Thread: Linearly independent

I got stuck on showing that the set is linearly independent in the continuous function space. Suppose there exist such that for all x, I need to show that . 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 , which is infinite, so somehow it must be the case that
Any help is appreciated.

if the polynomial has at most n roots 0, but the polynomial is given to be 0 for EVERY x, what does this tell us about the ?

to make it sharper, suppose is not the 0-polynomial. then we can find some x for which 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 is a basis for R[x], and we can extend any basis of a subspace to a basis for the enitre space. then must be linearly independent, being a subset of a basis.

Originally Posted by Deveno

if the polynomial has at most n roots 0, but the polynomial is given to be 0 for EVERY x, what does this tell us about the ?

Thank you for your help. This is the question I have above too. I don't know what this tells us about the . I understand the alternative proof, but I'd like to be able to prove this using the above idea too.

Suppose for all x. Then, in particular, it is 0 for x= 0. That means that . Now, you can proceed in either of two ways. The more "sophisticated" is to say that since we now know that , we know that the polynomial is really and so . And, since P(x) is 0 for all x, it is a constant, so its derivative is identically 0 also. That is, for all x. Again, take x= 0 to get . Now we know that and, again, that is 0 for all x. It is a constant so its is identically equal to 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 for all x, we must have that for all x. Since x is 0 only for x= 0, we must have 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, for all x. In particular, for x= 0, so we must have . Continue in the same way.

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