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I am using "Linear Algebra Done Right" as a self study guide, and was confused by the following question in the text:

For $m$ a nonnegative integer, let $P_m(F)$ denote the set of all polynomials with coefficients in $F$ and degree at most $m$. You should verify that $P_m(F)$ is a subspace of $P(F)$ [where $P(F)$ is the space of all polynomials with coefficients in $F$]

The book's defines degree as:

A polynomial $p \in P(F)$ is said to have degree $m$ if there exists scalars $a_0, a_1,...a_m$ with $a_m \neq 0$ such that $p(z)=a_0+a_1z+...+a_mz^m$

For $P_m$ to be a subset of $P(F)$, $P_m$ must contain the additive inverse of $P(F)$, which is the polynomial with all 0 coefficients (according to the book). If my understanding of all this is correct, then isn't $P_0$, and subsequently any $P_m$, not a subspace of $P(F)$ because the definition of polynomials of degree $m$ requires at least one non-zero coefficient?

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Yes, a subspace must contain the additive identity (not inverse). The degree of the zero polynomial isn't really something people agree on, unfortunately. So you should interpret $P_m(F)$ as "all polynomials of degree $m$ or less, and also the zero polynomial". –  Henry Swanson May 26 at 2:46
    
Ah sorry, I mistyped. Thanks, I wasn't sure if I was misunderstanding something simple. –  bsm May 26 at 4:22

4 Answers 4

up vote 1 down vote accepted

The zero vector indeed should be the polynomial with all zero coefficients. It is common to introduce the convention that this polynomial has degree $-\infty $, and thus it is included in every $P_n$.

Following the definition strictly, everything is ok. Ask yourself: is the zero polynomial excluded from $P_n$? To answer affirmatively, you will have to show that the zero polynomial has degree larger than $m$. But that is not the case!

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Not so: the definition as stated does not assign any degree to the zero polynomial. –  vadim123 May 26 at 2:51
    
This commenter is correct, the book defines the zero polynomial's degree as he does –  bsm May 26 at 4:23

Strictly speaking, you're correct that the zero polynomial does not have a degree. But it is conventional to think of it as having degree "$-\infty$". With this convention, then, it is in $P_m$ for each $m$.

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You are right: the definition of degree you quoted does not allow for $0$ to be considered a degree-zero polynomial. (In fact many authors adopt the convention that the degree of $0$ is actually $- \infty$.)

But the question is, what is the book's definition of "polynomial"? You quoted the definition of degree, but that definition takes for granted that we already know what polynomials are, and are defining what the degree of a polynomial is. Somewhere prior to that there ought to be a definition of "polynomial", and the definition ought to be such that every $a \in F$, including $a=0$, is a polynomial.

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We may all agree that the zero function is a polynomial, but the above definition does not assign a degree to all polynomials. –  vadim123 May 26 at 2:52
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And some would even disagree that polynomials are functions. $x^2 = x$ for all $x \in \mathbb Z/2\mathbb Z$, but $x^2 \neq x$ as elements of $\mathbb Z/2\mathbb Z[x]$. –  kahen May 26 at 2:54
    
I am with you, @kahen. Polynomials are one thing, polynomial functions are something else. –  mweiss May 26 at 2:56
    
@vadim123 You are quite correct. –  mweiss May 26 at 3:02

The additive inverse of a polynomial is not the $0$ polynomial, but the negative of the original polynomial. The additive identity is the $0$ polynomial.

As $-p(x) + p(x) = 0$, and both $\pm p(x)$ are polynomials of the same degree, we see that $P_m$ does contain additive inverses.

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