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
$$V=\{p\in P_n(\mathbb{R}):p(\alpha)=0\}$$
Let V be a subspace of $P_n(\mathbb{R})$.
Task: Find the dimension of V.
Here is what I've tried. And I was told is incorrect.
With $L:P_n(\mathbb{R}) \mapsto (\mathbb{R})$ defined by $L(p)=p(\alpha)$.
We use the formula: $Dim(V)=Dim(Ker(L))+Dim(L(v))$
We know $V=Ker(L)$. So we have $Dim(V)=Dim(V)+Dim(L(v))$.
We know that $im(L)=\mathbb{R}$ so $Dim(\mathbb{R})=1$. And $Dim(P_n(\mathbb{R}))$ is spanned by $\{{1,x,x^2,...,x^n}\}$. Therefore $dim(P_n(\mathbb{R})=n+1$. So we get
$1+Dim(V)=n+1=>Dim(V)=n$. (I was told this is incorrect).
I was given the hint that $Dim(P_n(\mathbb{R}))$ is in fact spanned by $\{{1,x,x^2,...,x^{n-1}}\}$. So is it simple that
$dim(P_n(\mathbb{R})=n$ and therefore we have
$1+Dim(V)=n=>Dim(V)=n-1$
Yes, $\dim (V)=n-1$ is correct.