# Dimension of a subspace of a vector space

Let $$V$$ = $$P_{n}(\Bbb{R})$$ be a vector space of polynomials with real coefficients up to degree $$n$$.

Let $$W = \{ p(x)\in V\mid p(a) = p'(a) = p''(a)=\ldots=p^{(r)}(a) = 0 \}$$

What is the dimension of $$W$$?

I can notice that if $$p(x)$$ belongs to $$W$$ then $$x-a$$ will be a factor of each of $$p(x),p'(x),p''(x),\ldots, p^{(r)}(x)$$ but still I am unable to explicitly write this polynomial to find the dimension of the subspace.

• Indeed, we have the stronger statement that $(x-a)^r$ will be a factor of $p(x)$ ... does that help? Commented Jun 13, 2020 at 20:29
• (n+1)-(r+1)?..isn't it? Commented Jun 13, 2020 at 20:37

The space $$V$$ consists of those polynomials $$p(x)$$ of the form $$(x-a)^{r+1}q(x)$$, with $$\deg q(x)\leqslant n-(r+1)$$ (I am assuming that $$r). Therefore, $$\dim V=n-r$$.

• How do we get that factor? Commented Jun 13, 2020 at 20:33
• @Bernard Right! I've edited my answer. Thank you. Commented Jun 13, 2020 at 20:33
• @Gitika: With Taylor's formula at order $n$, which is an exact formula for polynomials. Commented Jun 13, 2020 at 20:37
• @Gitika Write your $p(x)$ as a polynomial in $x-a$:$$p(x)=a_0+a_1(x-a)+a_2(x-a)^2+\cdots+a_n(x-a)^n.$$Then $p(a)=0\iff a_0=0$, $p'(a)=0\iff a_1=0$ and so on. So,\begin{align}p(x)&=a_{r+1}(x-a)^{r+1}+\cdots+a_n(x-a)^n\\&=(x-a)^{r+1}(a_{r+1}+\cdots+a_n(x-a)^{n-r}).\end{align} Commented Jun 13, 2020 at 20:38
• Why is p(x) of that form? Commented Jun 13, 2020 at 20:43

hint

Assuming $$r, by Taylor-Lagrange formula,

$$p(x)=\sum_{k=0}^n\frac{(x-a)^k}{k!}p^{(k)}(a)$$

$$=\sum_{k=r+1}^n\frac{(x-a)^k}{k!}p^{(k)}(a)$$

Here is an ad hoc way:

Let $$b_k(x) = (x-a)^k$$, $$k=r+1,...,n$$, note that the $$b_k$$ are linearly independent.

Note that $$b_k \in W$$ for $$k=r+1,...,n$$ hence $$\dim W \ge n-r$$.

Now suppose $$p \in W$$. Using synthetic division is it straightforward to show that $$p(x) = (x-r)^{r+1} g(x)$$. Since $$p \in P_n(\mathbb{R})$$ we see that $$\partial g \le n-r-1$$ and we can write $$g(x) = g_0+g_1(x-r)+\cdots + g_{n-r-1}(x-a)^{n-r-1}$$ and so $$p = g_0 b_{r+1}+g_1 b_{r+2} + \cdots +g_{n-r-1}b_n$$. Hence $$W = \operatorname{sp} \{ b_{r+1}, \cdots , b_n \}$$ and so $$\dim W = n-r$$.