Showing that a Vector Space is Isomorphic Let $F = \mathbb{R}, V = \mathbb{C}, W = \{f \in P_2(\Bbb R) \mid f(3) = 0\}.$ Determine whether or not that these vector spaces are isomorphic.
Workings:
I know that $V \cong W$ iff $\dim(V) = \dim(W)$
So $\dim (V) = 2$ since the basis of $\mathbb{C}$ is $\{1,i\}$
And the $\dim(P_2(\mathbb{R})) = 3$
So as $\dim(V) \neq \dim(W)$. They are not isomorphic.
I'm not sure if I did this right. Any help will be appreciated.
 A: You didn't compute $\dim(W)$, you computed $\dim(P_2(\mathbb{R}))$. These are not the same.
The evaluation map $e_3(f) : f\mapsto f(3)$ is a linear map. It is a clearly surjective map from $P_2(\mathbb{R})\to \mathbb{R}$.
$W=\ker e_3$. As you mentioned $\dim P_2 = 3$, and $\dim \mathbb{R}=1$.
Thus $\dim W + 1 =3$ by rank-nullity, or $\dim W =2$. Thus $W\cong V$ since they have the same dimension.
A: If $f(x)$ is a polynomial and $f(3)=0$ then $f(x)=(x-3)(\cdots\cdots)$, where $\text{“}\cdots\cdots\text{''}$ is a polynomial.
If $\deg f(x)\le 2$ then $\deg(\text{“}\cdots\cdots\text{''})\le1$.  So you've got $f(x) = (x-3)(ax+b)$ for some scalars $a$ and $b$ (in this context you're construing "scalar" to mean a real number; in some other contexts it means a complex number).
So it take two scalars $a$ and $b$ to specify a member of the space of polynomials that vanish at $3$.  That suggests that space is $2$-dimensional.  Let's find a basis of it.  How about this:
$$
\Big\{ (x-3)x,\  (x-3) \Big\}
$$
So $\{1,i\}$ is a basis of $\mathbb C$ and here we have a basis of this other space.
Here is an isomorphism:
\begin{align}
1 & \mapsto (x-3) \\
i & \mapsto (x-3)x \\
\Big(\text{any linear combination of $1$ and $i$}\Big) & \mapsto \Big(\text{the same linear combination of $(x-3)$ and $(x-3)x$}\Big)
\end{align}
