4
$\begingroup$

As it states in the title, I'd like to prove that $1, x, x^2, \ldots , x^n$ are linearly independent in $C[-1,1]$.

Should I use an induction argument or integrate for $x^m$ and $x^n$ with cases $m=n$ and $m \neq n$? The inner product is $$ \langle f,g \rangle = \int_{-1}^1 f(x)g(x)dx.$$ Do both methods work?

$\endgroup$
3
  • $\begingroup$ If you just prove for $x\to x^m$ and $x\to x^n$, you will only have proven that $x\to x^m$ and $x\to x^n$ are linearly independent. $\endgroup$
    – Git Gud
    Jun 3, 2013 at 19:45
  • $\begingroup$ A more general version of this question has been answered in math.stackexchange.com/q/364784/73324 $\endgroup$
    – vadim123
    Jun 3, 2013 at 19:47
  • 4
    $\begingroup$ @vadim123 Isn't easier than that? We don't need to use the inner product at all here. $\endgroup$ Jun 3, 2013 at 19:49

5 Answers 5

9
$\begingroup$

Suppose they aren't linearly independent in $[-1,1]$. Then $a_0+a_1x^1+\cdots+a_nx^n=0$ for some set of coefficients, where not all of them are zero. But an $n$ degree polynomial can have at most $n$ roots, but this one has infinitely many, a contradiction.

$\endgroup$
2
  • $\begingroup$ Infinitely many because it is identically zero on the interval (in fact on the whole of $\mathbb{R}$)? $\endgroup$
    – Jared
    Jun 3, 2013 at 20:09
  • $\begingroup$ Yes, since every point in $[-1,1]$ is a root for this polynomial. $\endgroup$
    – JLA
    Jun 3, 2013 at 20:10
8
$\begingroup$

Hint: Write down the definition of linear independence and use the fact that two polynomials are equal iff their coefficients are all the same.

Note that we don't need to use the inner product at all. Linear independence is a property of a set in a vector space.

$\endgroup$
7
  • $\begingroup$ Does this still require an induction argument? Or just the observation that none of these polynomials have any coefficients in common and call it quits? $\endgroup$
    – Jared
    Jun 3, 2013 at 19:55
  • $\begingroup$ If you believe that "two polynomials are equal iff their coefficients are all the same", then there is certainly no induction necessary. $\endgroup$ Jun 3, 2013 at 19:57
  • $\begingroup$ Isn't this preempting the result? $\endgroup$
    – Did
    Jun 3, 2013 at 19:59
  • $\begingroup$ @MarkMcClure If you're using $\cdots$, then induction is needed. $\endgroup$
    – Git Gud
    Jun 3, 2013 at 19:59
  • $\begingroup$ Yes, isn't this in a sense assuming what the OP is trying to prove? $\endgroup$
    – JLA
    Jun 3, 2013 at 20:05
7
$\begingroup$

Let $\lambda_0,\lambda_1,\cdots,\lambda_n$ $n+1$ real such that $$\lambda_0+\lambda_1 x+\cdots +\lambda_n x^n=0\tag{1}$$ by derivative of $(1)$ succesively $k$ times $k=n,\ldots,1$ we find that $\lambda_n=0$ then $\lambda_{n-1}=0$ and so on hence we conclude

$\endgroup$
2
  • $\begingroup$ Hey, that doesn't use the inner product either! $\endgroup$ Jun 3, 2013 at 19:58
  • $\begingroup$ Yes the inner product isn't necessary. $\endgroup$
    – user63181
    Jun 3, 2013 at 20:00
3
$\begingroup$

Set $f_n(x)=x^n$; you want to see that if $$ \alpha_0f_0+\alpha_1f_1+\dots+\alpha_nf_n=0 $$ then $\alpha_0=\alpha_1=\dots=\alpha_n=0$. The hypothesis means that, in particular, \begin{align} &\alpha_0c_0^0+\alpha_0c_0^1+\dots+\alpha_nc_0^n=0\\ &\alpha_0c_1^0+\alpha_0c_1^1+\dots+\alpha_nc_1^n=0\\ &\dots\\ &\alpha_0c_n^0+\alpha_0c_n^1+\dots+\alpha_nc_n^n=0 \end{align} where $c_0$, $c_1$, $\dots$, $c_n$ are pairwise distinct points in $[-1,1]$. Therefore you have a homogeneous linear system with matrix $$\begin{bmatrix} 1 & c_0 & c_0^2 & \dots & c_0^n \\ 1 & c_1 & c_1^2 & \dots & c_1^n \\ \dots & \dots & \dots & \dots & \dots \\ 1 & c_n & c_n^2 & \dots & c_n^n \end{bmatrix}\,.$$ This is a Vandermonde matrix, therefore invertible.

$\endgroup$
0
$\begingroup$

A more verbal solution. Assume that one of the monomials, say $x^p$, linearly depend on the other monomials. Then it would be possible to write $x^p$ as linear combination of several $x^q$, where $q\ne p$. But this is clearly impossible and the contradiction proves that $x^p$, for any $p$, does not depend on the set of remaining $x^q$.

$\endgroup$
1
  • 5
    $\begingroup$ "Clearly impossible" doesn't prove it. $\endgroup$
    – JLA
    Jun 3, 2013 at 20:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.