I have a homework assignment, that I can't quite wrap my head around, let me state the question, and show my progress.
Let $n$ be a natural number, and let $\alpha \in \mathbb{R}$ be a real scalar. Consider the subset \begin{equation*} V=\{p\in P_n(\mathbb{R}):p(\alpha)=0\}, \end{equation*} of the real vector space $P_n(\mathbb{R})$.
First I need to show, that $V$ is a subspace of $P_n(\mathbb{R})$, and as far as I figure, $V$ is the set containing all polynomials of degree $\leq n$ where no matter the variable $\alpha$ the polynomial equals $0$.
I've then showed that since all elements are $0$ it's closed under addition, scalar multiplication and it ofcourse contains $0$ itself. I assume what I have done so far is correct, but am unsure.
Next up I need to decide the dimension of $V$ and I have been hinted that I need to use the formular $\text{dim}(v)=\text{dim}(\text{ker}(L))+\text{dim}(L(V))$ on the imaging $L:P_n(\mathbb{R})\rightarrow\mathbb{R}$ defined as $L(p)=p(\alpha)$.
I assume $\ker(L)=0$ since all coefficients in the polynomial has to be $0$, and therefore $\dim(\ker(L))=0$ as well. Same goes for $L(V)=0$ and $\dim(L(V))=0$, so is the answer to this question just $0$?
The reason why I'm unsure of myself is because I feel it's almost too easy with all these zeroes.
Sorry if some of the terminology is wrong, english is not my first language.
By Bézout's Theorem we have $$p\in V\iff p(x)=(x-\alpha)q(x),$$ where $q\in P_{n-1}(\Bbb R).$ Then $\dim V=n$ (the same as the dimension of $P_{n-1}(\Bbb R)).$ Why $V$ is a subspace? Because it is easy to see that $ap+bq\in V$ whenever $p,q\in V$ and $a,b\in\Bbb R.$