Take $P_{n+1}(\Re)$ to be the space of polynomials with coefficients in the real numbers, with the degree of all polynomials $\le n+1$. I want to show that, given any $n$ points $a_1, a_2, ..., a_n \in \Re,$ $\exists f \in P_{n+1}(\Re), f\ne0,$ s.t. the function is zero at all of those points $a_i$, but the sum of the derivatives at those points is $0$.
I don't know where to start. The hint in the textbook says to use the dimension theorem, but I don't see how that can be used here.
On
Another hint:
Consider the map \begin{align} \varphi:P_{n+1} &\longrightarrow \bf R^{n+1}\\ f&\longmapsto\begin{bmatrix}f(a_1)\\f(a_2)\\\vdots\\f(a_n)\\\sum\limits_{i=1}^n f'(a_i)\end{bmatrix} \end{align} You have to check that $\varphi$ is a linear map, and that $\;\dim(\ker\varphi)>0$.
On
So I'm not sure which dimension theorem you are referring to, but there is a dimension counting argument to get the answer.
The key thing to note is that "evaluation at $a_i$" $f\rightarrow f(a_i)$ is a linear map from your space of polynomials to the reals.
Then, note that differentiation is also a linear map $P_{n+1}(\Re)\rightarrow P_{n+1}(\Re)$, so composing and summing these gives that the "evaluate derivative at $a_i$ and sum" operator is linear, from $P_{n+1}(\Re)$ to $\mathbb{R}$.
So then we have $n+1$ linear maps from $P_{n+1}(\Re)$ to $\mathbb{R}$, the evaluations, and the "derivative then evaluate then sum". Then note, your space is $n+2$ dimensional.
At this point, you should be able to use a dimension theorem to give the final result, that there is a common zero to all of them. (though note that as stated, the zero map also works as a solution to the question).
Hint:
Define the mapping from $P_{n+1}(\mathbb R)$ to $\mathbb R^{n+1}$, defined as
$$p\mapsto \left[p(a_1), p(a_2),\dots, p(a_n), \sum_{i=1}^n p'(a_n)\right]$$
and notice that it is a linear mapping.