# Linearly independent

• Sep 28th 2011, 07:02 AM
jackie
Linearly independent
I got stuck on showing that the set \$\displaystyle S=\{ x^1, x^2,...,x^n\}\$ is linearly independent in the continuous function space. Suppose there exist \$\displaystyle a_1,a_2,...,a_n\$ such that \$\displaystyle a_1x_1+a_2x^2+...+a_nx^n=0\$ for all x, I need to show that \$\displaystyle 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 \$\displaystyle x\$, which is infinite, so somehow it must be the case that \$\displaystyle a_1=...=a_n=0\$
Any help is appreciated.
• Sep 28th 2011, 09:16 AM
Deveno
Re: Linearly independent
if the polynomial \$\displaystyle 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 \$\displaystyle a_j\$?

to make it sharper, suppose \$\displaystyle a_1x + a_2x^2 + \dots + a_nx^n\$ is not the 0-polynomial. then we can find some x for which \$\displaystyle 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 \$\displaystyle \{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 \$\displaystyle \{x,x^2,\dots,x^n\}\$ must be linearly independent, being a subset of a basis.
• Sep 28th 2011, 09:58 AM
jackie
Re: Linearly independent
Quote:

Originally Posted by Deveno
if the polynomial \$\displaystyle 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 \$\displaystyle 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 \$\displaystyle a_j\$. I understand the alternative proof, but I'd like to be able to prove this using the above idea too.
• Sep 28th 2011, 11:35 AM
HallsofIvy
Re: Linearly independent
Suppose \$\displaystyle 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 \$\displaystyle 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 \$\displaystyle a_0= 0\$, we know that the polynomial is really \$\displaystyle P(x)= a_3x^3+ a_2x^2+ a_1x\$ and so \$\displaystyle 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, \$\displaystyle P'(x)= 3a_3x^2+ 2a_2x+ a_1= 0\$ for all x. Again, take x= 0 to get \$\displaystyle P'(0)= 3a_3(0^2)+ 2a_2(0)+ a_1= a_1= 0\$. Now we know that \$\displaystyle 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: \$\displaystyle 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 \$\displaystyle P(x)= a_3x^3+ a_2x^2+ a_1x= 0\$ for all x, we must have that \$\displaystyle x(a_3x^2+ a_2x+ a_1)= 0\$ for all x. Since x is 0 only for x= 0, we must have \$\displaystyle 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, \$\displaystyle a_3x^2+ a_2x+ a_1= 0\$ for all x. In particular, for x= 0, \$\displaystyle a_3(0^2)+ a_2(0)+ a_1= a_1= 0\$ so we must have \$\displaystyle a_1= 0\$. Continue in the same way.
• Sep 28th 2011, 11:56 AM
jackie
Re: Linearly independent
Thank you, HallsofIvy. I think I can continue from your big hint.