a vector space!
  • naturally (almost by the same \(0v=0\) proof), linear maps take \(0\) to \(0\)
• Product of Linear Maps is just composition. These operations are:
  • associative
  • distributive
  • has an identity

Results and Their Proofs

• technically a result: any polynomial map from Fn to Fm is a linear map
• basis of domain of linear maps uniquely determines them

Questions for Jana

• why does the second part of the basis of domain proof make it unique?