---
title: "Axler 2.A"
source: https://www.jemoka.com/posts/kbhaxler_2_a/
---

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&mdash;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 + &ldquo;spans&rdquo; 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 &ldquo;obvious way&rdquo; to change Linear Dependence Lemma&rsquo;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&rsquo;s worth something right
