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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 $i<n$?

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  • $\begingroup$ It would be much easier to write down the matrix of the linear transformation from the canonical basis to this set and prove that it is invertible. It is an upper triangular matrix with diagonal entries $1$. $\endgroup$ – user258700 Jan 23 '16 at 20:01
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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.

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Let $b_k$ be the basis elements (that is, $b_0(t) = 1, b_1(t) = t+1,...$). Suppose $f=\sum_{k=0}^{n-1} \alpha_k b_k = 0$.

Then ${d ^n f(0) \over dt^n } = n! \alpha_{n-1} = 0$.

Hence $f=\sum_{k=0}^{n-2} \alpha_k b_k = 0$ and so ${d ^{n-1} f(0) \over dt^{n-1} } = (n-1)! \alpha_{n-2} = 0$.

Etc, etc.

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If a polynomial is identically equal to zero, meaning it takes on the value $0$ for all possible values of $t$, as is the case in your step for linear independence, then all the coefficients must be zero.

Another way to argue is that $\{1,t,t^2,\ldots , t^{n}\}$ forms a basis, hence linearly independent, so the only linear combination resulting in zero polynomial must be the zero combination. Thus you have,

$$c_{i}+c_{i+1}=0 \qquad \forall 0 \leq i < n$$ And $c_n=0$.

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