Linear Independence of Linear Transformations Define $V=P(\mathbb{R})$ and for $j\geq 1$ we define $T_j(f(x))=f^{(j)}(x)$ (jth derivative of $f$). We want to show that the subset $\{T_1, T_2,...T_n\}$ of the vector space $L(V)$ of linear transformations from a vector space V to itself is linearly independent. 
My question is why can we say this set is linearly independent when any g(x) with $deg(g(x))\leq n-2$ will produce the zero polynomial if plugged in to $T_i$ for $i\geq n-1$ and thus we can set the coefficient $a_i=0$ of any $T_i$ for $1\leq i\leq n-2$  and use any nonzero value of $a_i$ for $i=n-1$ and $i=n$ and still obtain the zero transformation as a result?
Thanks in advance.
 A: Suppose that $$\sum_{i=1}^n a_i T_i = 0$$
This means that for any polynomial $f$ you have
$$a_1f'+a_2f''+ \cdots+ a_nf^{(n)} \equiv 0$$ is the constant polynomial $0$.
Now consider the polynomial $f=x^{n+1}$. It is clear that
$$\sum_{i=1}^n a_i T_i(x^{n+1}) = \sum_{i=1}^n ((n+1) \cdots (n-i+2))a_i x^{n+1-i} = 0$$
implies that all coefficients are $0$, so all $a_i$ are $0$.
A: The space $V=P(\mathbb R)$ contains polynomials of arbitrary degree, while you have only considered polynomials of degree $\le n-2$. So what you showed is the following: Let $$U = \{\, f \in V : \deg f \le n-2\,\}$$ be the subspace of polynomials of degree $\le n-2$, then $$\{\left.T_1\right|_U, \dots, \left.T_n\right|_U\}$$ is a linear dependent subset of $L(U)$. This is of course true, since $\left.T_{n-1}\right|_U = \left.T_n\right|_U = 0$ and any collection of vectors containing $0$ is linearly dependent.
However, considering $T_{n-1}$ and $T_n$ on the space of all polynomials over $\mathbb R$, these operators are not the zero operators. Hence your argument fails.
