Question regarding the simple proof involving dual space and annihilator 
If $\eta$ is a suspace in a finite-dimensional vector spce $v$, then
$\eta^{00}=\eta$
Proof: Observe that we use here the convention, established at the end
of 16, that identifies $v$ and $v''$. By definition, $\eta^{00}$ is
the set of all vectors $x$ such that $[x,y]=0$ for all $y$ in
$\eta^{0}\dots$

16 explains that in convention, we can identify $v$ with its double dual space $v''$, because they are isomorphic. But I'm not sure how the second sentence in the above proof follows.
From my understanding,
by definition, $\eta^{00}$ is the set of all vectors $x$ in $v''$ such that $[y, x]=0$ for every $y\in\eta^{0}$. I'm not sure how we can switch the position of $x$ and $y$ from the isomorphism of $v$ and $v''$.
 A: I believe it is probably this. The Convention you are talking about is to identify $x\in v$ with a functional $\overline{x}:v'\to k$ such that $\overline{x}(y)=y(x)$ for every $y\in v'$. That means that
$$[y,\overline{x}]=\overline{x}(y)=y(x)=[x,y]$$
Now, $\eta^{00}$ is the set of all vectors $\overline{y}\in v''$ such that $[x,\overline{y}]=0$ for all $x\in\eta^0$. (Note we can always see those vectors as being of the form $\overline{y}$ for some $y\in v$, because the correspondence $v\mapsto v''$ happens to be "onto".)
Swap the role of the letters $x$ and $y$, and you get that $\eta^{00}$ is the set of all vectors $\overline{x}\in v''$ such that $[y,\overline{x}]=0$ for all $y\in\eta^0$.
Now, identify $\overline{x}$ with $x$ and you get that $\eta^{00}$ is the set of all vectors $x\in v$ such that $[x,y]=0$ for all $y\in\eta^0$.
A: The identification of $V''$ with $V$ is confusing. If you're interested, I have a long answer about just this. With that in mind, I will first answer without the $[\cdot,\cdot]$ notation. For any element $x \in V$, I will use $\alpha_x$ to denote the associated element of $V''$. Note that $\alpha_x:V' \to \Bbb F$ is defined by
$$
\alpha_x(f) = f(x).
$$
Proof: By definition, $\eta^{00}$ is the set of all elements $\alpha \in V''$ for which $\alpha(f) = 0$ holds for all $f \in \eta^0$. Note that every $\alpha \in V''$ is identified with the unique $x \in V$ for which $\alpha = \alpha_x$. Thus, the set $\eta^{00}$ is identified with the set of all $x \in V$ such that for all $f \in \eta^0$,
$$
\alpha_x(f) = 0 \implies f(x) = 0.
$$
Now, by the definition of $\eta^0$, we can see that if $x \in \eta$, then the element of $V''$ associated with $x$ is an element of $\eta^0$. That is, $\eta \subseteq \eta^{00}$. On the other hand, with finite-dimensionality we have
$$
\dim(\eta^{00}) = \dim(V) - \dim(\eta^0) = \dim(V) - [\dim(V) - \dim(\eta)] = \dim(\eta).
$$
Thus, we indeed have $\eta = \eta^{00}$.

So what does this look like in terms of the brackets? The point is that $[x,y]$ is really $\alpha_x(y)$, and $[y,x] = y(x)$. The identification of $V$ with $V''$ is defined such that $\alpha_x(y) = y(x)$, which is to say that $[x,y] = [y,x]$.
