# Linear Independency of the List $[p_0,p_1,p_2,…,p_m]$

Original Question: My understanding from the book

Consider the following list

$$L=[p_0,p_1,p_2,...,p_m]$$

where $m$ is some nonnegative integer and $p_i$ is a function defined by

$$p_i=\{\,(z,z^i):z\in \Bbb{F}\,\}, \text{ for i \in \Bbb{W}}$$

that $\Bbb{F}$ is an arbitrary field and $\Bbb{W}=\{0,1,2,3,...\}$. Also, we have defined

$$z^i= \left\{ \begin{array}{l} z^{i-1}\cdot z & i \ne 0 \\ 1& i=0 \end{array} \right.$$

where $1$ is the multiplicative identity of the field $\mathbb{F}$.

Consider that $\mathcal{P}(\mathbb{F})$ is the set of all polynomials with coefficients in $\mathbb{F}$. In fact, $\mathcal{P}(\mathbb{F})$ is a subspace of the space of all functions $f:\mathbb{F} \to \mathbb{F}$.

How can I prove that the list $L$ in the linear space $\mathcal{P}(\mathbb{F})$ is linearly independent?

Original Definition and Examples in the Book: Linear Algebra Done Right by Sheldon Axler

• "Linear independent"...in what vector space and over what field? Because $\;\Bbb F\;$ is a linear space over itself but of dimension $\;1\;$ ... – DonAntonio Aug 2 '16 at 15:47
• Are you sure that $z$ is an element of $\Bbb F$? If so this set is obviously not independent. Now, if $z$ was in indeterminate over $\Bbb F$, that is if we were considering the polynomials $1,\dots,z^m$ that would be different. – David C. Ullrich Aug 2 '16 at 15:54
• Ah. What you say in your comment is entirely different from what you said in your post! Again it raises a question: Are you sure the question is about the space of polynomial functions? Because if you're actually talking about the space of polynomial functions, that set may or may not be independent, depending on $\Bbb F$. On the other hand if you're talking about the space of polynomials over $\Bbb F$ then the set is independent. (Somewhere earlier in the book there should be an explanation of the difference between "polynomial" and "polynomial function"...) – David C. Ullrich Aug 2 '16 at 15:57
• Example: Say $\Bbb F=\Bbb Z_2$, the field with two elements. If we're talking about polynomial functions then $z+z^2=0$, so those functions are not independent. But if we're talking about polynomials then those poynomials are independent. ($z+z^2$ is a non-zero polynomial. Not that that proves independence.) – David C. Ullrich Aug 2 '16 at 16:03
• I just noticed that you've said you think you're talking about polynomials, and you've also said you think you're talking about polynomial functions. In case you've missed it: When you check, check very carefully! Because polynomials and polynomial functions are totally different things. – David C. Ullrich Aug 2 '16 at 16:15

There is something wrong (in the book):

1. The Definition of polynomial is not what is commonly called a polynomial; normally, this would rather be called a polynomial function.
2. The polynomial functions $z\mapsto 1, z\mapsto z, \ldots, z\mapsto z^m$ are not necessarily linearly independent.

To see the second point, consider the case that $\Bbb F$ is the field of two elements (or any finite field and adjust the following numbers). Then there are only four functions $\Bbb F\to \Bbb F$ to begin with. Hence, $\mathcal P(\Bbb F)$, which is a subset of $\Bbb F^{\Bbb F}$ also has at most four elements. Thus for $m\ge 4$, the elements of the list $[p_0,\ldots, p_m]$ cannot even be distinct, let alone linearly independent.

However, if your field $\Bbb F$ has infinite cardinality, then yes, these functio0ns are linearly independent: Any linear dependence, i.e., a non-trivial linear combination of the monomial functions would result in a polynomial function that is identically zero on all of $\Bbb F$. That is, we have a polynomial with infinitely many zeroes. Only the zero polynomial has this property.

• In my book, the field $\mathbf{F}$ is explicitly defined to be either the field of real numbers or the field of complex numbers. Thus it’s possible to think of a polynomial as a function from $\mathbf{F}$ to $\mathbf{F}$, which has the pedagogical advantage of being a familiar object to students. – Sheldon Axler Aug 5 '16 at 5:14
• @SheldonAxler: At the end of section $1.A$ (Digression on Fields), you said that "In this book we will not need to deal with fields other than $\Bbb{R}$ or $\Bbb{C}$. However, many of the definitions, theorems, and proofs in linear algebra that work for both $\Bbb{R}$ and $\Bbb{C}$ also work without change for arbitrary fields. If you prefer to do so, throughout Chapters 1,2, and 3 you can think of $\Bbb{F}$ as denoting an arbitrary field instead of $\Bbb{R}$ and $\Bbb{C}$, except that some of the examples and exercises require that for each positive integer $n$ we have $1+1+...+1 \ne 0$." – H. R. Aug 5 '16 at 7:58
• @SheldonAxler: A suggestion: In page $45$, in the box, you mention a really nice point. "Thus when we talk about the dimension of a vector space, the role played by the choice of $\Bbb{F}$ cannot be neglected." I think this post mentioned here can be used to emphasize the role of $\Bbb{F}$ in the concept of Linear Dependence. :) – H. R. Aug 5 '16 at 8:08
• Let me clear up some misinformation, because one of the comments above states that my book Linear Algebra Done Right “does not restrict the discussion to $\mathbf{R}$ or $\mathbf{C}$ in the first three chapters.” In fact, starting with Chapter 2, the first page of every chapter in the book prominently states in a box that “$\mathbf{F}$ denotes $\mathbf{R}$ or $\mathbf{C}$.” The book includes a comment at the end of Chapter 1 that “many” of the results hold for arbitrary fields, but the book explicitly states that some of them require that the field have characteristic 0. – Sheldon Axler Aug 5 '16 at 19:20
• @SheldonAxler: You are right! :) But it will be much instructive and efficient if you highlight these examples and theorems in the text in the first three (or maybe two) chapters. The reason is that they are not too many and they can contain nice points like this one that emphasizes the role of $\Bbb{F}$ in the concept of linear dependence. Also, it can avoid confusions for those who may want to think of $\Bbb{F}$ as arbitrary fields. :) – H. R. Aug 5 '16 at 19:56