Show a subset is subspace of a vectorspace and find the dimension
Let $n>1$ be a natural number and let $\alpha\in\mathbb{R}$ be a real scalar. Let $V$ be a subset of the vector space $P_n(\mathbb{R})$. Define $V$ as
$$V=\{p\in P_n(\mathbb{R}):p(\alpha)=0\}$$
Task: show that V is a sub space of $P_n(\mathbb{R})$ and find the dimension of V.
I believe I have the correct definitions but I'm not sure if I've done this correctly.
I know V is a subspace if the following holds
1) Vector 0 is in the subset.
2) V is closed under addition.
3) V is closed under scalar multiplication.
For (1): I have a hard time showing but from the definition of V it only contains the zero vector. Meaning it satisfies (1).
For (2): Any real scalar multiplied with the zero vector gives the zero vector, which is in $V$. Therefore it satisfies (2).
For (3): If you add two zero vectors you will always get a the zero vector. Therefore it satisfies (3).
I'm uncertain how to find the dimension but I know how to define it using the dimension formula for V.
$L:P_n(\mathbb{R}) \mapsto (\mathbb{R})$ defined by $L(p)=p(\alpha)$.
We can use the formula: $Dim(V)=Dim(Ker(L))+Dim(L(v))$
Solution 1:
For 1), the zero polynomial evaluated at any point is $0$, hence $0\in V$.
For 2), if $f(\alpha)=g(\alpha)=0$ then $(f+g)(\alpha)=f(\alpha)+g(\alpha)=0+0=0$. Hence $f+g\in V$.
For 3), if $f(\alpha)=0$ then for any $c\in\mathbb{R}$ you have $(c\cdot f)(\alpha)=c\cdot f(\alpha)=c\cdot 0=0$. Hence $c\cdot f\in V$.
The condition that $p(\alpha)=0$ is a single linear condition on $P_n(\mathbb{R})$. Namely, if you write each polynomial as $p=a_n\cdot x^n+...+a_0$ then $p(\alpha)=0$ holds iff $\alpha^n\cdot a_n+...+a_0=0$. Hence the dimension drops by $1$, so the dimension of $V$ equals $n$. Alternatively, you can see that $p(\alpha)$ is always a scalar, i.e. in $\mathbb{R}$ and the map $L:P_n(\mathbb{R})\rightarrow\mathbb{R}$ is onto since the scalars in $P_n(\mathbb{R})$ maps to scalars, hence by rank-nullity theorem you see that the kernel $V$ has dimension $n$.