Subspace of polynomials vanishing to order $n$ 
Let $V\subset\mathbb{R}[x]$ be a vector space of dimension $k$. We say that a polynomial $f$ vanishes to order $n$ at $a\in\mathbb{R}$ if $f(a)=0$ and $n$ is the smallest positive integer such that $f^{(n)}(a)\not=0$.
a) Show that $V_n=\{f\in V|f\text{ vanishes to order}\geq n\text{ at }a\}$ is a subspace of $V$.
b) Let $a\in\mathbb{R}$. Show that $\dim(V_n)-\dim(V_{n+1})$ is either $0$ or $1$.
c) Conclude that there are precisely $k$ integers $n$ such that there exists a nonzero $f\in V$ that vanishes to order $n$ at $a$.

I have already proved part a. I think we can use rank-nullity on part b, but I am not sure how. I don't see how simply changing $a$ will cause the difference in dimensions to change. And how can I use this to prove part c? Any advice?
 A: Part b: Let $W = \Bbb R_m[x]$ denote the set of polynomials of degree at most $m$, where $m$ has been chosen so that $V \subset W$.  Note that $\dim(W_n) = m+1-n$, and $W_{n+1} \subset W_n$. Let $S_n$ denote a subspace of $W_n$ such that $W_{n} = S_n \oplus W_{n+1}$; note that $\dim(S_n) = 1$.
Now, we note that $V_n = W_n \cap V$, and that $V_{n+1} \subset V_n$. With that, we have $\dim(V_{n+1}) \leq \dim(V_n)$ and
$$
\begin{align}
\dim(V_{n+1}) &= \dim(W_{n+1} \cap V) 
\\ & \geq \dim([W_{n+1} \oplus S] \cap V)  - \dim(S \cap V)
\\ & = \dim(W_n\cap V) - \dim(S \cap V) = \dim(V_n) - \dim(S \cap V)
\\ & \geq \dim(V_n) - \dim(S) = \dim(V_n) - 1.
\end{align}
$$
The conclusion follows.
Regarding part c, it is helpful to note that $V_0 = V$, note that there exists an $n$ such that $V_n = \{0\}$, and consider the sequence $\dim(V_0),\dim(V_1),\dim(V_2),\dots$.
A: For $(b)$, let $\varphi \colon V_n \rightarrow \mathbb{R}$ be the linear functional given by $\varphi(f) = f^{(n)}(a)$. Note that
$$ V_n = \{ f \in V \, | \, f(a) = f'(a) = \dots = f^{(n-1)}(a) = 0 \}, \\ 
\ker(\varphi) = \{ f \in V_n \, | \, f^{(n)}(a) = 0 \} = \{ f \in V \, | \, f(a) = \dots = f^{(n-1)}(a) = f^{(n)}(a) = 0 \} = V_{n+1} $$
and so by rank-nullity or directly, there are two options:


*

*Either $\varphi = 0$ and then $V_{n+1} = \ker(\varphi) = V_n$.

*Or $\varphi \neq 0$ and then $\dim V_{n+1} = \dim \ker(\varphi) = \dim V_n - 1$.


For part $(c)$, let $M \geq k - 1$ be the maximal degree of an element in $V$ (this exists because $V$ is finite dimensional) and consider the sequence of descending subspaces
$$ V := V_0 \supseteq V_1 \supseteq V_2 \supseteq V_3 \supseteq \dots \supseteq V_{M+1}. $$
Prove that $V_{M+1}= \{ 0 \}$. Taking dimensions, we get
$$ k = \dim V \geq \dim V_1 \geq \dim V_2 \geq \dim V_3 \geq \dots \geq 0. $$
By part $(b)$, whenever we have a strict inequality the dimension decreases by $1$ and so there are precisely $k$ non-negative distinct integers $0 \leq n_1,\dots,n_k \leq M$ such that $\dim V_{n_i + 1} + 1 = \dim V_{n_i}$. Finally, note that $0\neq f \in V$ vanishes to order $n$ iff $f \in V_n \setminus V_{n+1}$ which implies that $\dim V_n > \dim V_{n+1}$ which happens iff $n = n_i$.
