# Lecture 16

## Chapter IV Polynomials

**$\mathbb{F}$ denotes $\mathbb{R}$ or $\mathbb{C}$**

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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)

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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 $b_k^2\leq 4c_k$
