1. ## Linearly independent

I got stuck on showing that the set $S=\{ x^1, x^2,...,x^n\}$ is linearly independent in the continuous function space. Suppose there exist $a_1,a_2,...,a_n$ such that $a_1x_1+a_2x^2+...+a_nx^n=0$ for all x, I need to show that $a_1=a_2=...=a_n=0$. From here I am not sure how to proceed. I'm tempted to say that the fundamental theorem of algebra says that the above equation has exactly n solutions, but this is true for all $x$, which is infinite, so somehow it must be the case that $a_1=...=a_n=0$
Any help is appreciated.

2. ## Re: Linearly independent

if the polynomial $a_1x + a_2x^2 + \dots + a_nx^n$ has at most n roots 0, but the polynomial is given to be 0 for EVERY x, what does this tell us about the $a_j$?

to make it sharper, suppose $a_1x + a_2x^2 + \dots + a_nx^n$ is not the 0-polynomial. then we can find some x for which $a_1x + a_2x^2 + \dots + a_nx^n$ is NOT 0, contradicting our choice of coefficients.

alternatively, note that R[x] is a subspace of the continuous function space (polynomials are continuous functions). the set $\{1,x,x^2, \dots, x^n\}$ is a basis for R[x], and we can extend any basis of a subspace to a basis for the enitre space. then $\{x,x^2,\dots,x^n\}$ must be linearly independent, being a subset of a basis.

3. ## Re: Linearly independent

Originally Posted by Deveno
if the polynomial $a_1x + a_2x^2 + \dots + a_nx^n$ has at most n roots 0, but the polynomial is given to be 0 for EVERY x, what does this tell us about the $a_j$?
Thank you for your help. This is the question I have above too. I don't know what this tells us about the $a_j$. I understand the alternative proof, but I'd like to be able to prove this using the above idea too.

4. ## Re: Linearly independent

Suppose $P(x)= a_3x^3+ a_2x^2+ a_1x+ a_0= 0$ for all x. Then, in particular, it is 0 for x= 0. That means that $P(0)= a_3(0^3)+ a_2(0^2)+ a_1(0)+ a_0= a_0= 0$. Now, you can proceed in either of two ways. The more "sophisticated" is to say that since we now know that $a_0= 0$, we know that the polynomial is really $P(x)= a_3x^3+ a_2x^2+ a_1x$ and so $P'(x)= 3a_3x^2+ 2a_2x+ a_1$. And, since P(x) is 0 for all x, it is a constant, so its derivative is identically 0 also. That is, $P'(x)= 3a_3x^2+ 2a_2x+ a_1= 0$ for all x. Again, take x= 0 to get $P'(0)= 3a_3(0^2)+ 2a_2(0)+ a_1= a_1= 0$. Now we know that $P'(x)= 3a_3x^2+ 2a_2x$ and, again, that is 0 for all x. It is a constant so its is identically equal to 0: $P''(x)= 6a_3+ 2a_2= 0$ for all x. Do you see how to continue in that way?

A slightly less sophisticated argument (it does not use the derivative directly but does use "continuity") is to argue that since $P(x)= a_3x^3+ a_2x^2+ a_1x= 0$ for all x, we must have that $x(a_3x^2+ a_2x+ a_1)= 0$ for all x. Since x is 0 only for x= 0, we must have $a_3x^2+ a_2x+ a_1= 0$ for all non-zero x. But since that is a polynomial, it is continuous and so it must also be 0 for x= 0. That is, $a_3x^2+ a_2x+ a_1= 0$ for all x. In particular, for x= 0, $a_3(0^2)+ a_2(0)+ a_1= a_1= 0$ so we must have $a_1= 0$. Continue in the same way.

5. ## Re: Linearly independent

Thank you, HallsofIvy. I think I can continue from your big hint.