Prove that every set formed by polynomials of different degrees is linearly independent How to prove that every set formed by polynomials of different degrees is linearly independent.
My problem is in how to make a general set of polynomials of different degrees. 
 A: Try an argument along the following lines:


*

*Let $A={f_1,...,f_n}$ be a set of polynomials, each of which has a different degree. Suppose that they are arranged in order so $f_1$ has the lowest degree, and $f_n$  the higest.

*Assume that they are not linearly indepedant. Then we can find constants which are not all zero such that $$c_1 f_1+...c_n f_n=0.$$

*what must $c_n $ be? Then what must $c_{n-1} $ be? Why does this lead to a contradiction?

A: A polynomial of finite degree $n$ can be written as $p(x) = \sum_{k=0}^n a_k x^k$ and can be represented as tupel of coefficients $(a_0, a_1, \dotsc, a_n)$. Polynomials of lesser degrees could be embedded in that $n+1$-dimensional vector space. The standard base would consist of the monomials $x^k$. Different degrees would mean different highest order monomials and thus linear independence.
A: I would order them in increasing degree.  Then, if $\langle S \rangle$ denotes the subspace generated by a subset $S$, I would use the following fact:
FACT:  If $u_1, \ldots , u_r$ are linearly independent and $u_{r+1}$ does not belong to $\langle u_1, \ldots, u_r\rangle$, then $u_1, \ldots , u_{r+1}$ are linearly independent.
This follows from considering a linear combination which yields $0$, i.e. $\sum_{i=1}^{r+1} \lambda_i u_i=0,$ and studying the cases $\lambda_{r+1}\neq 0$ and $\lambda_{r+1}=0$ separately.
Applying the above to our collection of polynomials with increasing degrees, the result follows.
