Lecture 21 - Polynomial Spaces (4.A)

Note for today that we denote:

T \circ T = T^{2}

and in general:

\underset{k times}{\underset{⏟}{T \circ T \circ \dots \circ T}} = T^{k}

But we can define a polynomial operator like:

T^{4} + T^{3} - 2 T^{2}

as some new operator:

(T^{4} + T^{3} - 2 T^{2}) v = T^{4} v + T^{3} v - 2 T^{2} v

We'll revisit these, but for this section, we'll care a lot about polynomials, and specifically how they factor, which will be important then.

Note

We go from right to left for our transformations. However, for polynomials since $T \circ T$ is commutative then for our polynomials they don't matter.

Properties of $C$

We talked about some properties of $C$ , most of which you know. Here $z = a + b i$ :

$z + \bar{z} = 2 R (z)$ , where $R$ denotes extracting the real part from $z$ .
$z - \bar{z} = 2 I (z)$ , where $I$ denotes extracting the imaginary part from $z$ .
$z \cdot \bar{z} = | z |^{2}$ , where $| z | = \sqrt{a^{2} + b^{2}}$ .
$\overset{―}{w + z} = \bar{w} + \bar{z}, \overset{―}{w z} = \bar{w} \bar{z}$
$\bar{\bar{z}} = z$
$| R (z) | \leq | z |, | I (z) | \leq | z |$
$| \bar{z} | = | z |$
$| w z | = | w | | z |$
(Triangle Inequality) $| w + z | \leq | w | + | z |$

We'll use these sporadically as we go on.

Properties of Polynomials

Division Algorithm for Polynomials

Given polynomials $p, s \in P (F)$ , there exist unique polynomials $q, r \in P (F)$ such that:

p = s q + r

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

Notice that this is the standard division algorithm except with polynomials. Just like how we did the division algorithm to generate $q, r$ , we can do polynomial division to generate our $q, r$ polynomials here.

Note

If $p (λ) = 0$ then $(x - λ)$ is a factor of $p$ , in other words:

p (x) = (x - λ) q (x)

This is actually the case from the division algorithm itself. Note that if $\deg s = 1$ then we are guaranteed that $deg (r) = 0$ so then $r$ is a constant polynomial (constant function).

Let's do an example. We can can $q, r$ through long division of $p = x^{4} + 3 x^{2} - x^{2} + 2 x + 2$ and $q = x^{2} - x + 1$ :

\begin{array}{r} x^{2} + 4 x \\ x^{2} - x + 1 x^{4} + 3 x^{3} - x^{2} + 2 x + 2 \\ x^{4} - x^{3} + x^{2} \\ 4 x^{3} - 2 x^{2} \\ 4 x^{3} - 4 x^{2} + 4 x \\ 2 x^{2} - 2 x \\ 2 x^{2} - 2 x + 2 \\ 0 \end{array} .

Thus here $q = x^{2} + 4 x$ , $s = x^{2} - x + 1$ while $r = 0$ . Hence:

s q + r = (x^{2} - x + 1) (x^{2} + 4 x) + 0

But we can now prove the division algorithm for polynomials:

Proof
Let $n = deg (p)$ and $m = deg (s)$ . If $n < m$ then $deg (p) < deg (s)$ so then $q = 0$ and $r = p$ via the algorithm above, which works:

p = s \cdot 0 + r = s \cdot 0 + p

Now suppose $n \geq m$ . Define $T_{s} : P_{n - m} (F) \times P_{n - 1} (F) \to P_{n} (F)$ by:

T_{s} (q, r) = s q + r

We can verify that it's a linear transformation pretty quickly:

Suppose $T_{s} (q, r) = 0$ . Then $s q + r = 0$ so then $q = 0, r = 0$ (notice if one wasn't zero, then we get an invalid inequality since $s q$ has a degree of $n \geq m$ while $r$ is at most degree $m - 1$ ).
Notice that $dim (null (T)) = 0$ so $T_{s}$ is injective, and furthermore, by the FTOLM, then we have $dim (range (T)) = dim (D)$ where $D$ is a shorthand for our domain. Here $dim (D) = (n - m + 1) + (m - 1 + 1) = n + 1$ which is the dimension of our RHS. Hence $P_{n} (F) = range (T)$ so then $T$ is surjective.
Hence, $T$ is bijective, and therefore $q, r$ are unique as we needed.
☐

Factorization of Polynomials Over $F$

Roots and Factors of Polynomials

Consider $p \in P (F)$ . A number $λ \in F$ is called a zero, or root, of $p$ if $p (λ) = 0$ .
A polynomial $s \in P (F)$ is called a factor of $p$ if there exists a polynomial of $q$ such that $p = s q$ .

Theorem

$λ$ is a zero of $p$ iff $(z - λ)$ is a factor of $p$ .

Proof
Consider $\leftarrow$ , so suppose $z - λ$ is a factor of $p$ . Thus $p = (z - λ) q$ for some $q$ . But then:

p (λ) = (λ - λ) q = 0 q = 0

thus $λ$ is a zero of $p$ .

Consider $\to$ , so suppose $λ$ is a zero of $p$ , so $p (λ) = 0$ . Using the division algorithm for polynomials, apply this to $p$ , where $p = p$ and $s = (z - λ)$ . What do we get? We get that there must exist some $q, r$ where $deg (r) < deg (s)$ such that:

p = (z - λ) q + r

we need to show that $r = 0$ . We know that $s = (z - λ)$ so then $deg (s) = 1$ so that forces that $deg (r) = 0$ so then $r = c \in F$ (ie: $r$ is a scalar). But notice that:

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

so then $r (λ) = c = 0$ so $r = 0$ as required. So then $p = s q$ so then $s$ is a factor of $p$ .
☐
Recall from our prior example that:

x^{4} + 3 x^{3} - x^{2} + x + 1 = (x^{2} + 4 x) (x^{2} - x + 1) + 0

so here $x = 0, x = - 4$ are zeroes of this original polynomial, as $x^{2} + 4 x = (x - 0) (x + 4)$ are factors of our original polynomial.

Fundamental Theorem of Algebra

For any $p \in P (F)$ with $deg (p) = m \geq 0$ , then $p$ has at most $m$ distinct zeroes in $F$ .

Theorem

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

We don't prove the latter theorem, but it introduces us factoring in $C$ :

Factoring in

C

Let $p \in P (C)$ is an non-constant polynomial with $deg (m)$ . Then $p$ can be factored uniquely (up to order $m$ ) as:

p (z) = c (z - λ_{1}) (z - λ_{2}) \dots (z - λ_{m})

where $c, λ_{1}, \dots, λ_{m} \in C$

But crazy enough, this doesn't work for $R$ !

Factoring with

R

coefficients.

Let $p \in P (C)$ , but only with real coefficients. Then $λ$ is a zero of $p$ iff $\bar{λ}$ is a zero of $p$ .

The proof of this is pretty simple.

Factorization

Suppose $b, c \in R$ . Then:

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

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

Factoring in

P (R)

Let $p \in P (R)$ is a non-constant polynomial. Then $p$ can be factored uniquely (up to order) as:

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

Properties of C

Properties of Polynomials

Factorization of Polynomials Over F

Factorization

Properties of $C$

Factorization of Polynomials Over $F$