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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.

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  • $\begingroup$ You will not get the zero transformation, since not all polynomials have degree $\leq n-2$. $\endgroup$ Feb 20 '15 at 8:48
  • $\begingroup$ I guess what is confusing me is that the set of all polynomials does include those have degree $n-2$, in which case the set does not seem to be linearly independent. $\endgroup$ Feb 20 '15 at 8:49
  • $\begingroup$ But just because some transformation is zero on that subset does not mean it is zero on the entire space. $\endgroup$ Feb 20 '15 at 8:51
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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 $0$.

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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.

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  • $\begingroup$ so the error in my reasoning is that because the zero vector for the vector space of all linear transformations from $V$ to itself is $T_{0}:V\longrightarrow \{0\}$, I would need to show that for every $f(x)\in P(\mathbb{R} )$,there should be at least one derivative $f^{(j)}(x)$ that equals the zero vector. In the case of $f(x)=x^{n+1}$, we can see that the resulting polynomials belongs to $P_{n}(\mathbb{R})$ and therefore shows that not every $f(x)$ is mapped to the zero polynomial. Correct? $\endgroup$ Feb 20 '15 at 18:55
  • $\begingroup$ Not sure I get what you're saying, the zero polynomials also belongs to $P_n(\mathbb R)$, so that's not the point. Your argument fails, because for example $T_n x^{n+1} = (n+1)! x\neq 0$, so the linear combination you propose isn't the zero-operator. $\endgroup$
    – Christoph
    Feb 20 '15 at 19:00
  • $\begingroup$ Sorry I guess I should have said that for any $x^{k}$ for $k\geq n+1$ no ith derivative for $1\leq i\leq n$ can be the zero polynomial and so the only way a combination of them can equal the zero polynomial is if all of the coefficients are zero. Sorry if what I said before is confusing I'm trying to give a correct explanation in my own words to ensure my own understanding. $\endgroup$ Feb 20 '15 at 20:21

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