For a fixed $a \in R$, determine the dimension of the subspace of $P_n(R)$ defined by ${f \in P_n (R): f(a)=0}$. Here is the question: For a fixed $a \in R$, determine the dimension of the subspace of $P_n(R)$ defined by ${f \in P_n (R): f(a)=0}$.
Since $f(a) =0$, we can write $a_n x^n+ a_n-1 x^{n-1}+... a_0 x^0$=0
How do I argue that for $P_n (R)$ in this case, there are $n$ dimensions?
 A: Each polynomial can be wroten as $p(x) = \sum_{i=0}^n a_i x^i$ where you can see that {$x^i$}$_i$ is a base of $P_n$. So, $P_n$ has n+1 dimentions (note that from $0$ to n we have n+1 terms).
If $p(a)=0$ we have that p(x) = (x-a)($\sum_{i=0}^{n-1}b_ix^i$) at least, by the fundamental theorem of algebra. So, the space of this functions have n dimentions (this new sum only have n elements).
A: Hint: What is the degree of the polynomial once you factor the root $(x-a)$ from  $f$? If $a$ is the only root, then $g(x)(x-a) = f(x)$ and $g(x)$ would be a basis.
The reason that the set $1, x, x^2, ..., x^n$ is a basis for $P_n(\mathbb{R})$ is because by definition a polynomial $p \in P_n(\mathbb{R})$ has the form: $$p(x)= a_n x^n+a_{n-1}x^{n-1}+...a_0$$ thus a linear combination of the selected basis hits all vectors in the space. We need not choose this set as a basis, however. Any linearly independent set of vectors which has a length corresponding to the dimension of the space will also be a basis.
