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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?

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morrowmh
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  • Regarding the fact that changing $a$ changes the difference in dimensions: note that we could have $V \subsetneq \Bbb R[x]$. – Ben Grossmann May 03 '20 at 00:11
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    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 $V_{n+1}$). – Arturo Magidin May 03 '20 at 00:15
  • You might find it helpful to consider an example. For instance, $V = \operatorname{span}({x^3,x+1})$. Consider what happens relative to $a = 0,1,2$. – Ben Grossmann May 03 '20 at 00:17

2 Answers2

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

Ben Grossmann
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  • 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 $D(x^3) = 3x^2\notin V$. – Arturo Magidin May 03 '20 at 00:25
  • @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 – Ben Grossmann May 03 '20 at 00:26
  • 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... – Arturo Magidin May 03 '20 at 01:00
  • @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. – Ben Grossmann May 03 '20 at 01:05
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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 $n = n_i$.

levap
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  • For point 2, did you mean $^\varphi=1^$ implies $\dim V_{n+1}=\dim\text{ker}(\varphi)=\dim V_n-1$? – morrowmh May 03 '20 at 18:46
  • @Sunyata: Sorry, I meant $\varphi \neq 0$. – levap May 03 '20 at 19:28
  • Last question: how do we know $V=V_0$? Is this vacuous? – morrowmh May 03 '20 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 $k$. – levap May 03 '20 at 21:43
  • Got it, thanks. – morrowmh May 03 '20 at 21:47