In this problem, let $n$ be a positive integer, and let $r$ and $q$ be distinct real numbers. We want to show that every polynomial $p \in P_{2n+1}(\mathbb{R})$ can be written as a sum $p = u + w$, where $u,w \in P_{2n+1}(\mathbb{R})$ and $u(r) = u'(r) = ...= u^n(r) = w(q) = w'(q) = ...=w^n(q) = 0$
My Attempt at a Solution
Let $U = \{u \in P_{2n+1}| u(r)=...=u^n(r) = 0\}, W = \{w \in P_{2n+1}|w(r)=...=w^n(r) = 0\}$.
We know that we can show that a subspace $U$ of a vector space $V$ is equivalent to V if and only if $\dim{U} = \dim{V}$. We know that $\dim{P_{2n+1}(\mathbb{R})} = 2n+2.$ We need to show that $\dim{P_{2n+1}(\mathbb{R})} = 2n+2 = \dim{(U+W)}$. To calculate the dimensions, employ the standard basis of $P_{2n+1}$ and differentiate it m times and input the value.
My Question
a). How do I compute the $\dim{U}$ and $\dim{W}$ and $\dim{(U \cap W)}$? I know I can start with the standard basis of $P_{2n+1}$ and perhaps repeatedly differentiate it, but I still have trouble in calculating.
b) Is my approach on the right track, or is there a better way to approach this problem?
Fix $p,q\in\mathbb{R}$, with $p\ne q$, and let $U,V$ be given by \begin{align*} U&=\{u \in P_{2n+1}(\mathbb{R}){\,\mid\,} u(p)=u'(p)=\cdots =u^{(n)}(p) = 0\}\\[4pt] V&=\{v \in P_{2n+1}(\mathbb{R}){\,\mid\,} v(q)=v'(q)=\cdots =v^{(n)}(q) = 0\}\\[4pt] \end{align*}
It's clear that $U,V$ are subspaces of $P_{2n+1}(\mathbb{R})$.
To show $U+V=P_{2n+1}(\mathbb{R})$, it suffices to show that $U+V$ contains a linearly independent subset of $2n+2$ elements.
Letting $f=(x-p)^{n+1}$, and $g=(x-q)^{n+1}$, we can recast $U,V$ as \begin{align*} U&=\{u\in P_{2n+1}(\mathbb{R}){\,:\,}f{\,|\,}u\}\\[4pt] V&=\{v\in P_{2n+1}(\mathbb{R}){\,:\,} g{\,|\,}v\}\\[4pt] \end{align*} Suppose $w\in U\cap V$.
Then from $w\in U$, we get $f{\,|\,}w$, and from $w\in V$, we get $g{\,|\,}w$.
Hence, since $p\ne q$, it follows that $fg{\,\mid\,}w$.
But then since $fg$ has degree $2n+2$, it follows that $w=0$.
Thus, $U\cap V=\{0\}$.
Let $A=\{x^kf{\;\mid\;}0\le k\le n\}$, and let $B=\{x^kg{\;\mid\;}0\le k\le n\}$.
Then $A\subset U$, hence $A\subset U+V$, and $B\subset V$, hence $B\subset U+V$.
Thus, $A\cup B\subset U+V$.
Since the polynomials $1,x,...,x^n$ are linearly independent, it follows that each of the sets $A,B$ is linearly independent, and moreover, since $U\cap V=\{0\}$, it follows that the set $A\cup B$ is linearly independent.
Thus, $A\cup B$ is a linearly independent subset of $U+V$, and has $2n+2$ elements.
This completes the proof.