Chapter 4 - Polynomials (short)

Again, we denote $F$ as $R$ or $C$ . Really, this can be any field, which you look at more in an Abstract Algebra course.

Complex Conjugate and Absolute Value

R (z)

I (z)

Suppose $z = a + b i$ where $a, b \in R$ . The real part of $z$ , denoted $R (z)$ , is defined by $R (z) = a$ . Similarly, the imaginary part of $z$ , denoted $I (z)$ , is defined by $I (z) = b$ .

So then:

z = R (z) + I (z) i

Complex Conjugate

\bar{z}

, absolute value

| z |

Suppose $z \in C$ .

The complex conjugate of $z \in C$ , denoted $\bar{z}$ , is defined by $\bar{z} = R (z) - I (z) i$ .
The absolute value of a complex number $z$ , denoted $| z |$ , is defined by $| z | = \sqrt{R (z)^{2} + I (z)^{2}}$ .

Properties of

C

$z + \bar{z} = 2 R (z)$ ;
$z - \bar{z} = 2 I (z) i$ ;
$z \bar{z} = | z |^{2}$ ;
$\bar{w + z} = \bar{w} + \bar{z}$ and $\bar{w z} = \bar{w} \bar{z}$ ;
$\bar{\bar{z}} = z$ ;
$| R (z) | \leq | z |$ and $| I (z) | \leq | z |$ ;
$| \bar{z} | = | z |$ ;
$| w z | = | w | | z |$ ;
$| w + z | \leq | w | + | z |$ (triangle inequality).

Uniqueness of Coefficients for Polynomials

p \in P (F)

where

p = \vec{0}

then all

a_{i} = 0

Suppose $a_{0}, . . ., a_{m} \in F$ . If:

a_{0} + a_{1} z + \dots + a_{m} z^{m} = 0

for every $z \in F$ then $a_{i} = 0$ for all $0 \leq i \leq m$ .

\begin{proof}
We'll prove the contrapositive. If not all the coefficients are 0, then by changing $m$ we can assume some $a_{m} \neq 0$ . Let:

z = \frac{| a_{0} | + \dots + | a_{m - 1} |}{| a_{m} |} + 1

Note here that $z \geq 1$ and thus $z^{j} \leq z^{m - 1}$ for all $j = 0, . . ., m - 1$ . Using the triangle inequality:

| a_{0} + a_{1} z + \dots + a_{m - 1} z^{m - 1} | \leq (| a_{0} | + | a_{1} | + \dots + | a_{m - 1} |) z^{m - 1} < | a_{m} z^{m} |

Thus $a_{0} + a_{1} z + \dots + a_{m - 1} z^{m - 1} \neq - a_{m} z^{m}$ . Hence $a_{0} + a_{1} z + \dots + a_{m - 1} z^{m - 1} + a_{m} z^{m} \neq 0$ .
\end{proof}

This lemma implies then that the degree of a polynomial is uniquely determined, namely by the first $a_{i} \neq 0$ . For notation, we say that $deg (p) = m$ for a polynomial like the one above. For simplicity, $deg (0) = - \infty$ , using the obvious arithmetic with $- \infty$ .

The Division Algorithm for Polynomials

Division Algorithm for Polynomials

Suppose that $p, s \in P (F)$ with $s \neq 0$ . Then there exist unique polynomials $q, r \in P (F)$ such that:

p = s q + r

where $deg (r) < deg (s)$ .

We actually did a proof of this in lecture via Lecture 21 - Polynomial Spaces (4.A)#^79ae01.

Zeros of Polynomials

zero of a polynomial, factor

A number $λ \in F$ is called a zero (or root) of a polynomial $p \in P (F)$ if:

p (λ) = 0

On the other hand, a polynomial $s \in P (F)$ is called a factor of $p$ if there exists a polynomial $q \in P (F)$ such that $p = s q$ .

Although zeroes and factors are different concepts, they are related by the following lemma:

Zeroes of a polynomial correspond to degree-1 factors

Suppose $p \in P (F)$ and $λ \in F$ . Then $p (λ) = 0$ iff there is a polynomial $q \in P (F)$ such that:

p (z) = (z - λ) q (z)

for every $z \in F$ .

Really, this is saying that if you know there's a zero, then you can factor it out, and if you have a factor, then it's a valid zero.

\begin{proof}
Going $\leftarrow$ is obvious, since if $p (z) = (z - λ) q (z)$ then $p (λ) = (λ - λ) q (λ) = 0 q (λ) = 0$ .

Going $\to$ , suppose that $p (λ) = 0$ . The polynomial $z - λ$ is degree 1. Because a polynomial with degree less than 1 is a constant function, using the Division Algorithm for Polynomials implies there's some polynomial $q \in P (F)$ and a number $r \in F$ such that:

p (z) = (z - λ) q (z) + r

for every $z \in F$ . Since $p (λ) = 0$ then plug $z = λ$ :

p (λ) = (λ - λ) q (λ) + r = 0 \to r = 0

thus $r = 0$ so then $p (z) = (z - λ) q (z)$ as required.
\end{proof}

A polynomial has at most as many zeroes as its degree

Suppose $p \in P (F)$ is a polynomial with degree $m \geq 0$ . Then $p$ has at most $m$ distinct zeros in $F$ .

\begin{proof}
If $m = 0$ then $p (z) = a_{0} \neq 0$ so $p$ has no zeros.

If $m = 1$ then $p (z) = a_{0} + a_{1} z$ where $a_{1} \neq 0$ . There's only one zero $λ = - a_{0} / a_{1}$ .

Suppose $m > 1$ . We use induction on $m$ , assuming that every polynomial with degree $m - 1$ has at most $m - 1$ distinct zeros. If $p$ has no zeros in $F$ then we are done. If $p$ has a zero $λ \in F$ then by Chapter 4 - Polynomials (short)#^b4d22e there's a polynomial $q$ such that:

p (z) = (z - λ) q (z)

for all $z \in F$ . Clearly $deg (q) = m - 1$ . The equation above shows that if $p (z) = 0$ then either $z = λ$ or $q (z) = 0$ . In other words, the zeros of $p$ consist of $λ$ and the zeros of $q$ . By our induction hypothesis, $q$ has at most $m - 1$ distinct zeros in $F$ . Thus $p$ has at most $m$ distinct zeros in $F$ .
\end{proof}

Factorization of Polynomials over $C$

Fundamental Theorem of Algebra

Every non-constant polynomial with complex coefficients has a zero.

We just won't prove it here, as the proof uses Liouville's theorem which we haven't proved. If we had to prove it we'd do the following. Let $p$ be a non-constant polynomial with complex coefficients. Suppose $p$ has no zeros. Then $1 / p$ is an analytic function on $C$ . Furthermore, $| p (z) | \to \infty$ as $| z | \to \infty$ , which implies that $1 / p \to 0$ as $| z | \to \infty$ . Thus, $1 / p$ is a bounded analytic function on $C$ . By Liouville's Theorem, every such function is constant, but if $1 / p$ is constant, then $p$ is constant, contradiction our assumption that $p$ is non-constant.

Factorization of a polynomial over

C

If $p \in P (C)$ is a non-constant polynomial, then $p$ has a unique factorization (except for the order of the factors) of the form:

p (z) = c (z - λ_{1}) \dots (z - λ_{m})

where $c, λ_{i} \in C$ for all $i$ .

See [[Year3/Winter2024/MATH306-LinearAlgebraII/2015_Book_LinearAlgebraDoneRight.pdf#page=125]] for the proof if you're curious. It's a relatively simple proof though, just long for a write-up here. Notice that both existence and uniqueness of these factors are here.

Factorization of Polynomials over $R$

We've seen how factoring over $C$ allows us to use $λ \in C$ , but what if we are restricted to factorizing over $R$ ?

Polynomials with real coefficients have zeros in pairs

Suppose $p \in P (C)$ is a polynomial with real coefficients. If $λ \in C$ is a zero of $p$ , then so is $\bar{λ}$ .

\begin{proof}
Let:

p (z) = a_{0} + a_{1} z + \dots + a_{m} z^{m}

where $a_{0}, . . ., a_{m}$ are real numbers. Suppose $λ \in C$ is a zero of $p$ . Then:

a_{0} + a_{1} λ + \dots + a_{m} λ^{m} = 0

Take the complex conjugate of both sides of this equation, obtaining:

a_{0} + a_{1} \bar{λ} + \dots + a_{m} {\bar{λ}}^{m} = 0

via the property of complex conjugate via Chapter 4 - Polynomials (short)#^4d08d5. Thus $\bar{λ}$ is a zero for $p$ .
\end{proof}

Factorization of a quadratic polynomial

Suppose $b, c \in R$ . Then there is a polynomial factorization of the form:

x^{2} + b x + c = (x - λ_{1}) (x - λ_{2})

where now $λ_{1}, λ_{2} \in R$ iff $b^{2} \geq 4 c$ .

We omit the proof here, which again is at [[Year3/Winter2024/MATH306-LinearAlgebraII/2015_Book_LinearAlgebraDoneRight.pdf#page=127]]. The proof is very derived from elementary algebra you've likely done before.

Factorization of a polynomial over

R

Suppose $p \in P (R)$ is a non-constant polynomial. Then $p$ has a unique factorization (except for the order of the factors) of the form:

p (x) = c (x - λ_{1}) \dots (x - λ_{m}) (x^{2} + b_{1} x + c_{1}) \dots (x^{2} + b_{M} x + c_{M})

where $c, λ_{1}, . . ., λ_{m}, b_{1}, . . ., b_{M}, c_{1}, . . ., c_{M} \in R$ with $b_{j}^{2} < 4 c_{j}$ for each $j$ .

Again, the proof is long and arduous. Look at [[Year3/Winter2024/MATH306-LinearAlgebraII/2015_Book_LinearAlgebraDoneRight.pdf#page=128]].