# 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?

• Regarding the fact that changing $a$ changes the difference in dimensions: note that we could have $V \subsetneq \Bbb R[x]$. May 3, 2020 at 0:11
• There is no “change in $a$.” You pick $a$, you get your $V_n$ and $V_{n+1}$ relative to that $a$, and the question is about those two. There is one fixed $a$ throughout. As to part (b), consider the differentiation map from $V_{n+1}$ to $V_n$ (or from $V$ to itself, and what it does when you apply it to an element of $V_{n+1}$). May 3, 2020 at 0:15
• You might find it helpful to consider an example. For instance, $V = \operatorname{span}(\{x^3,x+1\})$. Consider what happens relative to $a = 0,1,2$. May 3, 2020 at 0:17

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 $$\dim(V_0),\dim(V_1),\dim(V_2),\dots$$.

• Actually, it is not clear that $V$ is closed under $D$, is it? E.g., we could take the vector space spanned by $x^3$. Then $D(x^3) = 3x^2\notin V$. May 3, 2020 at 0:25
• @ArturoMagidin It is not. I'm sure there must be something involving the map $D$ here but having thought about it again I'm not sure what the exact connection is May 3, 2020 at 0:26
• Yup, I was just getting there. I was going to pick the basis $1,(x-a),(x-a)^2,\ldots,(x-a)^m$ and go from there... May 3, 2020 at 1:00
• @ArturoMagidin I have a feeling that there's a more elegant way to package this proof that I'm missing, though. At the very least, I could be more specific and take $S_n = \operatorname{span}\{(x-a)^n\}$, but that doesn't seem very illuminating on its own. May 3, 2020 at 1:05

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:

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

• For point 2, did you mean $^*\varphi=1^*$ implies $\dim V_{n+1}=\dim\text{ker}(\varphi)=\dim V_n-1$? May 3, 2020 at 18:46
• @Sunyata: Sorry, I meant $\varphi \neq 0$. May 3, 2020 at 19:28
• Last question: how do we know $V=V_0$? Is this vacuous? May 3, 2020 at 20:25
• @Sunyata: I defined it to be $V$. In fact, the problem as stated is wrong because you need to define that $f = 0$ vanishes to order $n$ for all $n \in \mathbb{N}_0$ (otherwise, $V_n$ isn't even a subspace with your given definition) and also say that $f$ vanishes to order zero at $a$ if $f(a) \neq 0$. Then, the correct conclusion is that there are precisely $k$ non-negative integers $k$ (you must include zero) such that there exists $f \neq 0$ which vanishes to order $k$. May 3, 2020 at 21:43
• Got it, thanks. May 3, 2020 at 21:47