# Proving that this set is linearly independent

Ler $V=P(\Bbb R)$ be the vector space of the polynomials with real coefficients, on the field of real numbers $\Bbb R$. For $i \geq 1$, let $T_i(f)=f^{(i)}$ the $i$th derivate of $f$. I have to show that for any $n \in \Bbb N$, $\{T_1, T_2,..., T_n\}$ is a linearly independent subset of $L(V)$ (the set of oprators of $V$.)

$\textbf{Attempt:}$ Suppose there exist $c_1,...,c_n \in \Bbb R$ such that $$c_1T_1 + \cdots c_n T_n = 0.$$ Then, for any $f \in P(\Bbb R)$, we have $$(c_1T_1 + \cdots c_n T_n)(f)=c_1 T_1 (f) + \cdots c_n T_n (f) = 0.$$

We must show that $c_1 = c_2 = \cdots = c_n =0.$ Any $f \in P(\Bbb R)$ is of the form

$$f=a_m x^m + \cdots + a_1 x + a_0.$$

for some $m \in \Bbb N$. Also, we have that

$$f^{(k)}=\sum_{i=k}^m \bigg( \prod_{j=0}^{k-1}(i-j) \bigg)a_i x^{i-k}.$$

If $m \geq n$, then $T_n(f)=f^{(n)}$ is a linear combination of $1, x, x^2, ..., x^{m-n}$, which is a linearly independent set, and then their coefficients are all zero. Then, we substitute this coefficients in $f^{(n-1)}$, which is a linear combination of $1, x, x^2, ..., x^{m-n}, x^{m-n+1}$ and we obtain that the coefficient of $x^{m-n+1}$ is zero. We continue with this process and we obtain $c_1=c_2=\cdots =c_n = 0$.

I think this proof is not complete or even correct because I only handled the case $m \geq n$, and I don't know what to do with the case $m<n$, and also $m-n$ could be greater than $n$ and then we'd have that the coefficients of $1, x, x^2, ..., x^{m-n}$ are more than $n$ coefficients, so I'm really confused and I don't know what else I could do. I hope you can help giving some idea to solve this problem. Thank you in advance.

A slightly simpler proof than the others is to evaluate $(c_1T_1+\cdots+c_nT_n)=0$ at $x^n$. You get the polynomial $$c_1nx^{n-1}+c_2n(n-1)x^{n-2}+\cdots+c_nn!$$ Since this is the zero polynomial, all the coefficients are zero in a single step.

A much more direct proof is to just consider a few specific polynomials, namely $x^n$ for various $n$. We see that $x' = 1$ and $x'' = 0$, so

$$(c_1 T_1 + ... + c_n T_n) x = c_1 \cdot 1 + 0 + ... + 0 = 0$$

implies that $c_1 = 0$.

Next, we have

$$(c_1 T_1 + c_2 T_2 + ... + c_n T_n) x^2 = 0 + c_2 (x^2)'' + c_3 (x^2)''' + ... + c_n (x^2)^{(n)} = 2c_2 = 0$$

So $c_2 = 0$. Now proceed similarly to show that each $c_k$ is $0$.

• $4$ seconds. ${}{}{}$ – Pedro Tamaroff Oct 12 '13 at 1:59
• @PedroTamaroff Ha, indeed. – user61527 Oct 12 '13 at 2:00

I think this is simpler. Suppose that $$\alpha_1T_1+\cdots+\alpha_nT_n=0$$

Then evaluate at $P(X)=X$. Since $P'=1$, $P''=0$, we get that $$\alpha_1=0$$

Now evaluate at $P(X)=X^2$. Then $P''=2$; but $P'''=0$, so...?

(Observe you have a proof that $(\Bbb R[X])^\ast$ cannot be finite dimensional!)