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Linear Factors Of A Polynomial

Written By Saturday, November 5, 2022 Add Comment Edit

This lecture presents some facts about polynomials that are often used in linear algebra.

Table of Contents

Tabular array of contents

  1. Fields

  2. Integer powers

  3. Definition of polynomial

  4. Leading coefficient

  5. Monic polynomial

  6. Root of a polynomial

  7. Roots and factors

  8. Upper bound on the number of roots

  9. Polynomial of null degree

  10. Nix polynomial

  11. Linear independence of powers

  12. Spaces of polynomials

  13. Uniqueness of degree

  14. Fundamental theorem of algebra

  15. Factorization of complex polynomials

  16. Factorization into linear factors

In what follows nosotros are going to use the concept of a field, which was previously defined in the lecture on vector spaces.

All that we need to know is that a field is a set equipped with two operations (addition and multiplication) that satisfy a number of backdrop. The latter are the usual properties satisfied by the improver and multiplication of real numbers, which nosotros studied when we were in school. Importantly, these properties are likewise satisfied past the addition and multiplication of complex numbers. Thus, both the ready of real numbers R and the prepare of complex numbers $U{2102} $ , equipped with their usual operations, are fields.

When nosotros bargain with a field F , nosotros can take non-negative integer powers of the elements of F by repeatedly multiplying them: if $m$ is a positive integer and $zin F$ , then [eq1]

Nosotros adopt the convention that [eq2] where 1 is the multiplicative identity of the field.

We can now define polynomials.

Definition Allow F be a field. Permit $m$ be a non-negative integer. A office $p:F ightarrow F$ is called a polynomial of degree $m$ if and only if, for any $zin F$ , [eq3] where [eq4] vest to F and $a_{m} eq 0$ .

The elements [eq4] are chosen coefficients of the polynomial.

In the above definition $m$ is causeless to be a non-negative integer. If [eq6] (i.e., the coefficients are all equal to nil), then the degree of p is conventionally fix to $-infty $ .

Example Let us consider the field of existent numbers R . The part [eq7] that satisfies, for whatever $zin U{211d} $ , [eq8] is a polynomial of caste $2$ .

Example The function [eq9] that satisfies, for any $zin U{211d} $ , [eq10] is a polynomial of degree 0 .

The coefficient of the highest power of the polynomial (i.e., that defining the caste of the polynomial) is called leading coefficient.

Example The leading coefficient of the polynomial [eq11] is $a_{3}$ (provided $a_{3} eq 0$ ).

A polynomial whose leading coefficient is equal to 1 (the multiplicative identity of the field F ) is called a monic polynomial.

Example The polynomial [eq12] is monic.

Example The polynomial [eq13] is not monic considering its leading coefficient is $5$ .

We now introduce the concept of a root.

Definition Let F exist a field and $p:F ightarrow F$ a polynomial of caste $mgeq 1$ . We say that $lambda in F$ is a root of p if and but if [eq14]

Much of the theory of polynomials is concerned with studying roots and their backdrop.

Instance Consider the polynomial [eq15] defined by [eq16] And then, $lambda =1$ is a root of the polynomial because [eq17]

If we know a root of a polynomial p , so we can use information technology to factor p into simpler polynomials.

Proposition Permit F be a field and $p:F ightarrow F$ a polynomial of degree $mgeq 1$ . Then, $lambda $ is a root of p if and only if, for whatsoever $zin F$ , [eq18] where $q:F ightarrow F$ is a polynomial of degree $m-1$ .

Proof

Let us prove the "only if" function, starting from the hypothesis that $lambda $ is a root of p . Note that, for whatsoever integer $kgeq 1$ and $z,lambda in F$ , we have [eq19] Define [eq20] Note that $z^{k-1}$ has coefficient $lambda ^{0}=1$ . Thus, [eq21] is a polynomial of degree $k-1$ and [eq22] Since p is of degree $m$ , we have [eq23] Since $lambda $ is a root of p , we take [eq24] By subtracting the latter equation from the quondam, nosotros obtain [eq25] The polynomial [eq26] is of degree $m-1$ because the highest power of $z$ it contains is $a_{m}z^{m-1}$ (with $a_{m} eq 0$ by the assumption that p is of degree $m$ ). Let usa now prove the "if" part, starting from the hypothesis that [eq27] Past setting $z=lambda $ , we obtain [eq28] As a consequence, $lambda $ is a root of $pleft( z ight) $ .

Thanks to the previous factorization theorem, nosotros can put an upper bound on the number of roots of a polynomial.

Proffer Let F exist a field and $p:F ightarrow F$ a polynomial of degree $mgeq 1$ . Then, p has at most $m$ distinct roots.

Proof

The previous proffer does not encompass the case $m=0$ , in which [eq32] and $a_{0} eq 0$ . In this example, there are no roots.

No polynomial of positive degree tin be identically equal to zippo, provided that its underlying field has a sufficient number of members.

Proposition Permit F exist a field and $p:F ightarrow F$ the polynomial defined past [eq33] If F has at to the lowest degree $m+1$ members and [eq34] for any $zin F$ , so [eq35]

Proof

The requirement that the field F has at least $m+1$ members is always satisfied for the field R of real numbers and the field $U{2102} $ of complex numbers, which have infinitely many members.

The previous proffer can be seen as a result stating that the polynomials [eq37] are linearly contained: the merely fashion to linearly combine them so equally to get the zero polynomial as a event is to gear up all their coefficients equal to zero.

If you are wondering why we are speaking about polynomials using "vector space language" and, in particular, the concept of linear independence, y'all might want to revise the lectures on vector spaces and coordinate vectors, where we have discussed the fact that the set of all polynomials of degree $m$ is a vector space.

Since any polynomial of caste $m$ has the form [eq38] the infinite of all polynomials of degree $m$ is spanned past the polynomials [eq39] . Nosotros have merely demonstrated that the latter are linearly independent. Therefore, they are a basis for the space being discussed.

Since the representation in terms of a basis is unique, at that place is no other way to linearly combine the ground so as to obtain $pleft( z ight) $ . In other words, in that location is but one way to obtain a given polynomial by taking linear combinations of the functions $z^{k}$ . Every bit a consequence, the degree of a polynomial is unique.

The next proposition is known as the Key Theorem of Algebra.

Proposition Permit [eq40] be a polynomial of degree $mgeq 1$ . Then, p has at least one root.

Proof

This is a deep result in complex analysis, which we leave without a proof.

In other words, when nosotros are working with the field of complex numbers, then nosotros are guaranteed to find a root of a given polynomial.

By combining the Fundamental theorem of algebra and the factorization theorem, we obtain the post-obit important suggestion.

Proffer Let [eq41] be a polynomial of degree $mgeq 1$ . And then, there be circuitous numbers [eq42] such that [eq43] for whatsoever $zin U{2102} $ . The numbers [eq44] are unique up to a permutation of [eq45] .

Proof

We first demonstrate the being of [eq46] . Past the Primal Theorem of Algebra (FTA), p has at least one root. Announce information technology by $lambda _{1}$ . Then, we can factorize p equally [eq47] where $q$ is a polynomial of degree $m-1$ . If $m-1=0$ , then [eq48] and we are done. If $m-1geq 1$ , the FTA guarantees the existence of a root $lambda _{2}$ of $q$ . Then we have [eq49] where $r$ is a polynomial of degree $m-2$ . If $m-2=0$ , then [eq50] and we are washed. Otherwise, nosotros proceed past factoring out other terms, until nosotros get the desired outcome. Nosotros at present evidence uniqueness. By conveying out the multiplication of the factors of p , nosotros get [eq51] By the uniqueness of the representation of polynomials of caste $m$ in terms of the basis [eq52] , nosotros have that $c$ is unique. Suppose there is another factorization [eq53] We can write [eq54] where nosotros have divided both sides by $c$ (which is different from nil considering p is of degree $m$ ). Note that the latter equation holds for any $z $ . When nosotros set $z=lambda _{1}$ on the right-hand side, one of the factors on the left-hand side must be equal to zilch. We can suppose without loss of generality that it is $z-mu _{1}$ (if information technology is not, we can re-order the roots $mu _{j}$ ). Thus, [eq55] . We so dissever everything past [eq56] and obtain [eq57] By the same line of reasoning as earlier, we obtain [eq58] , mayhap afterwards re-ordering the roots $mu _{j}$ . We proceed in this way until we have proved that [eq59] for $j=1,ldots ,m$ .

A polynomial of degree 1 such as [eq60] is often called a linear gene.

Thus, the previous proposition shows that whatsoever circuitous polynomial can be written every bit a product of linear factors.

Moreover, the linear factors expose all the roots $lambda _{j}$ of the polynomial.

Delight cite as:

Taboga, Marco (2021). "Polynomials in linear algebra", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/polynomials-in-linear-algebra.

Linear Factors Of A Polynomial,

Source: https://www.statlect.com/matrix-algebra/polynomials-in-linear-algebra

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