Let $n$ be a positive integer and $V_n$ be the vector space over $\mathbb{R}$ which consists of all the polynomials in the variable $t$ of degree at most $n$ with real coefficients. Show that $\{1,1+t,t+t^2,t^2+t^3,\dots , t^{n-1}+t^n\}$ is a basis of $V_n$.
My proof so far:
Let $V_n$ be defined as above and $X=\{1,1+t,t+t^2,t^2+t^3,\dots , t^{n-1}+t^n\}$.
First show that $X$ is linearly independent. Consider $f(t)=$
$c_0+c_1(1+t)+c_2(t+t^2)+c_3(t^{2}+t^3)+\cdots + c_{n-1}(t^{n-2}+t^{n-1}) + c_n(t^{n-1}+t^n) =$
$ (c_0+c_1)+(c_1t+c_2t)+(c_2t^2+c_3t^2)+\cdots + (c_{n-1}t^{n-1} + c_nt^{n-1})+c_nt^n= $
$ (c_0+c_1)+(c_1+c_2)t+(c_2+c_3)t^2+\cdots + (c_{n-1} + c_n)t^{n-1}+c_nt^n= 0$.
Here
Now show that span$_\mathbb{R}X=V_n$. Let $f(t)\in V_n$. Then $f(t)=(c_0+c_1)+(c_1+c_2)t+(c_2+c_3)t^2+\cdots + (c_{n-1} + c_n)t^{n-1}+c_nt^n$ for $c_i\in\mathbb{R}$. So $V_n=$span$_\mathbb{R}X$. Thus $X$ is a basis of $V_n$.$\square$
The Here is where I am getting tripped up. I know that the derivative can be used to show that all the coefficients must be 0, but I am not sure how. If I take the $n^{th}$ derivative: Then $f^{(n)}(0)=c_nn!=0$. Thus $c_n=0$. But how can I show that $c_i=0$ for $i<n$?
Best Answer
Like you said it's simple enough to show that $c_{n}=0$. Now knowing that when we examine $f^{(n-1)}(t)$ we see that $c_{n-1}=0$. Knowing this we can show that $c_{n-2}$ is also $0$ by the same logic.
We can intuitively see that this process carries all the way down to finding $c_{0}=0$. To formally show it just use induction.