I am reading about Basis Function Sets with polynomial functions, and the set $\{1, x, x^2, x^3\}$ form a basis set for polynomial functions and so must be linearly independent.

The proof from the definition of linearity, that there must be some set of non-zero coefficients such that $a_0 + a_1x + a_2x^2 + a_3x^3 = 0$ is makes sense 'since a polynomial is zero if and only if its coefficients are all zero' from [1] (this itself is discussed in [2]).

However, I have also seen independence defined that one basis vector cannot be written as a linear combination of the others. This is clearly true with vectors i = [0, 1], j = [1, 0] but I cannot see why this is true for the polynomial basis:

e.g. $x^2 = a_0 + a_1x$ where $a_0=0$ and $a_1=x$. This example seems wrong but not sure why!

[1] https://ltcconline.net/greenl/courses/203/Vectors/basisDimension.htm

[2] Polynomial $p(x) = 0$ for all $x$ implies coefficients of polynomial are zero


1 Answer 1


This is linear independence over the field of real numbers. The $a_i's$ have to be real numbers. These are not functions depending on $x$.

  • $\begingroup$ Thanks for this, of course x is not a single number. For this to be true a1 would need to take on a different value for each value of x. In general, x can be thought of as the reals -inf to inf? (unless there are some specified bounds on the domain?). Just trying to make sure I have a solid conceptualisation of this as not to make the same mistake again. $\endgroup$
    – Joseph
    Feb 16, 2022 at 14:05
  • 1
    $\begingroup$ Yes, you are correct, the problem makes sense only if the domain of definition of the polynomials is specified. If not specified, it is usually assumed that this is ${\bf R}$. Note that if the polynomials are defined on an interval $[a,b]$, $a<b$, the result also holds, because a non-zero polynomial can have only finitely many roots, the number of roots being bounded by the degree of the polynomial. $\endgroup$
    – coudy
    Feb 16, 2022 at 14:10

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