Dgdrxrt

I'm trying to solve this question here:

Let $V = F_n-1left [ x right ]$ over some field F (i.e. V is the vector space of all polynomials with degree smaller or equal to n-1), and $x_1, x_2, ..., x_n in F$ be n different scalars. Assume $A_1, ... , A_k$ are disjoint sets, such as $left x_1,...,x_n right = sqcup _i=1^k A_i$.
Define $V_i = left p(x)in V mid forall x_j notin A_i, p(x_j) = 0 right $.

We need to prove that $V = oplus _i=1^k V_i$.

I managed to prove this in the case in which $k=n, A_i = left x_i right $, but I'm having trouble how to use this in order to prove the general case.

The nice thing about this question is you can be extremely concrete. Taking the $k=n$, $A_i=x_i$ case, we see that we have a basis $e_i(x)=prod_ineq j frac(x-x_j)x_i-x_j$ of $V$. (Clearly each $e_iin V_i$, $e_i(x_i)=1$, and moreover if $sum_j a_j e_j(x)=0$, then evaluating at $x_i$, $a_i=sum a_j e_j(x_i)=0$. Since we have $n$ linearly independent vectors in an $n$-dimensional space, it is a basis.)

Now we apply this to the general case by expressing everything in this basis. First, we have $V_i=bigoplus_x_jin A_i F e_j(x)$. Indeed, clearly each $e_jin V_i$ if $x_jin A_i$, and conversely any $p(x)in V_i$ can be written $p(x)=sum a_i e_i(x)$, and satisfies $0=p(x_j)=a_j$ for $x_jnotin A_i$, hence must be in the span of the $e_j$ with $x_jin A_i$.

Finally, note that the span of the $V_i$ contains the span of the basis, hence $sum V_i=V$, and since $V_i$ and $V_j$ contain distinct basis elements when $ineq j$ (due to the disjointness of $A_i$ and $A_j$), $V_icap V_j=emptyset$. The result follows.

For each $jin1,2,ldots,k$, $dim V_j=n-(n-#A_j)=#A_j$. And it is not hard to prove that $ineq jimplies V_icap V_j=0$. So,beginaligndimleft(bigoplus_j=1^kV_jright)&=sum_j=1^kdim V_j\&=sum_j=1^k#A_j\&=nendalignand therefore $bigoplus_j=1^kV_j=V$.