Prove that $1, x, x^2, \dots , x^n$ are linearly independent in $C[-1,1]$ As it states in the title, I'd like to prove that $1, x, x^2, \ldots , x^n$ are linearly independent in $C[-1,1]$.
Should I use an induction argument or integrate for $x^m$ and $x^n$ with cases $m=n$ and $m \neq n$?
The inner product is $$ \langle f,g \rangle = \int_{-1}^1 f(x)g(x)dx.$$ Do both methods work?
 A: Suppose they aren't linearly independent in $[-1,1]$. Then $a_0+a_1x^1+\cdots+a_nx^n=0$ for some set of coefficients, where not all of them are zero. But an $n$ degree polynomial can have at most $n$ roots, but this one has infinitely many, a contradiction.
A: Hint: Write down the definition of linear independence and use the fact that two polynomials are equal iff their coefficients are all the same.
Note that we don't need to use the inner product at all.  Linear independence is a property of a set in a vector space.
A: Let $\lambda_0,\lambda_1,\cdots,\lambda_n$ $n+1$ real such that
$$\lambda_0+\lambda_1 x+\cdots +\lambda_n x^n=0\tag{1}$$
by derivative of $(1)$ succesively $k$ times $k=n,\ldots,1$ we find that $\lambda_n=0$ then $\lambda_{n-1}=0$ and so on hence we conclude
A: Set $f_n(x)=x^n$; you want to see that if
$$
\alpha_0f_0+\alpha_1f_1+\dots+\alpha_nf_n=0
$$
then $\alpha_0=\alpha_1=\dots=\alpha_n=0$. The hypothesis means that, in particular,
\begin{align}
&\alpha_0c_0^0+\alpha_0c_0^1+\dots+\alpha_nc_0^n=0\\
&\alpha_0c_1^0+\alpha_0c_1^1+\dots+\alpha_nc_1^n=0\\
&\dots\\
&\alpha_0c_n^0+\alpha_0c_n^1+\dots+\alpha_nc_n^n=0
\end{align}
where $c_0$, $c_1$, $\dots$, $c_n$ are pairwise distinct points in $[-1,1]$.
Therefore you have a homogeneous linear system with matrix
$$\begin{bmatrix}
1 & c_0 & c_0^2 & \dots & c_0^n \\
1 & c_1 & c_1^2 & \dots & c_1^n \\
\dots & \dots & \dots & \dots & \dots \\
1 & c_n & c_n^2 & \dots & c_n^n
\end{bmatrix}\,.$$
This is a Vandermonde matrix, therefore invertible.
A: A more verbal solution. Assume that one of the monomials, say $x^p$, linearly depend on the other monomials. Then it would be possible to write $x^p$ as linear combination of several $x^q$, where $q\ne p$. But this is clearly impossible and the contradiction proves that $x^p$, for any $p$, does not depend on the set of remaining $x^q$. 
