Trying to visualize and understand double dual space Currently I am reading "Finite-dimensional vector spaces" by Paul Halmos. I would have a question regarding the theorem on page 25. It says:

If $V$ is a finite-dimensional vector space, then corresponding to every  linear functional $z_{0}$ on $V'$ there is a vector $x_{0}$ in $V$ such that $z_{0}(y) = [x_{0}, y] = y(x_{0})$ for every $y$ in $V'$; the correspondence $z_{0}\rightleftarrows x_{0}$ between $V''$ and $V$ is an isomorphism.

I tried to visualize the theorem and hoped to make it more clear for me. So I defined the variable the following way:
$z_{0} \in V'': z_{0}(y) = 2*y$ which would be $2*(4*\frac{3}{2} + 3*2) = 2* 12 = 24$  
$y \in V': y(x_{0}) = 4*\frac{3}{2} + 3*2 = 12 $
$x_{0} \in V: x_{0} = \begin{pmatrix}\frac{3}{2}\\2\end{pmatrix}$
I picked $z_{0}$ as a functional that doubles the functional $y$. The functional I picked for is $y(\begin{pmatrix}w_{1}\\w_{2}\end{pmatrix}) = 4*w_{1} + 3*w_{2}$, although it works for every $y \in V'$ as it is stated in the theorem. I picked $z_{0}$ such that I get a whole number as the scalar.
Somehow, as you see, I get $z_{0}(y) = 24 \neq 12 = y(x_{0})$ which is false. What did I do wrong or is my example just somehow flawed?
Here is the picture I tried to visualize my vector space and the element if the dual space which is responsible for $x_{0}$. Also is there a way to visualize the double dual space? 
visualization of dual space
Thanks for any answers!
 A: Visualization can only get you so far. While I'm not aware of a convenient way to visualize dual spaces, the theorem you mention may be conceptualized by saying that it exhibits a symmetry of the pairing of a vector space with its dual: not only can an evil element of the dual eat a vector to yield a scalar, but the vector itself can turn into a dual-vector eating monster. Therefore neither vector nor covector ought to be considered an evil overlord terrorizing the other; rather, both may come to the mutual agreement that joining forces to produce a scalar might be in their best interest. Note that when the vector space has extra structure afforded by an inner product, the pairing becomes canonical and given by the inner product. In this case, the annihilators of subspaces of $V$ are identified with the orthogonal complement of said subspaces.
Concerning your example, you picked $V= \mathbb{R}^2$ and $z_0:=2 \in V''$. This doesn't really make sense, as $z$ has to return a scalar when evaluated at some $y \in V'$. Therefore we should pick something like $z_0(y):=2y_1+3y_2$ instead. (I'm using subscripts to denote components in the dual basis of $V'$.) Now let's pick some $y \in V'$ to illustrate the theorem. Note that $y$ has to be linear; for instance, we could define $y(x) = x^1 - 2x^2$ (where now I'm using superscripts to denote components in the standard basis of $V$). This means that $y_1 = 1$, $y_2 = -2$. Now the theorem asserts the existence of a $x_0 \in V$, which is independent of $y$, such that $$-4 = 2y_1 + 3y_2 = z_0(y) = y(x_0) = x_0^1 - 2x_0^2, $$
and suggests that we take $x_0 := (2,3)$ (the components of $z_0$ in the double dual basis). And indeed, we find that this choice is works, and is the only one that actually works for any $y \in V'$.
