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Any finite-dimensional vector space is the dual space of another?

This is from a true/false section in my book and the statement is supposedly true. I can't say I see the reasoning behind this and any hint/direction is greatly appreciated. My intuition says it has something to do with the isomorphic relationship between finite-dimensional vector spaces and their duals.

Edit: Here is the question as exactly phrased.

Every vector space is the dual of some other space? True or False . . .

I don't know whether or not they mean isomorphic, equal to, etc...

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    $\begingroup$ By "is," do you mean "is equal to" or "is isomorphic to"? $\endgroup$ – Nishant Nov 6 '15 at 1:22
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    $\begingroup$ Perhaps you want to prove to yourself that every finite dimensional vector space is "reflexive". That is, the dual of the dual of $V$ is $V$ itself. $\endgroup$ – oxeimon Nov 6 '15 at 1:29
  • $\begingroup$ What is the name of the book? $\endgroup$ – Shahab Nov 6 '15 at 2:25
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    $\begingroup$ Linear Algebra by Friedberg, Insel, Spence $\endgroup$ – Aaron Zolotor Nov 6 '15 at 2:28
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    $\begingroup$ I stated above that its finite-dimensional. The directions include that assumption. $\endgroup$ – Aaron Zolotor Nov 6 '15 at 2:37
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For a finite dimensional inner-product space $V$, we can define any linear functional $f \in V^{\ast}$ as $f(w) = \langle v,w\rangle$ for a unique $v \in V$, this is often written as $f = \langle v,-\rangle$. For example, given the basis $\{e_1,\dots,e_n\}$ (the standard basis) we have the dual basis: $\{\pi_1,\dots,\pi_n\}$, where, if $w = w_1e_1 + \cdots w_ne_n$, then $\pi_j(w) = w_j$, so that if $v = v_1e_1 + \cdots + v_ne_n$ our linear functional $\langle v,-\rangle$ is: $v_1\pi_1 +\cdots + v_n\pi_n$ (we can add two linear functionals, and take scalar multiples of them using the operations of the underlying field $F$:

$(f+g)(v) = f(v) + g(v)\\ (cf)(v) = c(f(v)).)$

In the expression $f(w) = \langle v,w\rangle$, $f$ is the function, and $w$ is the variable, we might re-write this as $f = v^{\ast}$ ($f$ is clearly a linear functional "derived" from the vector $v$).

But if we pull a "switcheroo" (this is a technical term) letting $w$ be the function, and $v$ the variable, we can define an element of $(V^{\ast})^{\ast}$ like so:

$w^{\ast\ast}(v^{\ast}) = v^{\ast}(w)$.

The mapping $w \mapsto w^{\ast\ast}$ is a ($F$-linear) isomorphism. While describing $V^{\ast}$ forced us to choose a basis, and define a dual basis (to squeeze "numbers" (scalars) out of vectors, we usually need coordinates), the correspondence $w \mapsto w^{\ast\ast}$ is basis-free, which is what texts means by saying it is "canonical".

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