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

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In what follows we 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 properties. The latter are the usual properties satisfied by the addition and multiplication of real numbers, which we studied when we were in school. Importantly, these properties are also satisfied by the addition and multiplication of complex numbers. Thus, both the set of real numbers and the set of complex numbers , equipped with their usual operations, are fields.

When we deal with a field , we can take non-negative integer powers of the elements of by repeatedly multiplying them: if is a positive integer and , then

We adopt the convention thatwhere is the multiplicative identity of the field.

We can now define polynomials.

Definition Let be a field. Let be a non-negative integer. A function is called a polynomial of degree if and only if, for any ,where belong to and .

The elements are called coefficients of the polynomial.

In the above definition is assumed to be a non-negative integer. If (i.e., the coefficients are all equal to zero), then the degree of is conventionally set to .

Example Let us consider the field of real numbers . The function that satisfies, for any ,is a polynomial of degree .

Example The function that satisfies, for any ,is a polynomial of degree .

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

Example The leading coefficient of the polynomialis (provided ).

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

Example The polynomialis monic.

Example The polynomialis not monic because its leading coefficient is .

We now introduce the concept of a root.

Definition Let be a field and a polynomial of degree . We say that is a root of if and only if

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

Example Consider the polynomial defined byThen, is a root of the polynomial because

If we know a root of a polynomial , then we can use it to factor into simpler polynomials.

Proposition Let be a field and a polynomial of degree . Then, is a root of if and only if, for any ,where is a polynomial of degree .

Proof

Let us prove the "if" part, starting from the hypothesis that is a root of . Note that, for any integer and , we haveDefineNote that has coefficient . Thus, is a polynomial of degree andSince is of degree , we haveSince is a root of , we haveBy subtracting the latter equation from the former, we obtainThe polynomial is of degree because the highest power of it contains is (with by the assumption that is of degree ). Let us now prove the "only if" part, starting from the hypothesis that By setting , we obtainAs a consequence, is a root of .

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

Proposition Let be a field and a polynomial of degree . Then, has at most distinct roots.

Proof

The proof is by induction. For , we havewith . Therefore, has exactly one root . Thus, the claim is true for . Now, let us assume that it is true for a polynomial of degree . We need to show that the claim is true for a polynomial of degree . If has at least one root , thenwhere is a polynomial of degree . By the previous equation a root of different from must necessarily be a root of . But has at most distinct roots by the induction hypothesis. Therefore, has at most distinct roots.

The previous proposition does not cover the case , in whichand . In this case, there are no roots.

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

Proposition Let be a field and the polynomial defined byIf has at least members and for any , then

Proof

The proof is by contradiction. Suppose that at least one of the coefficients is different from zero. Then, the degree of is between and . As a consequence, can have at most distinct roots. But this contradicts the fact that the polynomial has at least distinct roots because has at least members and for any . Therefore, all the coefficients of the polynomial must be equal to zero.

The requirement that the field has at least members is always satisfied for the field of real numbers and the field of complex numbers, which have infinitely many members.

The previous proposition can be seen as a result stating that the polynomials are linearly independent: the only way to linearly combine them so as to get the zero polynomial as a result is to set 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, you 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 is a vector space.

Since any polynomial of degree has the formthe space of all polynomials of degree is spanned by the polynomials . We have just 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, there is no other way to linearly combine the basis so as to obtain . In other words, there is only one way to obtain a given polynomial by taking linear combinations of the functions . As a consequence, the degree of a polynomial is unique.

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

Proposition Let be a polynomial of degree . Then, has at least one root.

Proof

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

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

By combining the Fundamental theorem of algebra and the factorization theorem, we obtain the following important proposition.

Proposition Let be a polynomial of degree . Then, there exist complex numbers such thatfor any . The numbers are unique up to a permutation of .

Proof

We first demonstrate the existence of . By the Fundamental Theorem of Algebra (FTA), has at least one root. Denote it by . Then, we can factorize aswhere is a polynomial of degree . If , then and we are done. If , the FTA guarantees the existence of a root of . So we havewhere is a polynomial of degree . If , then and we are done. Otherwise, we proceed by factoring out other terms, until we get the desired result. We now prove uniqueness. By carrying out the multiplication of the factors of , we getBy the uniqueness of the representation of polynomials of degree in terms of the basis , we have that is unique. Suppose there is another factorizationWe can writewhere we have divided both sides by (which is different from zero because is of degree ). Note that the latter equation holds for any . When we set on the right-hand side, one of the factors on the left-hand side must be equal to zero. We can suppose without loss of generality that it is (if it is not, we can re-order the roots ). Thus, . We then divide everything by and obtainBy the same line of reasoning as earlier, we obtain , possibly after re-ordering the roots . We proceed in this manner until we have proved that for .

A polynomial of degree such asis often called a linear factor.

Thus, the previous proposition shows that any complex polynomial can be written as a product of linear factors.

Moreover, the linear factors expose all the roots of the polynomial.

Please cite as:

Taboga, Marco (2017). "Polynomials in linear algebra", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/polynomials-in-linear-algebra.

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