TR3.5

flo 11

Exr0n 2021-09-27 Mon 12:00

#flo

1 Polynomials

1.1 0 polynomial

  • Has degree \(-infty\)
  • Degrees are usually positive, except for the \(0\) degree
  • "that's too hard, and we're not going to do it here"

1.2 Identically zero

  • Like \(0\) or \(0 x^0\)
  • Most polynomials are sometimes zero, but polynomials that are "identically zero" means that it's always zero (instead of just sometimes zero)

1.3 \(\mathcal{P}_m(F)\)

  • Polynomials with coefficients in \(F\) whose highest degree is \(m\)
  • It can't be "whose degree is exactly \(m\)" because otherwise you won't have the identity and it won't be closed under addition (in the case where coefficient sum \(a_m + b_m = 0\))

1.3.1 It's a finite dimensional vector space

  • \[a_0z^0+...+a_mz^m + b_0z^0 + ... + b_mz^m = (a_0+b_0)z^0 + ... + (a_m+b_m)z^m\]

1.4 Proof of 2.16

  • Structure: proof by contradiction

2 Linear Independence

2.1 2.21 Linear Dependence Lemma 2.21

#toexpand - it's saying that any linearly independent list has a vector inside that doesn't "contribute anything", and that if you remove it you'l have the same span. Implicitly, maybe through induction?) if you remove a dependent vector enough times then you get a linearly independent list. - The list \((1, 1, 1), (2, 2, 2), (3, 3, 3)\) is really dependent, but \((0), (0), (0)\) is the most dependent (you have to remove all to get independence).

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3 Exercise 2.A.1

3.1 Lemma

If vectors \(v_1, v_2, v_3, v_4\) span \(V\), then the list \[v_1-v_2, v_2-v_3, v_3-v_4, v_4\] also spans \(V\).

3.2 Proof

We prove the lemma by showing that any vector \(v \in V\) can be written in the form \(a_1v_1 + a_2v_2 + a_3+v_3 + a_4v_4\) can also be written as a linear combination of the form \[ b_1 (v_1-v_2) + b_2 (v_2-v_3) + b_3(v_3-v_4) + b_4v_4 \]

If we set \[

\begin{aligned} b_1 &= a_1\\ b_2 &= a_1 + a_2\\ b_3 &= a_1 + a_2 + a_3\\ b_4 &= a_1 + a_2 + a_3 + a_4\\ \end{aligned}

\] then the two combinations will be equivalent:

\(\begin{aligned} &a_1(v_1-v_2) + (a_1+a_2)(v_2-v_3) + (a_1+a_2+a_3)(v_3-v_4) + (a_1+a_2+a_3+a_4)v_4\\ = &a_1v_1 \cancel{- a_1v_2 + a_1v_2} +a_2v_2 \cancel{- (a_1+a_2)v_3 + (a_1+a_2)v_3} +a_3v_3 - \cancel{(a_1+a_2+a_3)v_4 + (a_1+a_2+a_3)v_4} + a_4v_4\\ = a_1v_1 + a_2v_2 + a_3v_3 + a_4 v_4 \end{aligned}\)