126 lines
3.1 KiB
Markdown
126 lines
3.1 KiB
Markdown
# Lecture 16
|
|
|
|
## Chapter IV Polynomials
|
|
|
|
**$\mathbb{F}$ denotes $\mathbb{R}$ or $\mathbb{C}$**
|
|
|
|
---
|
|
|
|
Review
|
|
|
|
### Products and Quotients of Vector Spaces 3E
|
|
|
|
#### Theorem 3.107
|
|
|
|
Let $T\in \mathscr{L}(V,W)$, then define $\tilde{T}:V/null\ T\to W$, given by $\tilde{T}(v+null\ T)=Tv$
|
|
|
|
a) $\tilde{T}\circ \pi=T$ where $\pi: V/null\ T$
|
|
|
|
b) $\tilde{T}$ is injective
|
|
|
|
c) $range\ T=range\ \tilde{T}$
|
|
|
|
d) $V/null\ T$ and $range\ T$ are isomorphic
|
|
|
|
Example:
|
|
|
|
Consider $D:\mathscr{P}_M(\mathbb{F})\to \mathscr{P}_{m-1}(\mathbb{F})$ be differentiation map
|
|
|
|
$D$ is surjective by $D$ is not injective $null\ D=${constant polynomials}
|
|
|
|
$\tilde{D}:\mathscr{P}_M(\mathbb{F})/$ constant polynomials $\to \mathscr{P}_{m-1}(\mathbb{F})$
|
|
|
|
This map ($\tilde{D}$) is injective since $range\ \tilde{D}=range\ D=\mathscr{P}_{m-1}(\mathbb{F})$
|
|
|
|
$\tilde{D}^{-1}:\mathscr{P}_{m-1}(\mathbb{F})\to \mathscr{P}_M(\mathbb{F})/$ constant polynomials (anti-derivative)
|
|
|
|
---
|
|
|
|
New materials
|
|
|
|
### Complex numbers 1A
|
|
|
|
#### Definition 1.1
|
|
|
|
Complex numbers
|
|
|
|
$z=a+bi$ is a complex number for $a,b\in \mathbb{R}$, ($Re\ z=a,Im\ z=b$)
|
|
|
|
$\bar{z}=a-bi$ complex conjugate $|z|=\sqrt{a^2+b^2}$
|
|
|
|
#### Properties 1.n
|
|
|
|
1. $z+\bar{z}=2a$
|
|
2. $z-\bar{z}=2b$
|
|
3. $z\bar{z}=|z|^2$
|
|
4. $\overline{z+w}=\bar{z}+\bar{w}$
|
|
5. $\overline{zw}=\bar{z}\bar{w}$
|
|
6. $\bar{\bar{z}}=z$
|
|
7. $|a|\leq |z|$
|
|
8. $|b|\leq |z|$
|
|
9. $|\bar{z}|=|z|$
|
|
10. $|zw|=|z||w|$
|
|
11. $|z+w|\leq |z|+|w|$
|
|
|
|
### Polynomials 4A
|
|
|
|
$$
|
|
p(x)=\sum_{i=0}^{n}a_i x^i
|
|
$$
|
|
|
|
#### Lemma 4.6
|
|
|
|
If $p$ is a polynomial and $\lambda$ is a zero of $p$, then $p(x)=(x-\lambda)q(x)$ for some polynomial $q(x)$ with $deg\ q=deg\ p -1$
|
|
|
|
#### Lemma 4.8
|
|
|
|
If $m=deg\ p,p\neq 0$ then $p$ has at most $m$ zeros.
|
|
|
|
Sketch of Proof:
|
|
|
|
Induction using 4.6
|
|
|
|
### Division Algorithm 4B
|
|
|
|
#### Theorem 4.9
|
|
|
|
Suppose $p,s\in \mathscr{P}(\mathbb{F}),s\neq 0$. Then there exists a unique $q,r\in \mathscr{P}(\mathbb{F})$ such that $p=sq+r$, and $deg\ r\leq deg\ s$
|
|
|
|
Proof:
|
|
|
|
Let $n=deg\ p,m=deg\ s$ if $n< m$, we are done $q=0,r=p$.
|
|
|
|
Otherwise ($n\leq m$) consider $1,z,...,z^{m-1},s,zs,...,z^{r-m}s$. is a basis of $\mathscr{P}_n(\mathbb{F})$.
|
|
|
|
Then there exists a unique $a_1,...,a_n\in\mathbb{F}$ such that $p(z)=a_0+a_1z+...+a_{m-1}z^{m-1}+a_m s+...+ a_n z^{n-m}s=(a_0+a_1z+...+a_{m-1}z^{m-1})+s(a_m +...+a_n z^{n-m})$
|
|
|
|
let $r=(a_0+a_1z+...+a_{m-1}z^{m-1}), q=(a_m +...+a_n z^{n-m})$ then we are done.
|
|
|
|
### Zeros of polynomial over $\mathbb{C}$ 4C
|
|
|
|
#### Theorem 4.12 Fundamental Theorem of Algorithm
|
|
|
|
Every non-constant polynomial over $\mathbb{C}$ has at least one root.
|
|
|
|
#### Theorem 4.13
|
|
|
|
If $p\in \mathscr{P}(\mathbb{C})$ then $p$ has a unique factorization up to order as $p(z)=c(z-\lambda_1)(z-\lambda_m)$ for $c,\lambda_1,...,\lambda_m\in \mathbb{C}$
|
|
|
|
Sketch of Proof:
|
|
|
|
(4.12)+(4.6)
|
|
|
|
### Zeros of polynomial over $\mathbb{R}$ 4D
|
|
|
|
#### Proposition 4.14
|
|
|
|
If $p\in \mathscr{P}(\mathbb{C})$ with real coefficients, then if $p(\lambda )=0$ then $p(\bar{\lambda})=0$
|
|
|
|
#### Theorem 4.16 Fundamental Theorem of Algorithm for real numbers
|
|
|
|
If $p$ is a non-constant polynomial over $\mathbb{R}$ the $p$ has a unique factorization
|
|
|
|
$p(x)=c(x-\lambda_1)...(x-\lambda_m)(x^2+b_1 x+c_1)...(x^2+b_m x+c_m)$
|
|
|
|
with $b_k^2\leq 4c_k$
|