1. ## Functional problem

Let $V=R_n[x]$ and $\alpha_i(p)=p^i(0)$ be a functional over $V$( $p^i$ is the number i derivative of p)
Show that $\{\alpha_i\}_\forall\i$ is a basis of a dual space $V^*$

What I first should do is prove that $\alpha_i \forall i$ are linearly independent.
How should I do it?
Any ideas?

2. Originally Posted by wicked
Let $V=R_n[x]$ and $\alpha_i(p)=p^i(0)$ be a functional over $V$( $p^i$ is the number i derivative of p)
Show that $\{\alpha_i\}_\forall\i$ is a basis of a dual space $V^*$

What I first should do is prove that $\alpha_i \forall i$ are linearly independent.
How should I do it?
Any ideas?
Hint: notice that $\alpha_i(x^k)$ is $k!$ if i=k, and 0 if i≠k.

3. Originally Posted by Opalg
Hint: notice that $\alpha_i(x^k)$ is $k!$ if i=k, and 0 if i≠k.
$\alpha_0,....,\alpha_n$are independent iff
$\sum\lambda_i\alpha_i=0$

So if $p=a_0+a_1x+....+a_nx^n$
$\alpha_0(p)=(a_1+2a_2+...+na_nx^(n-1))(0)=a_0$
.
.
.
$\alpha_n(p)=(0+0+...+n!a_nx^0)(0)=n!a_n$
right?
Then $\forall a_1....a_n$
$\lambda_0\alpha_0+....+\lambda_n\alpha_n=0$

and therefore it's obvious that $\lambda_0=\lambda_1=.....=\lambda_n=0$

And therefore $dimV^*=dimV=n+1$ => $\{\alpha_i\}$ is basis.
This is more or less is the answer right?