Question: I have a very simple question but can't seem to find the answer for it.  

If I have an equation like so:

$\|z-w\|^2 \ge (\|z\|-\|w\|)^2$, where $z$ and $w$ are complex numbers and $\|\|$ is the modulus

Would the inequality (bigger or equal) stay true if I root them both? So would 

$\|z-w\| \ge \|z\|-\|w\|$ still hold?

Answer: 
Let $e\_1, \dots, e\_m$ a basis of $F^m$. 

Let $f:F^m \to F$ be a linear form, that is an element of the dual space. 

Then for any $v= a\_1e\_1 + \dots + a\_me\_m$ we have $f(v)= f(a\_1e\_1 +\dots + a\_me\_m) = a\_1f(e\_1) +\dots + a\_mf(e\_m)$. Thus $f$ is uniquely determined by 
$f(e\_1), \dots, f(e\_m)$ that is an $m$ elements from $F$, viz. an element of $F^m$. 

Conversely for every $(b\_1, \dots, b\_m) \in F^m$ we can define an associated linear form via defining $g(e\_i)=b\_i$ and thus $g(a\_1e\_1 +\dots + a\_me\_m) = a\_1b\_1 +\dots + a\_mb\_m$.

It follows the elements of the dual space are in bijection with the elements of $F^m$. 

It is not hard to show that this bijection is linear, and thus an isomorphism.

Note: the bijection depends on the basis. This is why one says there is no canonical isomorphism, but there still is an isomorphism. 

However the spaces are not literally equal. The one is made up of $m$-tuples of elements of $F$ the other of linear maps from $F^m$ to $F$, that is certain subsets of the Cartesian product of $F^m \times F$. 

However, when somebody says "the dual of $V$ is {something}", then usually this {something} is just isomorphic to the dual not literally the sets of linear maps from $V$ to $F$. 

---

For the supplementary question that motivated the question: 

Suppose the function $\Gamma$ is injective. This means that the only form that vanishes on all of $v\_1, \dots , v\_m$ is the form that maps the whole space to $0$. But this means that $v\_1, \dots, v\_m$ must span $V$, for if not then we could define a form by saying it is $0$ on the span of $v\_1, \dots, v\_m$ and $1$ for $w\_1, \dots, w\_n$ where $w\_1, \dots, w\_n$ is an independent set that completes $v\_1, \dots, v\_m$ to a generating set of the space.
Given a vector space $\mathbf F^m$, this space has its double dual space ${({\mathbf F^m})^{\ast\ast}} = \mathcal L(({\mathbf F^m})^{\ast},\mathbf F),$ the set of linear functionals on $({\mathbf F^m})^{\ast}$. This double dual is *naturally isomorphic* to $\mathbf F^m$. Let $v \in \mathbf F^m$, $\varphi \in ({\mathbf F^m})^{\ast}$ and define $\operatorname{ev}\_v:({\mathbf F^m})^{\ast}\to\mathbf F$, the *evaluation at $v$ functional on $({\mathbf F^m})^{\ast}$* by
$$
\operatorname{ev}\_v(\varphi) = \varphi(v) \in \mathbf F^m.
$$
Note that $\operatorname{ev}\_v$ is an element of the double dual of $\mathbf F^m$. The association of $\operatorname{ev}\_v$ with $v$ is so natural, they can be thought of as the same element and $\operatorname{ev}\_v\mapsto v$ is an isomorphism ${({\mathbf F^m})^{\ast\ast}}\cong \mathbf F^m$. In this sense, the dual space of $\mathbf F^m$ is $({\mathbf F^m})^{\ast}$ since the association $\operatorname{ev}\_v\mapsto v$ is so natural. This isomorphism $\operatorname{ev}\_v\mapsto v$ is *natural* in the sense that it does not depend on our choice of basis for $\mathbf F^m$.

However, a vector space $V$ can never truly be *equal* to its dual for the reason that $V^\ast$ is the set of linear functionals on $V$ and $V$ is itself a set of vectors. This association of $V$ with its double dual via the evaluation map is the closest thing to a vector space *equalling* its dual.