3

I would like to prove that the set of monomials is linearly independent in the complex linear space $C(\mathbb R)$. I understand the definition of linear independence and I'm stuck on how to prove linear independence for all subsets. Where would one start with such a proof?

leo
  • 10,433
Tito
  • 421
  • This is pretty much by definition. How do you define a monomial ? – Ewan Delanoy Sep 30 '13 at 14:46
  • Perhaps the terminology you would use depends on the space in question. Are you dealing with polynomials (formal), $C[a,b]$, or anything similar? – Jonathan Y. Sep 30 '13 at 14:53
  • The set is in the space C($\mathbb R$) - updated question. – Tito Oct 02 '13 at 10:42
  • A set is linearly independent if and only if all it's finite subsets are linearly independent. If by the set of monomials you mean ${1,x,x^2,\ldots,}$ then yes, they are linearly independent. To prove it you can proof this more general statement: Let $S$ be a set of non zero polynomials over a field $F$. If there are no two polynomials with the same degree, then $S$ is a linearly independent subset of $F[x]$. – leo Oct 07 '14 at 02:28

2 Answers2

5

One can prove this by showing the following statement for all $n$:

The functions represented by $1,\,x,\,x^2,\ldots,\,x^n$ are linearly independent.

Note that, if adding a single element to a linearly independent set creates a linearly dependent set, then that new element must be writable as a linear combination of those initially in the set. Thus, all we must prove is:

The function $x^n$ cannot be written as a linear combination of $1,\,x,\,x^2,\ldots,\,x^{n-1}$.

But this isn't so hard: Note that $\lim_{x\rightarrow\infty}\frac{x^m}{x^{n}}=0$ for $m<n$. We can therefore deduce that, for any $P(x)$ writable as a linear combination of such $x^m$ we have $\lim_{x\rightarrow\infty}\frac{P(x)}{x^n}=0$ which implies $P(x)$ is not $x^n$, completing the proof. The linear independence of the whole set follows from the independence of these subsets, as any linear dependence among the whole set could only involve finitely many monomial.

Milo Brandt
  • 60,888
1

A poly nomial is, by definition, a finite linear combination of monomials (i.e., of the form $\sum_{i=1}^n a_i m_i$ with the $a_i$ elements of the underlying field and the $m_i$ (distinct) monomials). A polynomial is $0$ if and only if all its coefficients are $0$ and that's exactly the claim that the set of monomials is linearly independent.

Magdiragdag
  • 15,049
  • 1
    Note that if one thinks of polynomials as functions (i.e., not simply elements of $\mathbb{F}[x]$), one needs to evoke the fact that a polynomial has only as many zeros as its degree; This doesn't change the equivalency you've noted, but it does affect how it's to be proven (as 'formal' polynomials, that's simply the definition of an additive unit). (Unless one deals with $C(\Omega)$ where $\Omega$ is finite, in which case the monomials are no longer linearly independent.) – Jonathan Y. Oct 01 '13 at 11:21
  • 2
    If you identify polynomials with the functions they represent, then the set of monomials is not necessarily linearly independent anymore. So I'm guessing the OP is seeing them as elements of ${\mathbb F}[x]$. – Magdiragdag Oct 01 '13 at 11:27
  • Agreed; in the absence of clarification (I did ask), I just wanted to be explicit. – Jonathan Y. Oct 01 '13 at 11:33
  • @Magdiragdag That's the case if the coefficients of the polynomials come from a finite field. OP is considering polynomial functions over $\Bbb R$ or $\Bbb C$ (question is not very clear). – leo Oct 07 '14 at 03:00