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?

  • $\begingroup$ Regarding the fact that changing $a$ changes the difference in dimensions: note that we could have $V \subsetneq \Bbb R[x]$. $\endgroup$ May 3, 2020 at 0:11
  • 1
    $\begingroup$ 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 $V_{n+1}$). $\endgroup$ May 3, 2020 at 0:15
  • $\begingroup$ You might find it helpful to consider an example. For instance, $V = \operatorname{span}(\{x^3,x+1\})$. Consider what happens relative to $a = 0,1,2$. $\endgroup$ May 3, 2020 at 0:17

2 Answers 2


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

  • $\begingroup$ 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 $D(x^3) = 3x^2\notin V$. $\endgroup$ May 3, 2020 at 0:25
  • $\begingroup$ @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 $\endgroup$ May 3, 2020 at 0:26
  • $\begingroup$ 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... $\endgroup$ May 3, 2020 at 1:00
  • $\begingroup$ @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. $\endgroup$ 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 $n = n_i$.

  • $\begingroup$ For point 2, did you mean $^*\varphi=1^*$ implies $\dim V_{n+1}=\dim\text{ker}(\varphi)=\dim V_n-1$? $\endgroup$ May 3, 2020 at 18:46
  • $\begingroup$ @Sunyata: Sorry, I meant $\varphi \neq 0$. $\endgroup$
    – levap
    May 3, 2020 at 19:28
  • $\begingroup$ Last question: how do we know $V=V_0$? Is this vacuous? $\endgroup$ May 3, 2020 at 20:25
  • $\begingroup$ @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 $k$. $\endgroup$
    – levap
    May 3, 2020 at 21:43
  • $\begingroup$ Got it, thanks. $\endgroup$ May 3, 2020 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.