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Math Help - Linearly independent

  1. #1
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    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.
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  2. #2
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    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.
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  3. #3
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    Re: Linearly independent

    Quote Originally Posted by Deveno View Post
    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.
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  4. #4
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    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.
    Last edited by HallsofIvy; September 28th 2011 at 04:19 PM.
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  5. #5
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    Re: Linearly independent

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