If $T$ is the derivative operator and $(v_0,…,v_m)$ is a standard basis, how can I prove that $Tv_k \in {\rm span}(v_0,…,v_k)$?

If I have that:

$p \in \mathbb{P}_m(\mathbb{C})$ is the set of all polynomials with complex coefficients with degree less than or equal to m.

And that $T$ is the differentiation operator:

$Tp=p'$

with $(v_0,...,v_m)=(1,x,x^2, ...,x^m)$ being a standard basis, how can I show that

$Tv_k \in {\rm span}(v_0,...,v_k)$ for $k=0,1,...,m$?

My work:

$Tv_k = T(a_01+a_1x+a_2x^2+...+a_kx^k)=a_0T(1)+a_1T(x)+...+a_kT(x^k)$ for $a_0,...,a_k \in \mathbb{C}$

$=a_1+2a_2x+...+a_kkx^{k-1}$ which is obvious $\in {\rm span}(v_0,...,v_k)$.

My main confusion here is if I needed to let

$Tv_k = T(a_01+a_1x+a_2x^2+...+a_kx^k+...+a_mx^m)$

instead of only up to $k$ terms as I did right under the bolded "My work" part. Would it make a difference?

Thanks!

• The question should be "...show that $Tv_k\in{\rm span}(v_0,\ldots,v_k)$", should it not? – David Feb 15 '14 at 7:28
• ah, just changed it, thanks! – user123276 Feb 15 '14 at 7:30
• I think you changed it in the title but not actually in the question ;-) – David Feb 15 '14 at 7:33

What you have done is more or less right, but way more work than you actually need. The question only asks for $T(v_k)$, so you don't need to look at $T$ of any old polynomial. All you need is $$T(v_k)=T(x^k)=kx^{k-1}\ ,$$ and this is in ${\rm span}(v_0,v_1,\ldots,v_k)$ because. . . ?
Note that $v_k=x^k$, so $\def\sp{\operatorname{span}}T(v_0)=0\in\sp(v_0)$ and $T(v_k)=kv_{k-1}\in\sp(v_0,\ldots,v_k)$ for $k>0$.
It is curious that the question asks about $\sp(v_0,\ldots,v_k)$ rather than $\sp(v_0,\ldots,v_{k-1})$ which would have sufficed, or even $\sp(v_{k-1})$ when $k>0$.